The hoax that is the 1.41% house advantage on pass line bets

Mosca: "Bothered" is not the correct terminology. Concerned is much more appropriate. The idea is to win and win often and playing the PL with the associated FO bet will not cut it.

The false promise of the 1.41% HA gives a perception of easy wins and lots of them. If a player steps up to the table and wins four or five PL bets and is ahead, he sticks around, does the same thing, and then loses the winnings and usually lots more. Players are heavily invested in the PL/FO betting because that is all they know or care to know and their losses are consistently high. So I am concerned and hope that some will listen and change their thoughts and approaches to the game we play often and love.

tuttigym

Sigh.

OK. Then tell me. How should a player bet, absent a crystal ball, and with the understanding that the next roll doesn't know any of the previous rolls, or any of the future rolls, or any of the rolls at any of the other tables all over the world?

Are you implying that you can use what you've been writing here (frankly, it looks like nonsense to me, but I'll put that aside for now) to predict the next roll of the dice?

Quote: tuttigym

darnits: The mathematical and statistical proof is always right in front of you when you are at the tables. Just look around, check out the number of players playing the PL/FO bets, and then take note of the number of winners and losers. Choppy and cold tables produce losers big time, and "hot" tables and shooters do win, but they do tend to push their "luck" and stick around too long.

Do you suppose that a casino manager would allow someone to review all the video recordings for a given period of time of craps play in order to chart the PL outcomes? Probably not.

tuttigym

So I guess since we can't look at all of the video recordings, we could maybe set up our own experiments and do it ourselves...ahhh, wait a second we can do a computer simulation, simulating 495 PL outcomes, 10000 times and averaging the number of PL wins and PL losses.

As many people have done before and the results shown...the average of the PL wins is 245 and losses is 251.

You say that the PL HA of 1.41% is a hoax. I think I see where you are coming from when you say people come up to a table win some, think the game is easy and then lose alot back. Yes sure, for them the HA is greater than 1.41%, but statistically it is around 1.41%.

Statistically the HA is 1.41%, but I think you are saying when the psychological effect of winning a lot, thinking you can win, and then losing more, pushes the HA higher.

In a theoretically perfect craps world, the HA is 1.41%. However, people aren't perfect. They will play more cause they think the game is in their favor and will eventually lose it back because of the HA.

Mosca: Absolutely NOT. I do not try to predict what the dice are going to do. You said it correctly -- the dice have NO memory, and their future is totally random.

My play almost always starts after the point is established. I then place wagers which allow me more ways to win than to lose. After a certain number of rolls ( just guessing here), if there is no premature 7 out, and if I get some hits (wins), I reduce my bets in half. If all of that occurs, I can not lose. I do not play every shooter or every point, and I try to be flexible with the table. The simplistic math previously posted works, for me.

Have I lost? Absolutely!

tuttigym

tuttigym:

The HA is an average.

A V E R A G E

Quote:

Average: In mathematics, the average is the central tendency of a data set and is a measure of the "middle" or "expected" value of the data set.

Your way of looking at it sounds to me like this:

• "If you come away with a loss, the HA is 100%."
• "If you break even, the HA is 0%."
• "If you come away with a win, the HA is -100%."

You've had people on Yahoo! Answers try to explain it to you. You've had people on Wizard of Vegas try to explain it to you. Your questions, statements, and facts do not mesh. We've analyzed it on all sides, successfully. We've given you counter-examples.

What will appease you? Actual, factual data from a casino, giving live-action results?
If that is the case, maybe you should start with an analysis of Roulette, since most casinos provide tote boards with the previous dozen spins or so.

darnits: Actually not. Computer simulations are just that. They can NEVER achieve the true randomness of the game itself. The computer is held hostage by the programmer and can only reflect the programmer's bias.

Consider this - with all the knowledge and information that is contained in all of our programs and computers how come we cannot ask a COMPUTER to answer my odds question about the 1.41% HA on PL bets? Shouldn't we just be able to ask a computer a simple math question and then get an answer?

tuttigym

Quote:

6 or 8 -- five ways to win and six ways to lose - HA or edge is 16 2/3%
5 or 9 -- four ways to win and six ways to lose - HA or edge is 33%
4 or 10 -- three ways to win and six ways to lose - HA or edge is at 50% or 2 to 1

Wrong: 6 or 8 -- 5 ways to win and six ways to lose: HA is 5 x 5/11 - 6 x 6/11 = 9.091%
5 or 9 -- 4 ways to win and six ways to lose: HA is 4 x 4/10 - 6 x 6/10 = 20%
6 or 9 -- 3 ways to win and six ways to lose: HA is 3 x 3/9 - 6 x 6/9 = 33.333%


It is not 1/6, 2/6 and 3/6, your denominator is wrong. There are 11 outcomes to a six and eight, not 6; 10 outcomes to a five and nine, not 6, and 9 outcomes to a four and ten, not 6.

Excel throws with neither hand.

Quote: tuttigym

Mosca: Absolutely NOT. I do not try to predict what the dice are going to do. You said it correctly -- the dice have NO memory, and their future is totally random.

My play almost always starts after the point is established. I then place wagers which allow me more ways to win than to lose. After a certain number of rolls ( just guessing here), if there is no premature 7 out, and if I get some hits (wins), I reduce my bets in half. If all of that occurs, I can not lose. I do not play every shooter or every point, and I try to be flexible with the table. The simplistic math previously posted works, for me.

Have I lost? Absolutely!

tuttigym

So you're guessing. But if you place all those wagers and have a premature 7 out, you lose them ALL, negating any perceived advantage. The house edge on ALL those wagers is higher than it is on the come-out roll and odds.

JUST BECAUSE YOU SAY IT IS DIFFERENT, OR BELIEVE THAT IT IS DIFFERENT, DOES NOT MAKE IT SO.

Wait a second. that's not right... let me try again...

JUST BECAUSE YOU SAY IT IS DIFFERENT, OR BELIEVE THAT IT IS DIFFERENT, DOES NOT MAKE IT SO.

There. That's better.

Okay, going back to your original post

Quote: tuttigym

The house advantage (HA) or edge on Pass Line (PL) bets is unequivically stated at 1.41%. The formula for that HA is also readily available, and broken down to its simplest form, states that out of 495 PL plays there will be 244 PL wins against 251 PL losses.

Just to be clear with the last sentence, it should read: Out of the 495 possible PL outcomes, 244 are PL wins and 251 are PL losses. I think this is the crux of your problem tuttigym. The statement of 495 outcomes above does not pertain to any random 495 rolls in the "real world". It applies to looking at each of the 495 different possible plays.

By accepting that each one of those 495 plays are equally possible to come up in random play, it would be fair to say that the bet has a house edge equal to the average result of all 495 possible outcomes, which happens to be 1.41% in favor of the house.

No mathematician worth his salt has EVER seriously stated that "Any 495 PL plays, random or not, MUST produce 244 PL wins and 251 PL losses." Which is how you are incorrectly interpreting your initial statement.

The other main complaint you're making seems to be "I accept that the HA is 1.41%, but it only comes up 3.5% (such a low percentage) of the time so it's misleading!" Guess what, it's a GOOD thing (for you and I as a gambler) that its not 100%.

You want a game where the HA is a guaranteed 1.41% for every 495 rolls? Roll a 6 sided die, if it comes up 1,2,3,4,5 or 6, 98.59% of your bet is returned to you. I'd bet you'd love that game.

I think tuttigym's an Iron Crosser. For those not familiar with the bet, it's placing the 5, 6 and 8 and a bet on the field. On all throws but the seven, you win.

Just like when Tutti says that the "house advantage" on a 4 and 10,(3-6)/6 = 50%, we can calculate the "house advantage" on the iron cross.

There are 30 ways to win and 6 ways to lose. By Tutti's math (taking the ways to win - the ways to lose) / ways to lose), the "HA" (as I believe he calculates it) is (30-6)/6 = "Player Advantage" of 400%!!!

And actually, on a table where the 12 pays 3:1, this isn't an terrible bet with an actual house advantage of 1.136% (per roll). (If the 12 pays 2:1, the house advantage is 1.768%, per roll).

Here's the real math to calculate the house advantage based on a $5 minimum ($5 on the field, $6 on the 6 and 8, $5 on the 5, total $22 bet):

HA = ($10 x 1/36 + $15 x 1/36 + $2 x (4 + 5 + 5)/36 + $5 x (2 + 3 + 4 + 3 + 2)/36 - $22 x 6/36) / $22
= ($10 + $15 + $28 + $70 - 132) / (36 x $22)
= -$9 / $792
= -.01136 = 1.136% House Advantage per roll

Similarly, the per roll House Advantage on a six or eight is ($7 * 5/36 - $6 x 6/36)/$6 = -.004630% = 0.463% House Advantage per roll (1.5152% x 11/36)
The per roll House Advantage on a five or nine is 1.4 x 4/36 - 1 x 6/36 = 1.1111% (4% x 10/36)
The per roll House Advantage on a four or ten is 1.8 x 3/36 - 1 x 6/36 = 1.6667% (6.6667 % x 9/36).

Just trying to make sure I understand...

1.41% is the HA on a RESOLVED pass line bet. Would the HA on a $100 resolved bet be $1.41?

1.136% is the HA on an "Iron Cross" bet as posted above. So if the average shooter rolls a seven on about the 6th roll, would that mean the total HA on a $100 set of bets would be $6.82?

Quote: tuttigym

darnits: Actually not. Computer simulations are just that. They can NEVER achieve the true randomness of the game itself. The computer is held hostage by the programmer and can only reflect the programmer's bias.

Consider this - with all the knowledge and information that is contained in all of our programs and computers how come we cannot ask a COMPUTER to answer my odds question about the 1.41% HA on PL bets? Shouldn't we just be able to ask a computer a simple math question and then get an answer?

tuttigym

So now you are saying that simulations cannot be used to convince you of the truth because you will assume that results are biased? Is this based on the fact that computer generated random numbers are not truly random? (I believe another thread has explored this issue.) Or are you saying that whoever is creating the simulation will purposely skew the results to support what they are trying to prove?

I'm not sure what you mean by "how come we cannot ask a COMPUTER?" That is exactly what we are doing when we run simulations, yet you just claimed that simulations are unreliable.

This thread has really gone on far too long. You are not going to convice the majority, if not all, of the readers that your arguments are correct. Obviously, neither are we going to convince you that you are wrong.

You state that that exact result of the "perfect 495" is rare, and you are correct. But if it weren't for the variance illustrated by the binomial distribution curve posted by boymimbo on page 10, I doubt most people would play the game. We are hoping that when we play the wins are tending to the right side of that curve, with more wins than losses.

Quote: tuttigym

My play almost always starts after the point is established. I then place wagers which allow me more ways to win than to lose. After a certain number of rolls ( just guessing here), if there is no premature 7 out, and if I get some hits (wins), I reduce my bets in half. If all of that occurs, I can not lose. I do not play every shooter or every point, and I try to be flexible with the table. The simplistic math previously posted works, for me.

I posted a table (reproduced here) way back on page 4 that shows the house advantage on your assumed method of betting. This method gives you the 5:1 "advantage" that you claim - one roll of the dice, 30 ways to win, 6 ways to lose. But, based on the amount bet and the actual money returned that one roll has a 2.3% house advantage. Sure, you will win more times than you lose, but when you lose, you lose big.

I might have missed it, but I never saw a response to this.

Quote: me

I did not see where you explained your strategy which gives you 30 chances to win and 6 to lose. One way would be to bet across and on the horn after the point is established and take all place bets down after one roll.

This yields a HA = 2.3% which is effective on the larger amount you would be betting - $36 vs $5 for the pass line - so this method will give the house more of your money over the same period of time.

number	bet	win	return	# rolls out of 36	weighted return
4	5	9	41	3	123
5	5	7	39	4	156
6	6	7	39	5	195
8	6	7	39	5	195
9	5	7	39	4	156
10	5	9	41	3	123
2	1	30	63	1	63
3	1	15	48	2	96
11	1	15	48	2	96
12	1	30	63	1	63
7	0	0	0	6	0
				total return	1266
				total bet	1296
				loss	30
				HA	2.3%

You can make similar calculations for whatever strategy you employ.

Continued discussion seems pointless because neither side is going to convince the other of the validity of their postion. Throughout this thread there is much repetition and variations of the same arguments. Some of us have become frustrated, resorting to shouting and name-calling.

You should keep playing craps the way you believe is correct. Most of us will thank you because we believe that your higher house advantage bets are helping to pay to keep the tables open for those of us who stick to the passline with odds.

Quote: boymimbo

I think tuttigym's an Iron Crosser. For those not familiar with the bet, it's placing the 5, 6 and 8 and a bet on the field. On all throws but the seven, you win.

Just like when Tutti says that the "house advantage" on a 4 and 10,(3-6)/6 = 50%, we can calculate the "house advantage" on the iron cross.

There are 30 ways to win and 6 ways to lose. By Tutti's math (taking the ways to win - the ways to lose) / ways to lose), the "HA" (as I believe he calculates it) is (30-6)/6 = "Player Advantage" of 400%!!!

And actually, on a table where the 12 pays 3:1, this isn't an terrible bet with an actual house advantage of 1.136% (per roll). (If the 12 pays 2:1, the house advantage is 1.768%, per roll).

Here's the real math to calculate the house advantage based on a $5 minimum ($5 on the field, $6 on the 6 and 8, $5 on the 5, total $22 bet):

HA = ($10 x 1/36 + $15 x 1/36 + $2 x (4 + 5 + 5)/36 + $5 x (2 + 3 + 4 + 3 + 2)/36 - $22 x 6/36) / $22
= ($10 + $15 + $28 + $70 - 132) / (36 x $22)
= -$9 / $792
= -.01136 = 1.136% House Advantage per roll

Similarly, the per roll House Advantage on a six or eight is ($7 * 5/36 - $6 x 6/36)/$6 = -.004630% = 0.463% House Advantage per roll (1.5152% x 11/36)
The per roll House Advantage on a five or nine is 1.4 x 4/36 - 1 x 6/36 = 1.1111% (4% x 10/36)
The per roll House Advantage on a four or ten is 1.8 x 3/36 - 1 x 6/36 = 1.6667% (6.6667 % x 9/36).

There it is, right there; the "5 to 1 player advantage".

But it overlooks the fact that if you win a bet, you only win one of those bets; if you lose a bet, you lose ALL of the bets. So it's a 5 to 1 if you win, but 1 to 5 if you lose, and the difference between true odds and payout is the house advantage, which others have calculated.

Quote: RonC

Just trying to make sure I understand...

1.41% is the HA on a RESOLVED pass line bet. Would the HA on a $100 resolved bet be $1.41?

1.136% is the HA on an "Iron Cross" bet as posted above. So if the average shooter rolls a seven on about the 6th roll, would that mean the total HA on a $100 set of bets would be $6.82?

All house advantages as stated are on "Resolved" bets. The EV (Expected Value) on a $100 resolved bet is $98.59 or a $1.41 expected loss.

Since there is a resolution on every roll (the field bet) of an iron cross, the HA must be expressed of a function of every roll, not before a seven out.

The average number of throws to resolve a pass line is 3.376, so the per roll HA on a pass line is 1.414% / 3.376 = 0.4189%.

To get to 3.376 average rolls on a come out:

12 come out rolls resolved in 1 bet. 24 other rolls are a point bet (which counts as one roll)
- 6 of those 24 point bets are 4 and 10 and are resolved in 4 rolls (9 resolutions / 36 combintions).
- 8 of those 24 point bets are 5 and 10 resolved in 3.6 rolls (10 resolutions / 36 combos). Of of point being a five or nine is 33.333%
- the other 10 of those 24 point bets are 6 and 8 resolved in 3.273 rolls (11 resolutions / 36 combos). Odds of point being a six or eight is 41.667%.

Add it all up. (12 x 1 + 6 x (4+1) + 8 x (3.6+1) + 10 x (3.273+1) / 36 = 3.376 rolls.

This matches the Wizard's results in (Craps Appendix 2)

You all seem to have such negative replies because you are so heavily invested in the PL/FO strategies.

Try this as it happens most times and that is why the false premise.

Player A puts $10 on the PL; shooter comes out with a 9; A FO bet of $50 plus $12 on each of the 6 and 8.

Player B (after the point is established) wagers $64 across the board including the point.

Shooter then tosses: 5; 6; 10; 8; 7 out

Player A wins $28 on the 6 & 8 and then loses all the bets still on the table - net loss $56.00

Player B wins $60 on the 5; 6; 8; & 10 and loses the bets on the table - net loss $4.00.

PL/FO players risk much more on one number which is less likely to win simply because there are more ways to lose than to win. When that same player posts more bets, he must hit more times to cover those bets.

You seem to believe that if I lose with a 7 out after the point is established, somehow the PL/FO wagers are safe. Anytime I lose so does the PL/FO bettor. If the point is converted, I still win, but the win, of course, is not as great as the PL/FO winner unless the shooter should happen to throw 5 or 6 or more covered numbers then my win will exceed the PL/FO player.

The PL/FO player needs point conversions, if the table is cold, I am still going to be there and the PL/FO player's bankroll will be gone. One roll 7 outs hurts both players equally.

So you all keep clinging to that flawed premise and the false promise that is the 1.41% HA, and I will continue to outlast most all PL/FO players at the tables. By the way, if there should happen to be a "hot" shooter, I would enevitably make more.

Hey Goat are these folks really inflamed or what ??

tuttigym

Quote: tuttigym

goat: Is it OK to call you "goat"? I mean I have no problem with you call me tutti which means we may be getting close to becoming friends rather than just acquaintences.

JB came up with the 3.5% which is what I have been referring to. I did not say that I accepted that calculation, but it does convey how remote the possibilities are of the perfect 495. To toss a 12 or 2 or a Hard number is a 36 to 1 shot, so I have a real problem wraping my arms around a 28 to 1 shot with 495 possible outcomes. Look at it this way, in the infiniteness of time, it could be possible to have 494 winners vs one loser, right? What did Einstein call it, "thought......" something or another? So to put the other comment to rest about the 49 vs 51 -- for me, that is also a false premise giving rise to the false promise of the same 1.41% HA.

tuttigym

Since I sign my name, you could call me "Alan", or "goat".

Do you consider the probability of rolling a 2 or 12 remote? Actually it's 35 to 1, not 36 to 1, since one of the 36 possibilities is the result we're talking about, isn't it?

You have trouble "wrapping your arms" around a 28-to-1 shot for 244-251 because you are not understanding that, on either side of that result the probabilities are going down. Look here:

from 0-495 up to 177-318, the probability is less than .000000001
178-317 .000000001
...
190-305 .000000249
...
200-295 .000013829
220-275 .003496865
230-265 .016261084
240-255 .033614965
243-252 .035705505
244-251 .035847716
245-250 .035701357
247-248 .034563938
248-247 .033599964
250-245 .030992935
260-235 .012753375
etc., etc.

In any binomial situation like this, the number of trials determines how many different possible outcomes there are, in this case 495, i.e. winning anywhere from 0 to all 495 bets. The probability of any particular outcome occurring is determined by the number of ways the wins and losses can be arranged, and the probability of winning. For example, there is only one way to "arrange" 495 losses or 495 wins, but there are many, many, many ways to arrange 244 wins and 251 losses. The maximum number of "ways" is at 247-248 or 248-247, but because of the less-than-50% probability of winning any bet, the maximum probability is reached at 244-251. You can picture this by looking at the graph from Excel that someone posted.

"Thought experiment" is the term Einstein used. The probabilities at either end of that distribution are theoretically possible but so unlikely that we can safely ignore them. However, when you get up to 285-210, the odds against that particular result are "only" 25,108 to 1. Although that result is unlikely to occur in any given set of 495 passline decisions, it is a virtual certainty to occur at some point and even money (.5 probability) to occur in 17,403 sets of 495 decisions. Consider how many passline decisions occur every day in all the world's casinos.
Cheers,
Alan Shank

Quote: tuttigym

goatcabin:
Now to answer those pesky questions:

1. Winning PL bets along with their FO partners is just simple arithmetic and players actually win some of those bets.

Come Out naturals - one roll - Player advantage 2 to 1 or 8 ways to win and 4 ways to lose.

Point Conversions - players actually win these bets too and with the added bonus of the FO play.

6 or 8 -- five ways to win and six ways to lose - HA or edge is 16 2/3%

5 or 9 -- four ways to win and six ways to lose - HA or edge is 33%

4 or 10 -- three ways to win and six ways to lose - HA or edge is at 50% or 2 to 1.

These figures of course are incorrect, as someone else has pointed out. You have stated before that you understand how the 1.4% is calculated, yet you do not demonstrate such knowledge here.

Quote: tuttigym

Notice here I deal only with the simplicity of the math and the consistency of winning and losing based not on probabilities or odds, but strictly on the unforgiving reality of what I might be faced with as a player at the table.

But the probabilities or odds flow from the number of ways to win or lose a bet, don't they? They are just different ways of expressing the same thing.

Quote: tuttigym

Questions 2 & 3 are basically the same about winning multiple times and getting ahead: Absolutely yes. The problem: craps players have a mind set that they must play every point with every shooter, and they play the same way virtually every time. The only way they win with PL/FO is with the OCCASIONAL "hot" shooter converting multiple points and throwing lots of numbers w/o a quick 7 out. They fall into a casino and craps trap that says if you play the PL you MUST also wager that FO to get that big win. It just does NOT happen often enough to get ahead and stay there. The simplistic math above absolutely reeks havoc with the bankroll.

tuttigym

You did not answer those questions, did you? I will repeat them:

At what number of bets do you believe it is not possible for a player to still be ahead?
What do you believe is the MOST LIKELY outcome for a player making 495 passline bets?
Please provide a rationale for your answer.
Cheers,
Alan Shank(Goat)

Quote: tuttigym

Mosca: "Bothered" is not the correct terminology. Concerned is much more appropriate. The idea is to win and win often and playing the PL with the associated FO bet will not cut it.

The false promise of the 1.41% HA gives a perception of easy wins and lots of them.

Not to anyone who understands what the 1.41% HA actually means. The "promise" of the 1.41% HA is only that the player has close to a 50% chance to win any single bet, and that the volatility is very low. No one posting to this forum, except you, thinks it means you will win 244 out of every 495 bets.

Quote: tuttigym

If a player steps up to the table and wins four or five PL bets and is ahead, he sticks around, does the same thing, and then loses the winnings and usually lots more.

That is a gross generalization and bears little relationship to reality. Of course, some players get ahead and then end up losing for that session; however, others take their winnings and walk, or continue to win even more.

Quote: tuttigym

Players are heavily invested in the PL/FO betting because that is all they know or care to know and their losses are consistently high. So I am concerned and hope that some will listen and change their thoughts and approaches to the game we play often and love.
tuttigym

Your statement "their losses are consistently high" is totally unsupported. The issue of getting ahead and then losing the winnings is really nothing to do with the passline/odds betting as opposed to other kinds of bets. It could just as well apply to place betting, field betting, whatever. You are muddying the waters here.

Cheers,
Alan Shank

Quote: boymimbo

I think tuttigym's an Iron Crosser. For those not familiar with the bet, it's placing the 5, 6 and 8 and a bet on the field. On all throws but the seven, you win.

But, of course, tuttigym is not saying how he bets. All you have to do to get more ways to win than to lose on any given roll is place any two numbers (well, if you place the 4 and 10 it would be equal ways). If you place the 6 and 8 you have 10 ways to win versus 6 to lose on any roll. This reminds me of the guy on rec.gambling.craps who asserted that the house had an advantage on the rightside odds because there were more ways to lose than to win, ignoring the payoff as a factor. Of course, since you lose both bets on a seven, there is still a house advantage, and it's a bit higher than that on the passline.

I wonder what tuttigym thinks the house advantage is on his method?
Cheers,
Alan Shank

Quote: tuttigym

You all seem to have such negative replies because you are so heavily invested in the PL/FO strategies.

Try this as it happens most times and that is why the false premise.

Player A puts $10 on the PL; shooter comes out with a 9; A FO bet of $50 plus $12 on each of the 6 and 8.

Player B (after the point is established) wagers $64 across the board including the point.

Shooter then tosses: 5; 6; 10; 8; 7 out

Player A wins $28 on the 6 & 8 and then loses all the bets still on the table - net loss $56.00

Player B wins $60 on the 5; 6; 8; & 10 and loses the bets on the table - net loss $4.00.

PL/FO players risk much more on one number which is less likely to win simply because there are more ways to lose than to win. When that same player posts more bets, he must hit more times to cover those bets.

You seem to believe that if I lose with a 7 out after the point is established, somehow the PL/FO wagers are safe. Anytime I lose so does the PL/FO bettor. If the point is converted, I still win, but the win, of course, is not as great as the PL/FO winner unless the shooter should happen to throw 5 or 6 or more covered numbers then my win will exceed the PL/FO player.

The PL/FO player needs point conversions, if the table is cold, I am still going to be there and the PL/FO player's bankroll will be gone. One roll 7 outs hurts both players equally.

So you all keep clinging to that flawed premise and the false promise that is the 1.41% HA, and I will continue to outlast most all PL/FO players at the tables. By the way, if there should happen to be a "hot" shooter, I would enevitably make more.

Hey Goat are these folks really inflamed or what ??

tuttigym

I know I just said it was pointless to continue this discussion, but you just hit one of my biggest pet peeves that people using flawed logic throw out to support poor strategy.

You can always construct a specific scenario that shows one strategy is better than another. The problem is not taking into account the relative frequency that the various scenarios occur. (Which ultimately boils down to the mathematically calculated house advantages for each strategy.) For example, an alternate outcome to the situation you set up: with the same betting structure and a point of 9, the shooter hits the 9, then re-establishes the 9, player A bets the same and player B's place bets are still up. Shooter 7's out.

Player A won 10+75 and lost 10+50 - 24 for a net win of 1
Player B won 14 and lost 64 for a net loss of 50

What is more likely to happen? A point of 9 hits 40% or the time and loses 60% of time, so winning one and then losing one is 40% * 60% = 24%. The odds of hitting one of the numbers (except the point, as your sequence depicts) before the 7 are approximately 20 out of 30, or 67%, and hitting 4 of them and then the 7 is 67%^4 * 6/30 = 4%.

HA of placing across until a point or a 7 hits is 1.25%. HA of a pass line bet with 5x odds is 0.33%. The pass line is the better bet, but you will never believe this.

I have played many sessions where I have outlasted place bettors, but such comparisons are meaningless because we don't know the relative sizes of bankrolls or tolerance for loss.

I realize this is slightly off-topic, but I think it fits nicely in this lively discussion.

MS has casinos with 10x odds, vig on win on the 4,5,9,10, and 3x on the field 12.

We also know that 1 out of every 6 rolls is a 7.

I am hoping one of you math gurus can help me arrive at my probable outcome for the following strategy (strategy being a very loose term here):

It is the Iron Cross, placed after the first point, but take down all bets after the 4th roll.

Buy__ 5 for $30- pays $44 ______________ HE 2.0%
Place 6 for $30- pays $35 ______________ HE 1.52%
Place 8 for $30- pays $35 ______________ HE 1.52%
Field__ for $30- pays $35, $60, or $90____ HE 2.78%

I started to do the math on the actual HA, but gave up. I'm guessing it is right around 2.0%

However, wouldn't that only be applicable if the bets stayed up to the doomed conclusion?
The Total Bet is $120
The weighted average profit per roll is $32.35
The weighted average profit for the 4 rolls is $129
Outcome if a 7 out occurs on the shooter's roll after the point (using weighted avg for the win):
1 rolls (120)
2 rolls (88)
3 rolls (55)
4 rolls (23)
At this point bets are OFF, and $129 profit is locked in (actual profit could be as low as $20 for 4 six or eights, to as high as $360 for 4 twelves).


I think what I need to know for this 'strategy' is, how likely is the 7 to come up on the shooter's:
1st roll after the point?
2nd roll after the point?
3rd roll after the point?
4th roll after the point?
5th roll after the point?
6th roll is the 7

With this information, can't I then weight my expected loss scenario's, to give me a realistic idea of what my real potential losses will look like?

Remembering my sessions, I know there are always a few point-7 outs, and a couple 2 roll 7 outs, but it seems to me that most shooters manage to throw the dice 4 or 5 times, even if some of the throws are the junk (2,3,11,12). With the above 'strategy', I could care less what they throw, just as long as they can make 4 throws. what percentage of shooters get past 4 rolls?

If 1 out of every 3 shooters gets me to 4 throws, I should still make money, assuming the 2 losers don't 7 out on throw 1.

Please point out the pitfalls I am missing. How bad is this play? I realize discipline to pull the bets down after roll 4 is critical, and that I could be caught watching a hot table, but that may be acceptable for a 'winning' strategy. I am not delusional, no strategy can overcome the HA, but hopefully this would smooth out the variance.

Sigh...

The math on the above system:

You win $30 on 14/36 (3,4,9,10,11)
You win $60 on 1/36 (2)
You win $90 on 1/36 (12)
You win (44-30) $14 on 4/36 (5)
You win (35-30) $5 in 10/36 (6,8)
You lose $120 on 6/36 (7)

Add it all up: (420 + 60 + 90 + 56 + 50 - 720)/36 = -44/36 = -$1.222 / roll.

Per roll house advantage = $1.222 / $120 bet = 1.019% (per roll).

Compare that to craps appendix 2, where the pass line is .42%/roll, don't pass is .40%/roll, and the 6/8 (Place) is .493%.

Alan: Terrific post although the numbers are mind numbing. When you stated the odds at 25,108 to 1 on the numbers 285 - 210, is that wins to losses or the opposite? Also, are you saying that based on the perfect math of 244/251, it will be a virtual "certainty," and it will require a minimum of 17,403 sets of 495 PL outcomes or within any given 17,403 sets of 495?

Isn't the 495 PL outcome a virtual number without reality? It is a continuum - when does it start and when does it end? I mean, what if the first seven shooters establish a point and then 7 out? Does the 495 re-start?

I will post this and go back to some of the other entries to comment.

I was not evading your questions, I misinterpreted what you were asking.

tuttigym

There is no strategy. The seven has a 1/6 probability of coming out at any time. Locking in and unlocking betting systems do nothing but preserve the state at the table. Every time you have that $120 bet up in that configuration, you will, on average, and over time, lose money, just like the pass line, just like every bet at the table.

There is no system, no betting configuration, and no way to beat a craps table. Just like every other game at the casino, it is designed for you to lose. The only exception is the very difficult skill of controlled shooting.

Alan: There is no answer to either question that would provide certainty with one exception. A player that loses his entire bankroll regardless of the number of bets can never recover.

I have a friend who would go to the table and buy in for $5,000. His modest goal was to win between $750 and $900, period. He would play the PL/FO plus three numbers and his table bet after the point was established was $225 on any given point dispersed into the five bets. He would go to the casino at least once a month and play only one session.

For just under two years, he played with the same $5,000. During that time he won about $18,000.
Most times when he played, it took between 2 to 5 hours to win, and there were times he got down by as much as $3,000 only to come back and recover and eventually win that $750 or so. As you can see, he had a pretty good run.

Then three times in a row, he lost the $5,000. He just quit playing. He was not broke; he could afford the losses; in fact, he was still up about $3,000, but he just did not want to play craps. In three sessions, the "system" of PL/FO betting just beat him down. He still goes to the casino, but he plays "21" or roulette using a modified Martingale for each game. I am hoping that one day soon, he will rekindle his enthusiasm for craps.

Anecdotally, I have watched players win and lose with the PL/FO betting strategy. It really sickens me when they continue to throw their money on the table win some and then lose it all or most of it and their buy in. While I recognize many make stupid sucker bets, the real culprit and the big losses come from the PL/FO or the Come/FO. They quote the low HA and rely on it for big time wins. It is truly a false promise.

tuttigym

I am a lazy gambler.

Can someone give me the cliff notes on this thread? Thank you. :)

Quote: tuttigym

You all seem to have such negative replies because you are so heavily invested in the PL/FO strategies.

Try this as it happens most times and that is why the false premise.

Player A puts $10 on the PL; shooter comes out with a 9; A FO bet of $50 plus $12 on each of the 6 and 8.

Player B (after the point is established) wagers $64 across the board including the point.

Shooter then tosses: 5; 6; 10; 8; 7 out

Player A wins $28 on the 6 & 8 and then loses all the bets still on the table - net loss $56.00

Player B wins $60 on the 5; 6; 8; & 10 and loses the bets on the table - net loss $4.00.

Finally, tuttigym gets down to specifics! Excellent!
1) He puts forward a single scenario, alleging that it happens "most times". Actually, a point of nine is expected to occur just 11.1% of passline comeouts.
2) He carefully picks a series of rolls which result in a significantly better outcome for his method.
3) He has the pass/odds player risking $84, $20 more than his $64. Note also that he still has the pass/odds player making place bets as well.

This reminds me of John Patrick.

How about this: same bets, point is 9.
shooter tosses 9, 7
Player A wins $95 ($85 on the 9, $10 on the winner seven) and walks
Player B has no decision

Anyone can come up with a scenario where his/her method works better, because every time of betting is chasing a certain pattern of rolls. A flat bettor is concerned only with overall wins vs. losses; a progressive bettor is chasing a streak. A place bettor (across, inside, whatever) is chasing a bunch of point numbers before a seven. A pass/odds bettor is chasing a certain number before a seven. Every bet is independent, so when tuttigym places $64 across, he's paying 4% on the 5/9, 1.52% on the 6/8 and 6.7% on the 4/10 (although maybe he buys the 4/10).

Of course, tuttigym conveniently ignores the comeout, a common oversight among those touting the place bets.

Quote: tuttigym

PL/FO players risk much more on one number which is less likely to win simply because there are more ways to lose than to win.

Once again, you generalize. PL/FO players do not necessarily "risk much more". In your example the PL player is taking 5X odds, which isn't even possible most places on the 9 (3, 4, 5X is the most prevalent odds maximum now, I believe). Assuming that's a $10 minimum table, suppose the pass bettor bet $10 on the line and took single odds. That would be $20 risked. In order to cover all the points, you have to bet $64. The pass/odds bettor can bet anything from zero to $50 on the odds.

Quote: tuttigym

You seem to believe that if I lose with a 7 out after the point is established, somehow the PL/FO wagers are safe.

I don't see any reason to think that anyone believes that.

You seem to be unaware that the across bettor is fighting a house advantage on every one of those bets, while your PL/FO bettor has $50 of his $84 on a bet with no house advantage. If all of those bets are resolved, the ev for the PL/FO/6/8 bettor is -$.505, while the ev for the $64-across bettor is -$2.50 if you place the 4/10; if the casino allows you to buy the 4 and 10 $10 each for one $1 vig and collects the vig only on a win, the ev is still -$1.50.
Cheers,
Alan Shank

Quote: tuttigym

Alan: Terrific post although the numbers are mind numbing. When you stated the odds at 25,108 to 1 on the numbers 285 - 210, is that wins to losses or the opposite?

That is 285 wins, 210 losses.

Quote: tuttigym

Also, are you saying that based on the perfect math of 244/251, it will be a virtual "certainty," and it will require a minimum of 17,403 sets of 495 PL outcomes or within any given 17,403 sets of 495?

I am saying that is a virtual certainty to have occurred at some point in some casino, because there are so many casinos offering craps 24/7. The 17,403 is the number of sets of 495 passline decisions that have to take place before there is a .5 probability (50% chance) that one of those sets came out 285-210.

Quote: tuttigym

Isn't the 495 PL outcome a virtual number without reality? It is a continuum - when does it start and when does it end? I mean, what if the first seven shooters establish a point and then 7 out? Does the 495 re-start?

495 is just a number chosen because it's the fewest decisions where you can express the probabilities of different outcomes in whole numbers, as in:

110 comeout wins
55 comeout losses
134 point wins
196 point losses
---
495 (110 + 134 = 244 wins; 55 + 196 = 251 losses)
244/495 = .4929
251/495 = .5071

Those are the smallest whole numbers that reflect the actual relative probabilities of those outcomes. There is nothing magic about the number 495. I actually use 1980 because then you can express the probabilities of winning or losing each different point number.

Suppose you take the number 200; if you apply the known probability of winning a passline decision to that number, you get 98.5858. Well, you can't actually win .5858 of a passline decision, can you? 495 is used because you can actually win 244 and lose 251 and that W-L expresses the actual probabilities of winning and losing.

Whether you use whole numbers, fractions, percentages, decimal fractions, the numbers are all based on the well-known "pyramid" of dice results, with the seven in the middle, then 6/8, 5/9, 4/10, 3/11, 2/12 on either side.

comeout win: 6 sevens plus 2 elevens = 8 / 36 = .22222
comeout loss: 1 two plus 1 twelve plus 2 threes = 4 / 36 = .11111
point 6 : 5 sixes 5 / 36 = .138889
: 5 sixes win, 6 sevens lose, so p(win) = 5/11 = .454545
: p(win on 6) = .138889 * .454545 = .0631313
: p(lose on 6)= .138889 * .545454 = .0757571
etc. etc.

If you go through that exercise, and you add up all the win probabilities and all the loss probabilities, you get .492929 and .5070707; the difference between .492929 and .5070707 is .01414. It means that, if you bet passline for thousands and thousands of decisions, the percentage of wins will be very close to 49.3%. It's really no different from saying that, if you roll a pair of fair dice thousands and thousands of times, the percentage of sevens will be very close to 16.7%. One figure flows from the other. The dice probabilities drive the passline probability; the passline probability drives the probabilities of different W-L record for any number of decisions, etc. etc.

Cheers,
Alan Shank

Quote: boymimbo

The math on the above system:

You win $30 on 14/36 (3,4,9,10,11)
You win $60 on 1/36 (2)
You win $90 on 1/36 (12)
You win (44-30) $14 on 4/36 (5)
You win (35-30) $5 in 10/36 (6,8)
You lose $120 on 6/36 (7)

Add it all up: (420 + 60 + 90 + 56 + 50 - 720)/36 = -44/36 = -$1.222 / roll.

Per roll house advantage = $1.222 / $120 bet = 1.019% (per roll).

Compare that to craps appendix 2, where the pass line is .42%/roll, don't pass is .40%/roll, and the 6/8 (Place) is .493%.

Thanks. That was a much easier way of going about it.

I guess where I am fooling myself is with the concept of taking down the bet after 4 rolls, and what affect that has on the outcome. I understand the HA of 1.019% vs the .42%, but in order for that to be accurate, don't you have to play the same conditions? In other words, the bets are always in play, and eventually both PL and Iron Cross lose the total bet.
However, since I am taking down the bets after only 4 rolls, doesn't that have an affect on the calculations? Doesn't that have to be accounted for? Now the dispersion of the 7s becomes more critical, doesn't it? Since I have picked up all of my bets after roll 4, I have total profit and zero loss for every roll that goes longer than 4 throws.
If the bet stays up, I have to win rolls 5,6,7,and 8 to offset the $120 in bets I am going to lose. That now means the shooter has to beat the average of a 7 every 6th roll, in order to accomplish the same net amount that I win by picking up after roll 4.

Have I done these calculations correctly?
chance of rolling a 7 = 16.7%
chance of rolling a 4 or 10 = 8.3%
chance of rolling a 5 or 9 = 11.1%
chance of rolling a 6 or 8 = 13.9%
Chance of rolling 2/3/11/12 = 16.7%
chance of rolling anything but a 7 = 83.3%

Quote: gambler

I am a lazy gambler.

Can someone give me the cliff notes on this thread? Thank you. :)

Sorry - I just couldn't resist:

OP: The pass line is a sucker bet because the 1.41% house advantage is a hoax.

Everyone Else: Huh? Here's some math that proves the house advantage on the pass line bet is 1.41%

OP: You are all wrong. Here's some math that proves it.

Everyone Else: Your math is flawed. Here is some more math and results of computer simulations that prove that the house advantage is 1.41%

OP: No it's not. Your math is too complicated and your computer simulations are biased. I'm right and you are all wrong.

Everyone Else: Your math is still flawed and here's why.

OP: My math is not flawed. Here's a specific series of rolls that proves I'm right.

Everyone else: No it doesn't. Here's some more math.

...


Editor's note: this could go on forever because OP is convinced he is correct and that Everyone Else needs to be enlightened. Everyone Else keeps trying, to no avail, to point out the flaws in OP's logic.

^
^
^
Perfect.

Quote: tuttigym

Alan: There is no answer to either question that would provide certainty with one exception. A player that loses his entire bankroll regardless of the number of bets can never recover.

I have a friend who would go to the table and buy in for $5,000. His modest goal was to win between $750 and $900, period. He would play the PL/FO plus three numbers and his table bet after the point was established was $225 on any given point dispersed into the five bets. He would go to the casino at least once a month and play only one session.

For just under two years, he played with the same $5,000. During that time he won about $18,000.
Most times when he played, it took between 2 to 5 hours to win, and there were times he got down by as much as $3,000 only to come back and recover and eventually win that $750 or so. As you can see, he had a pretty good run.

Then three times in a row, he lost the $5,000. He just quit playing. He was not broke; he could afford the losses; in fact, he was still up about $3,000, but he just did not want to play craps. In three sessions, the "system" of PL/FO betting just beat him down.

Well, you have not described his betting. If he was making very large bets, then of course he could lose the $5000. But you can lose $5000 betting place bets, too. I can figure a "risk of ruin" for any betting system. You are mixing up the types of bets a person makes and the amount risked, it seems to me.

Quote: tuttigym

Anecdotally, I have watched players win and lose with the PL/FO betting strategy. It really sickens me when they continue to throw their money on the table win some and then lose it all or most of it and their buy in. While I recognize many make stupid sucker bets, the real culprit and the big losses come from the PL/FO or the Come/FO. They quote the low HA and rely on it for big time wins. It is truly a false promise.tuttigym

Well, all of our experience is anecdotal, because none of us has observed enough craps to draw valid conclusions. That is why simulation is so important. Here again, a place bettor is just as likely to "win some, lose it all or most of it and their buy in" as a PL/FO bettor. The low HA does not "promise" anything except a lower expected loss than a high HA applied to the same bet handle.

What you are getting at here, it seems to me, is volatility. A passline bet by itself has a very low volatility, because it's an even-money bet with an almost-50% chance to win it. If your friend had started with $5000 and made $100 passline bets, no odds, he would have had an extremely low probability of busting, along with about a 40% chance of winning at least $750 within about four hours of play. However, if he took 3,4,5X odds, he would have increased his chances of busting to about 1 in 6, while increasing his chances of winning the $750 to about 4 in 5. So, losing the $5000 three times in a row would be about as likely as three sevens in a row.

Taking odds add volatility, the more odds the more volatility. The more volatility, the higher probability of busting a given bankroll. However, place bets also have considerable volatility; after all, they are resolved under the same conditions as the odds bets and pay less-than-true odds. A player who splashes money across all the numbers is even more likely to bust $5000 than the PL/FO player with the same bet handle.

Quote: tuttigym

You all seem to have such negative replies because you are so heavily invested in the PL/FO strategies.

Try this as it happens most times and that is why the false premise.

Player A puts $10 on the PL; shooter comes out with a 9; A FO bet of $50 plus $12 on each of the 6 and 8.

Player B (after the point is established) wagers $64 across the board including the point.

Player A's expected value (per roll): 14 * 10/36 + 85 * 4/36 - 84 * 6/36 = -$.66667.
House advantage per roll: -$.66667/84 = 0.7937%

Player B's expected value (per roll): 6 / 36 * 18 + 8/36 * 14 + 10/36 * 14 - 6/36 * 64 = -$.66667.
House advantage per roll: -$.66667/64 = 1.0417%

To be fair, and to make things equal (bet size wise) I would only put triple odds on the 5 and 9 so that both A and B are betting $64.

Player A1's expected value (per roll): $14 * 10/36 + 55 * 4/36 - 64 * 6/36 = -$.66667.
House advantage per roll: -$.666667/64 = 1.0417%.

Now let's say the point is 4 or 10.

Player A2's expected value (per roll): $14 * 10/36 + 70 * 3/36 - 64 * 6/36 = -$.94444.
House advantage per roll: -$.944444/64 = 1.476%

And let's say the point is 6 or 8 and we're a dumb player and continue to place 3x odds and place the six or eight without backing up out point.

Player A3's expected value (per roll): $14 * 10/36 + 46 * 5/36 - 64 * 6/36 = -$.38889.
House advantage per roll: -$.388889/64 = 0.608%

All come out rolls established in a point: 8/24 * (-.66667) + 6/24 * (-.94444) + 10/24 * (-.388889) = -.62037. House advantage per roll = 0.969%. This is still better than placing all numbers.

Of course, we forget about the come out roll where we have an expected player advantage of $1.111 on a $10 bet. This is significant.

Quote: RaleighCraps

Quote: boymimbo

The math on the above system:

You win $30 on 14/36 (3,4,9,10,11)
You win $60 on 1/36 (2)
You win $90 on 1/36 (12)
You win (44-30) $14 on 4/36 (5)
You win (35-30) $5 in 10/36 (6,8)
You lose $120 on 6/36 (7)

Add it all up: (420 + 60 + 90 + 56 + 50 - 720)/36 = -44/36 = -$1.222 / roll.

Per roll house advantage = $1.222 / $120 bet = 1.019% (per roll).

Compare that to craps appendix 2, where the pass line is .42%/roll, don't pass is .40%/roll, and the 6/8 (Place) is .493%.

Thanks. That was a much easier way of going about it.

I guess where I am fooling myself is with the concept of taking down the bet after 4 rolls, and what affect that has on the outcome. I understand the HA of 1.019% vs the .42%, but in order for that to be accurate, don't you have to play the same conditions? In other words, the bets are always in play, and eventually both PL and Iron Cross lose the total bet.
However, since I am taking down the bets after only 4 rolls, doesn't that have an affect on the calculations? Doesn't that have to be accounted for? Now the dispersion of the 7s becomes more critical, doesn't it? Since I have picked up all of my bets after roll 4, I have total profit and zero loss for every roll that goes longer than 4 throws.
If the bet stays up, I have to win rolls 5,6,7,and 8 to offset the $120 in bets I am going to lose. That now means the shooter has to beat the average of a 7 every 6th roll, in order to accomplish the same net amount that I win by picking up after roll 4.

Have I done these calculations correctly?
chance of rolling a 7 = 16.7%
chance of rolling a 4 or 10 = 8.3%
chance of rolling a 5 or 9 = 11.1%
chance of rolling a 6 or 8 = 13.9%
Chance of rolling 2/3/11/12 = 16.7%
chance of rolling anything but a 7 = 83.3%

Those are correct. Actually, the average number of rolls for a shooter is 8.52, but that's including the comeout roll. Whether you pick your bets up after four rolls or not, the same probabilities apply when you make the bets again, unless, of course, you never play again. Also, you have a one-roll bet in there, so you can lose it up to four times if you replace it.

Here are some probabilities to consider:

after comeout:
p (seven on 1st roll) = .1667
p (seven on 2nd roll) = .8333 * .1667 = .1389
p (seven on 3rd roll) = .8333^2 * .1667 = .1158
p (seven on 4th roll) = .8333^3 * .1667 = .0965
p (no seven in 4 rolls) = .8333^4 = .4822

The delusion here is that there's any difference in their expected outcome between letting the bets resolve as opposed to taking them down, then putting them up for another four rolls later, etc. etc. What you are doing is spacing the bets out so that, per time, you bet less and therefore have less expected loss.
Cheers,
Alan Shank

Quote: goatcabin

Here are some probabilities to consider:

after comeout:
p (seven on 1st roll) = .1667
p (seven on 2nd roll) = .8333 * .1667 = .1389
p (seven on 3rd roll) = .8333^2 * .1667 = .1158
p (seven on 4th roll) = .8333^3 * .1667 = .0965
p (no seven in 4 rolls) = .8333^4 = .4822

The delusion here is that there's any difference in their expected outcome between letting the bets resolve as opposed to taking them down, then putting them up for another four rolls later, etc. etc. What you are doing is spacing the bets out so that, per time, you bet less and therefore have less expected loss.
Cheers,
Alan Shank

Thanks Alan.
Will you give me a bit of an explanation on the 'after comeout' probabilities above?
I would have expected the probability of a seven to get larger with each added roll, but it appears to me that it is going down.
Also, as I read it, there is a 48% chance of no seven in 4 rolls. Do I have that right ?

Since the HA only increases from .42% to 1% with this scheme, I am toying with trying it for 1 hour on my next trip, just to see how I fair. At 1%, it would still be an improvement from the current PL w/5x to 10x odds on any point, plus playing the buy 4,5,9,10 (vigs only on win), and place 6,8 that I currently do.

Thanks for the cliff notes Seattledice! I appreciate it.

.42% to 1% means that that for an average bet of $60/roll, you are losing an extra $.35, per roll. Over 60 rolls an hour, that's $21.

It's significant.

All: I will reply (group grooooaaaan!!!) in about 4 days as I am off to gamble at my closest available full service casino (6 hr drive).

I know you all wish me well, so thank you. From me to you all, stay healthy, alert, and feisty.

tuttigym

Quote: boymimbo

.42% to 1% means that that for an average bet of $60/roll, you are losing an extra $.35, per roll. Over 60 rolls an hour, that's $21.

It's significant.

I agree that it is a significant amount, if I had been only playing PL/FO and CB/FO. As with most craps players, I want more action than PL and 1 CB, so I ended up spreading $150 across, plus my PL/10x FO bet. Because of the vig on win buys on 4,5,9,10, my worst HA is the 2.0% on the 5,9. I am guessing my HA on this play is around 1.5%, depending on the point of course, which is greater than the 1% I would get with the Iron Cross.
I understand your point that I am not making the optimal play, and I do appreciate that advice. My reason for posting was to get help understanding exactly what the HA was in this system, and you and Alan have been very helpful. Thank you both.

I'm a bit worried that I will be playing the system when a couple of shooters throw 14x and I am on the bench for the last 10 throws. But, on the last trip, I remember a fairly long spell where no one could make a single point. They would establish a point, roll 2 to 4x, and then 7 out. The Iron Cross will limit my loss, and even get me a win, if they get to that 4th roll. This may help me last through the cold spells a bit better. When the table gets a roll going, switching to PL/FO and CB/FO is certainly the way to go.

"on a roll" really means nothing. True mathematicians will say that the odds of throwing a seven are always 1 in 6.

This is why I love craps. It is such a superstitious game: "That shooter is hot, bet on him"; "It's the first time that she's throwing, bet on her"; "The waitress came by - that's why he sevened out"; "his hands are in the tub - watch your hands!!!"; "he's a don't shooter - don't bet on him"; "you mentioned God - that's why the dice turned"; "he's setting dice - must be a good shooter"; "he doesn't care about his roll - don't bet on him"; "stick change - time to take down the bets!"; "he just threw a crap, all my bets are down!!".

In my opinion, the observance of hot and cold is just randomness at its beautiful work. So when you put up that bet, you are going up against the long term average of the house advantage, and the quality of that bet should be measured against the optimal bet which is the don't pass or come with as much odds as possible followed closely by the pass bet with as much odds as possible. You may come ahead of the PL/FO over some sessions, but in the long run, you are likely to lose more using your method because the long term house advantage > PL/FO.

In my opinion, placing a 4 and 10 is just about as bad as playing insurance in blackjack.

Quote: boymimbo

"on a roll" really means nothing. True mathematicians will say that the odds of throwing a seven are always 1 in 6.

This is why I love craps. It is such a superstitious game: "That shooter is hot, bet on him"; "It's the first time that she's throwing, bet on her"; "The waitress came by - that's why he sevened out"; "his hands are in the tub - watch your hands!!!"; "he's a don't shooter - don't bet on him"; "you mentioned God - that's why the dice turned"; "he's setting dice - must be a good shooter"; "he doesn't care about his roll - don't bet on him"; "stick change - time to take down the bets!"; "he just threw a crap, all my bets are down!!".

In my opinion, the observance of hot and cold is just randomness at its beautiful work. So when you put up that bet, you are going up against the long term average of the house advantage, and the quality of that bet should be measured against the optimal bet which is the don't pass or come with as much odds as possible followed closely by the pass bet with as much odds as possible. You may come ahead of the PL/FO over some sessions, but in the long run, you are likely to lose more using your method because the long term house advantage > PL/FO.

In my opinion, placing a 4 and 10 is just about as bad as playing insurance in blackjack.

The reason I love all this is that it is so metaphysical, so cosmological; it is tied into the concept of the arrow of time. The whole idea of "on a roll" is the attempt to try to see into the future, to predict the direction of the arrow. And there is a desire to be present at an extraordinary event. As a streak gets longer and longer, the tension rises; of course the chances on every roll are the same, but is this streak predestined? Looking forward it is random, but looking backward it has happened and is unreal! As it unfolds, do we dare that it will last longer? Do we risk our money (and the ire of the table) on betting that it turns?

It's so freaking awesome. It is a look into the very weave of the fabric of reality. But with stakes.

Mosca, you should write a book.

Quote: RaleighCraps

Quote: goatcabin

Here are some probabilities to consider:

after comeout:
p (seven on 1st roll) = .1667
p (seven on 2nd roll) = .8333 * .1667 = .1389
p (seven on 3rd roll) = .8333^2 * .1667 = .1158
p (seven on 4th roll) = .8333^3 * .1667 = .0965
p (no seven in 4 rolls) = .8333^4 = .4822

The delusion here is that there's any difference in their expected outcome between letting the bets resolve as opposed to taking them down, then putting them up for another four rolls later, etc. etc. What you are doing is spacing the bets out so that, per time, you bet less and therefore have less expected loss.
Cheers,
Alan Shank

Thanks Alan.
Will you give me a bit of an explanation on the 'after comeout' probabilities above?
I would have expected the probability of a seven to get larger with each added roll, but it appears to me that it is going down.
Also, as I read it, there is a 48% chance of no seven in 4 rolls. Do I have that right ?

Those probabilities are all based, as they must be, on a .1667 probability of a seven on ANY ROLL. The probability of rolling a seven NEVER CHANGES. Those probabilities are for the seven showing on a particular roll after a point is established. So, first time, it's just .1667. Second time, it's non-seven times seven, since the probability of two events occurring is the product of the individual probabilities. The third time, it's two non-seven event followed by a seven, etc. etc.

It's a common misconception that, because you expect one seven out of six rolls, each non-seven raises the probability of a seven showing. That is called the Gambler's Fallacy.

The probability of no seven for four rolls is .83333^4 = .4822.

Quote: RaleighCraps

Since the HA only increases from .42% to 1% with this scheme, I am toying with trying it for 1 hour on my next trip, just to see how I fair. At 1%, it would still be an improvement from the current PL w/5x to 10x odds on any point, plus playing the buy 4,5,9,10 (vigs only on win), and place 6,8 that I currently do.

That 1% is per roll. Another thing you may be overlooking is that, assuming you replace each bet that wins during those four rolls, when you win on the 5, 6 or 8, you also LOSE the field bet, so you net only $14 on the 5 and only $5 on the 6 or 8. Your overall HA is going to be over 2%, and that's assuming the most favorable possible conditions for the buy $5, that is, charging $1 vig for $30 buy each time and collecting the vig only on a win.

I went ahead and wrote a WinCraps auto-bet program for this method, with these rules:

$1000 bankroll
after comeout:
buy 5 for $30
place 6 and 8 for $30 each
Field for $30

replace winning bets and losing Field bets
if shooter makes his point, no field bet on next comeout, other bets are OFF
on fifth roll after the point is established, take all bets down (or call OFF), wait for seven-out

session stopping conditions:
1) bankroll less than $120
OR
2) reach 200 rolls AND a comeout roll

So this represents something like two hours' play for a session.
I ran 10,000 sessions, roughly 2 million dice rolls.

WinCraps keeps track of just about everything, rolls, number of bets of each type, W-L for each bet, bet handle and net outcome, on a session-by-session basis; it figures all the descriptive statistics for each and even draws you a graph of the distribution. It is truly an awesome program, which you can get at www.cloudcitysoftware.com. A demo is free, just $19.95 to register and get full functionality. BTW, I am not associated with Cloud City Software, but I have used WinCraps for years.

the average session:

191 rolls
186 total bets resolved
$5584 total bet handle
net outcome:
mean: -$121
median: -$114 (equal number above and below)
mode: from -$917 to -$889 (most frequently occurring value)
standard deviation: $544
house advantage: -2.17%

4158 sessions ended ahead
4 broke even
5838 ended behind

2 sessions lost the whole $1000
1620 sessions ended due to not having $120 left
2655 lost more than $500
4074 lost more than $250
2592 won more than $250
1355 won more than $500
568 won more than $750
194 won more than $1000
5 won more than $1500

That business with the weighted average of wins did not take into consideration that the Field loses when the 5, 6 and 8 win. That is a big problem. OTOH, if you don't replace the Field bet when it loses, then you don't have your 30 ways to win on the next roll.

As Roseanne Roseannadanna used to say, "It's always something!". >:-)
Cheers,
Alan Shank

Quote: goatcabin

I went ahead and wrote a WinCraps auto-bet program for this method, with these rules:

$1000 bankroll
after comeout:
buy 5 for $30
place 6 and 8 for $30 each
Field for $30

replace winning bets and losing Field bets
if shooter makes his point, no field bet on next comeout, other bets are OFF
on fifth roll after the point is established, take all bets down (or call OFF), wait for seven-out

session stopping conditions:
1) bankroll less than $120
OR
2) reach 200 rolls AND a comeout roll

So this represents something like two hours' play for a session.
I ran 10,000 sessions, roughly 2 million dice rolls.

WinCraps keeps track of just about everything, rolls, number of bets of each type, W-L for each bet, bet handle and net outcome, on a session-by-session basis; it figures all the descriptive statistics for each and even draws you a graph of the distribution. It is truly an awesome program, which you can get at www.cloudcitysoftware.com. A demo is free, just $19.95 to register and get full functionality. BTW, I am not associated with Cloud City Software, but I have used WinCraps for years.

the average session:

191 rolls
186 total bets resolved
$5584 total bet handle
net outcome:
mean: -$121
median: -$114 (equal number above and below)
mode: from -$917 to -$889 (most frequently occurring value)
standard deviation: $544
house advantage: -2.17%

4158 sessions ended ahead
4 broke even
5838 ended behind

2 sessions lost the whole $1000
1620 sessions ended due to not having $120 left
2655 lost more than $500
4074 lost more than $250
2592 won more than $250
1355 won more than $500
568 won more than $750
194 won more than $1000
5 won more than $1500

That business with the weighted average of wins did not take into consideration that the Field loses when the 5, 6 and 8 win. That is a big problem. OTOH, if you don't replace the Field bet when it loses, then you don't have your 30 ways to win on the next roll.

As Roseanne Roseannadanna used to say, "It's always something!". >:-)
Cheers,
Alan Shank

Holy craps Alan!
That was an awesome explanation, and a rousing endorsement for WinCraps. I had heard of the program, but did not realize it was that powerful. I believe I will order the full feature program later tonight.

I was pretty sure I had accounted for losing the $30 field bet on the 5,6,8, as I only counted $5 or $14 as the win (since I had to replace the $30).

Roll	Ways	Bet	Win	Freq	Per roll Profit	Pro-rated profit
2	1	$30	$60	2.78%	$60	$1.67
3	2	$30	$30	5.56%	$30	$1.67
4	3	$30	$30	8.33%	$30	$2.50
5	4	$30	$44	11.11%	$14	$1.50
6	5	$30	$35	13.89%	$5	$0.69
7	6			16.66%	($120)	($19.99)
8	5	$30	$35	13.89%	$5	$0.69
9	4	$30	$30	11.11%	$30	$3.33
10	3	$30	$30	8.33%	$30	$2.50
11	2	$30	$30	5.56%	$30	$1.67
12	1	$30	$90	2.78%	$90	$2.50
	36			100.00%		($1.27)

I see what I did wrong. I added up the per roll wins and divided by 10. I realize now that was not a weighted average.

Nice assumption on the session bankroll ;-)
Too bad the results are so depressing. Have I opened Pandora's box? Is it too late to unlearn the information I have been given? There is something to be said for mildly ignorant and blissful.

Thanks again for the fabulous analysis.

Quote: boymimbo

"on a roll" really means nothing. True mathematicians will say that the odds of throwing a seven are always 1 in 6.

This is why I love craps. It is such a superstitious game: "That shooter is hot, bet on him"; "It's the first time that she's throwing, bet on her"; "The waitress came by - that's why he sevened out"; "his hands are in the tub - watch your hands!!!"; "he's a don't shooter - don't bet on him"; "you mentioned God - that's why the dice turned"; "he's setting dice - must be a good shooter"; "he doesn't care about his roll - don't bet on him"; "stick change - time to take down the bets!"; "he just threw a crap, all my bets are down!!".

In my opinion, the observance of hot and cold is just randomness at its beautiful work. So when you put up that bet, you are going up against the long term average of the house advantage, and the quality of that bet should be measured against the optimal bet which is the don't pass or come with as much odds as possible followed closely by the pass bet with as much odds as possible. You may come ahead of the PL/FO over some sessions, but in the long run, you are likely to lose more using your method because the long term house advantage > PL/FO.

In my opinion, placing a 4 and 10 is just about as bad as playing insurance in blackjack.

Beautifully written. You have summed up craps in a nice paragraph. Hot and cold though is real, lol. Last trip we went 3x around the table and no one could make a single point. Then I had a 20 minute roll, followed by the dice spending over an hour on the other half of the table. It was 1.5 hours before I got to roll again. I have seen this happen over and over. I understand there is no scientific reason for it, but that doesn't make it not real ...or does it?

I agree with your assessment on placing the 4/10, but when they are only charging a vig on the win, the HA is 1.67%, making it just slightly worse than placing the 6/8 at 1.52%

The final consideration is, I know you are never to play for comps, BUT, placing and buying bets counts for good comps, whereas FO bets don't get you any credit. So you have the same amount of money at risk, but one way they are comping you heavily, and with FO, you only get credit for the PL or come bet. Getting the room and meals, and most times the flights, offsets the expected loss increase I have with the place/buy bets.

You have proved your point though, and I agree that playing this scheme all the time would eventually end up with my money adding to MGM or Harrah's bottom line.

I can't believe you guys have not figured out this is sevenshooter again. I'm sure he will have solid winning results from his gambling trip which will blow all your mathematical proof out of the water.

Quote: Headlock

I can't believe you guys have not figured out this is sevenshooter again. I'm sure he will have solid winning results from his gambling trip which will blow all your mathematical proof out of the water.

Different guys. The personalities and writing styles are different. It didn't take a lot of sleuthing to turn up different credible identities:

tuttigym

sevenshooter

Google is always the first try, and it is getting more and more powerful.

Neither are really bad guys, if you read between the lines.

Quote: RaleighCraps

Beautifully written. You have summed up craps in a nice paragraph. Hot and cold though is real, lol. Last trip we went 3x around the table and no one could make a single point. Then I had a 20 minute roll, followed by the dice spending over an hour on the other half of the table. It was 1.5 hours before I got to roll again. I have seen this happen over and over. I understand there is no scientific reason for it, but that doesn't make it not real ...or does it

Let's define "hot" and "cold", so we all know what we're talking about. Which definition is closer to what you mean?

1. A "hot" table is one where the last (insert some number) of post-comeout rolls has had relatively few sevens and lots of point numbers, resulting in long rolls for all or most shooters.

2. A "hot" table is one where the last (insert some number) of post-comeout rolls has had relatively few sevens and lots of point numbers, resulting in long rolls for all or most shooters, and there is some reason to believe that this condition will continue into the future.

Now, in the situation you described, the table was "cold" for 3X around the table, but then it turned "hot". If the "cold" condition was actually some repeatable condition, how did the table turn to "hot"?

A common misconception among people about random numbers is that they must not have any pattern to them, that the results should be scattered. But truly random numbers (as well as pseudo-random ones generated by an algorithm) are expected to show streaks of various kinds. The critical point is that an observed streak gives no information about the next roll or (insert number) of rolls. A "hot" table has no guarantee of continued "heat", nor does it imply that the table is "due" to turn cold. People are very good at detecting patterns in data; too good.
Cheers,
Alan Shank

Quote: goatcabin

Let's define "hot" and "cold", so we all know what we're talking about. Which definition is closer to what you mean?

1. A "hot" table is one where the last (insert some number) of post-comeout rolls has had relatively few sevens and lots of point numbers, resulting in long rolls for all or most shooters.

2. A "hot" table is one where the last (insert some number) of post-comeout rolls has had relatively few sevens and lots of point numbers, resulting in long rolls for all or most shooters, and there is some reason to believe that this condition will continue into the future.
.
.
.
A "hot" table has no guarantee of continued "heat", nor does it imply that the table is "due" to turn cold. People are very good at detecting patterns in data; too good.
Cheers,
Alan Shank

I define a "hot" table as one that is very favorable to the right side players. A shooter who makes 3 or 4 points, is "hot", even if they did it in as few as 12 rolls and then 7 out. A shooter who makes no point, but executes 30 rolls is also "hot". Even though my PL/FO bet lost, the money made from place/buy bets means a positive outcome.
From a purely arithmetical perspective, past rolls have no bearing on future outcomes. And I am sure we all have seen one shooter throw for 20 minutes, and then the next one 7 outs on the first roll of the point. But, my experience has been that it generally doesn't end that quickly, and the next shooter gets a couple of points, or at least hits enough place/buy bets to break even on the roll.
I have observed the following over my last 9 craps trips (each trip lasts 2 to 7 days, and I will average 4 to 7 hours a day). EVERY trip I have been at the tables when there has been a "hot" streak, meaning 5 or more shooters out of 8 have had exceptional rolls. On the trips where I have come out ahead, I have been actively playing when that streak occurs. On the trips I have lost money, I have usually been getting killed and have already dropped into some ultra-conservative mode (no PL - only place 6/8,,, or only PL/some odds, and no other bets), and thus missed out on the majority of the roll, and the chance to heal my bankroll. The continued bleeding eventually breaks my bankroll.

I realize there is no science to back this up, but the alternative is to concede to the fact that the house has an unbeatable edge that can never be overcome, and this is all futile. That is just too depressing to even consider. Instead, I believe that all of my sessions are just a blip on that long time frame that is required for the statistical averages to work out, so because I am dealing with short time frames, any outcome is possible.

As others have pointed out, I too view this as recreational spend. I hope I don't lose my bankroll, but if I do, that just makes it an expensive vacation.