Calculate 1.4% on pass line

I can’t figure out how to calculate the true odds of winning on a pass or come bet, ie bet $25 on pass. I know that the house edge is 251 to 244 so 1.41% but how is that calculated. Obviously it includes winning on 7/11 and losing on craps before a point. Help.

Quote: Bossdog

I can’t figure out how to calculate the true odds of winning on a pass or come bet, ie bet $25 on pass. I know I house edge is 251 to 244 so 1.41% but how is that calculated. Obviously it includes winning ob 7/11 and losing on craps before a point. Help.
link to original post

Are we doing homework? JK. It's fine; I wrote research papers in college for cash, anyway.

The first thing that we note is that we immediately win if the Come Out roll is either 7 or 11. Your example says bet $25, so we might as well.

(2/36 + 6/36) * 25 = 5.55555555556

The second thing we note is that we immediately lose if the Come Out roll is 2, 3 or 12:

-(1/36 + 1/36 + 2/36) * 25 = -2.77777777778

The remaining numbers are 4, 5, 6, 8, 9 and 10. If we roll a Point Number, then the only numbers a Pass Line bet cares about are that Point Number and 7. If you roll the Point Number again, baby gets a new pair of shoes; if you don't, baby is walking home barefoot.

ROLLS:

4 OR 10: 6/36

5 OR 9: 8/36

6 OR 8: 10/36

You can also sum these with the other Come Out probabilities from above to prove 36/36; I do not know how anal the teacher is. JK

At that point, fours and tens are 3/9 to win and 6/9 to lose. Remember, other results just mean we get to stand by a Craps Table longer. We should be very happy to stand there longer; you cannot see 18 Yo's in a row unless you stand by a Craps Table for a significant period of time.

(3/9 * 25) - (6/9 * 25) = -8.33333333333

Above reflects the expected loss from a Come Out Roll of either four or ten. What we have to do now is multiply that by the probability of such a CO. CO is also an abbreviation for Correctional Officer, but college students usually won't have to worry about those. At least...not for anything serious. We should all have a little fun from time to time.

-8.33333333333 * 6/36 = -1.38888888889

The next thing we have to look at is fives and nines. However, if you complete your Stats & Prob courses, as well as other courses, you might have a nice nine to five. That's one way to try to make a living. Dolly Parton has imparted this knowledge unto us; may we carry it with us and use the wisdom passed on by that large-breasted muse, for all time.

(4/10 * 25) - (6/10 * 25) = -5

-5 * 8/36 = -1.11111111111

Finally:

(5/11 * 25) - (6/11 * 25) = -2.27272727273

-2.27272727273 * 10/36 = -0.63131313131

We have already accounted for winning rolls during which we have established a point number. Our only remaining positive Expected Outcome is a Come Out winner, so we must subtract from its Expected Value the Expected Value of all of our barefoot outcomes.

5.55555555556 -2.77777777778 -1.38888888889 -1.11111111111 - 0.63131313131 = -0.35353535353

That means, for every $25 Pass Line bet, we expect to lose $0.353535353535353 cents. I hate change, so this is fine with me.

-0.35353535353/25 = -0.01414141414

Voila! This converts to a House Edge of 1.41414141414141414141414141414141414141414141%. Or, just 1.41%, if you prefer.

Hey, if you play Craps you might go home barefoot, but always give your dealers shoes on their toke bets.

If you consider 1980 then
Pr (7) =6/36 = 1/6 : 1980/6 = 330 (i.e. out of 1980 rolls you win on a 7 330 times)
Pr (11) = 2/36 = 1/18 : 1980/18 = 110
Pr (4) Pr (win4) = 3/36*1/3 = 1/36 : 1980/36 = 55
Pr (5) Pr (win5) = 4/36*4/10 = 16/360 : 88
Pr (6) Pr (win6) = 5/36 * 5/11 = 25/496 : 125
8-10 counts for 55+88+125 = 268
The total of this is 976
So you win 976 and lose 1004 (1980-976) which is simiilar to your figures.

Wow. Thank you. So where does 1980 come from or was that just your number to use. How did you account for 3 and 12 on first roll, plus you need to roll place number twice.

Thanks. No homework. Was at the table in Vegas last night doing the usual, pass line, then two come bets with full odds. Point was 8. Some guy came after the come out roll and $25 on the pass line with $125 odds, and I was trying to calculate whether that made sense or not, plus couldn’t figure out where the often quoted “251 to 244” comes from.

Quote: Bossdog

Wow. Thank you. So where does 1980 come from or was that just your number to use. How did you account for 3 and 12 on first roll, plus you need to roll place number twice.
link to original post

The 1980 is used to ensure all the numbers come out (no pun ntended) as whole numbers. it only looks at the win scenarios (as for a PASS or COME the rest are losers). When you finish the process the totals can be divided by 4, but the intermediary figures can't.

Simple example, out of 1980 rolls one in six will be a come-out 7, so this is 1980/6 = 330..
When you get to rolling a 6 on the come out and then making the point, 5 out of 36 rolls with be a 6, that is 275 out of 1980. Of these 5/11 win and 6/11 lose, that's 125 or 150.
You can use similar logic for the other ways to win (and lose)

Question as I try to learn from your calcs.

Pr (4) Pr (win4) = 3/36*1/3 = 1/36 : 1980/36 = 55
Pr (5) Pr (win5) = 4/36*4/10 = 16/360 : 88
Pr (6) Pr (win6) = 5/36 * 5/11 = 25/496 : 125

I understand the pr(4) 3/36 but where does the 1/3 come from. Same as the 4/10 and 5/11 for 6 and 6 respectively.

1/3 is the same as 3/9 (i.e. 3 ways to make 4 compared to 6 ways to make 7). Thus you have 3 chances from 9 to make the point, rather than 7-out. Similar logic for the others.

Quote: Bossdog

Wow. Thank you. So where does 1980 come from or was that just your number to use. How did you account for 3 and 12 on first roll, plus you need to roll place number twice.
link to original post

1980 is the smallest number where all of the other numbers in the calculations are integers.
The fractions are:
(4) 3/36 x 1/2 = 3/72 = 1/24
(5) 4/36 x 2/5 = 8/180 = 2/45
(6) 5/36 x 5/11 = 25/396
(7) 1/6
(8) 5/36 x 5/11 = 25/396
(9) 4/36 x 2/5 = 8/180 = 2/45
(10) 3/36 x 1/2 = 3/72 = 1/24
(11) 2/36 = 1/18
The probability = the sum of these numbers; the denominator has to be a multiple of 396, as otherwise the numerators for 6 and 8 are not integers, and it also has to be a multiple of 45, as otherwise the numerators for 5 and 9 are not integers; the smallest number that is a multiple of both is 1980, which also happens to be multiples of 24, 6, and 18.

If you like to simply things without losing much accuracy:

2/9 + 2/3 * 2/5 = 22/45 chance of winning 2 for 1, so house edge of 1/45 = 2.2%

This assumes every point has a 4/36 chance of being rolled. I’ve found this assumption gives accurate approximations for almost every multi-point craps calculation, while greatly simplifying them

I figured out the “always quoted” 251 to 244 true odds on pass or come bet without odds.

Basically, it is the first roll, and then a second (after that each roll without 7 is the same) so just need probability of winning on first and or making your point.

The probability of winning on first roll, either 7 (6/36) or 11 (2/36), is 8/36. The probability of losing on this one roll, 2 (1/36), 3 (2/36), 12 (1/36) is 4/36. So player has an advantage, thus the reason why you can’t take down a pass line bet AND why it makes no sense to make a pass line bet after the initial roll.

So we start with 8/36 chance of wining on first roll.

Next, it is the probability of establishing a point and then hitting it again before a 7. After the come out roll, only a 7 loses. If any number other than your point comes up, it doesn’t matter.

So, for 4 and 10. There are 3 ways to make each and 6 ways to make 7, so for either, 3/9 to win and 6/9 to lose. Thus the 2:1 payoff on odds.

5 and 9. There are 4 ways to make each and 6 ways to make 7, so for either, 4/10 to win and 6/10 to lose. Thus the 3:2 payoff on odds.

6 and 8. There are 5 ways to make each and 6 ways to make 7, so for either, 5/11 to win and 6/11 to lose. Thus the 6:5 payoff on odds.

Back to probability of winning a pass line or come bet.

The probability of making a point and then winning is the probability of making the point + probability of making point before 7.

For 4, 3/36 x 3/9 = 1/36
For 5, 4/36 x 4/10 = 2/45
For 6, 5/36 x 5/11 = 25/396
For 8, 5/36 x 5/11 = 25/396
For 9, 4/36 x 4/10 = 2/45
For 10, 3/36 x 3/9 = 1/36

Probability of winning on come out roll is sum of above + 8/36, which is 244/495. I used an online calculator as adding those fractions is brutal otherwise.

So probability of winning a pass or come bet is 244/495

Probability of losing is 251/495

Casino edge is 251/495 - 244/495 = 7/495 which is 1.41%

So never make a pass line bet if you come late to the game, and take full odds to bring the 1.41 even lower.

Quote: Bossdog

Casino edge is 251/495 - 244/495 = 7/495 which is 1.41%

So never make a pass line bet if you come late to the game, and take full odds to bring the 1.41 even lower.

Odds bets do not lower the house edge on the pass line bet you have made.

They are a separate, correlated, no house edge bet. I’ll let someone else explain in more detail if my simple explanation is not enough for you….

And welcome to the forum!

Agree, but once you hit a point, and you put down the maximum odds, it theoretically reduces the house edge on the whole bet, or you can view it as a no house edge separate bet. Kind of like getting $10 off when you spend $100 at an online store. You could say it reduces your $100 to $90, or view the $10 as a separate amount.

the way you can lower the HE of Craps is to put the greater portion of your total action on bets that have no edge

you can't lower the expected value of a line bet by adding *any* bet available though

when you don't play craps, you may not be able to square these two facts. It revolves around the stupidity of making a $50 line bet on a table with a $10 minimum. If $10 is all you care to bet, there's a problem, but it really shouldn't be if you spend hours at the table betting a total of several thousand dollars, recirculating making it possible, sure, but I hope you see what I mean

Quote: Mission146

Quote: Bossdog

I can’t figure out how to calculate the true odds of winning on a pass or come bet, ie bet $25 on pass. I know I house edge is 251 to 244 so 1.41% but how is that calculated. Obviously it includes winning ob 7/11 and losing on craps before a point. Help.
link to original post

Are we doing homework? JK. It's fine; I wrote research papers in college for cash, anyway.

The first thing that we note is that we immediately win if the Come Out roll is either 7 or 11. Your example says bet $25, so we might as well.

(2/36 + 6/36) * 25 = 5.55555555556

The second thing we note is that we immediately lose if the Come Out roll is 2, 3 or 12:

-(1/36 + 1/36 + 2/36) * 25 = -2.77777777778

The remaining numbers are 4, 5, 6, 8, 9 and 10. If we roll a Point Number, then the only numbers a Pass Line bet cares about are that Point Number and 7. If you roll the Point Number again, baby gets a new pair of shoes; if you don't, baby is walking home barefoot.

ROLLS:

4 OR 10: 6/36

5 OR 9: 8/36

6 OR 8: 10/36

You can also sum these with the other Come Out probabilities from above to prove 36/36; I do not know how anal the teacher is. JK

At that point, fours and tens are 3/9 to win and 6/9 to lose. Remember, other results just mean we get to stand by a Craps Table longer. We should be very happy to stand there longer; you cannot see 18 Yo's in a row unless you stand by a Craps Table for a significant period of time.

(3/9 * 25) - (6/9 * 25) = -8.33333333333

Above reflects the expected loss from a Come Out Roll of either four or ten. What we have to do now is multiply that by the probability of such a CO. CO is also an abbreviation for Correctional Officer, but college students usually won't have to worry about those. At least...not for anything serious. We should all have a little fun from time to time.

-8.33333333333 * 6/36 = -1.38888888889

The next thing we have to look at is fives and nines. However, if you complete your Stats & Prob courses, as well as other courses, you might have a nice nine to five. That's one way to try to make a living. Dolly Parton has imparted this knowledge unto us; may we carry it with us and use the wisdom passed on by that large-breasted muse, for all time.

(4/10 * 25) - (6/10 * 25) = -5

-5 * 8/36 = -1.11111111111

Finally:

(5/11 * 25) - (6/11 * 25) = -2.27272727273

-2.27272727273 * 10/36 = -0.63131313131

We have already accounted for winning rolls during which we have established a point number. Our only remaining positive Expected Outcome is a Come Out winner, so we must subtract from its Expected Value the Expected Value of all of our barefoot outcomes.

5.55555555556 -2.77777777778 -1.38888888889 -1.11111111111 - 0.63131313131 = -0.35353535353

That means, for every $25 Pass Line bet, we expect to lose $0.353535353535353 cents. I hate change, so this is fine with me.

-0.35353535353/25 = -0.01414141414

Voila! This converts to a House Edge of 1.41414141414141414141414141414141414141414141%. Or, just 1.41%, if you prefer.

Hey, if you play Craps you might go home barefoot, but always give your dealers shoes on their toke bets.
link to original post

This post is very impressive. How long did it take you to complete?

The relevant question here is: Is it REAL?

How many rolls of the dice will it take to complete the above set without an improper duplication of a hand, i.e., two come out 12's or better yet 6 come out 7's?

What are the ODDS that the equation can be performed?

My guess: Winning the lottery would be easier?

tuttigym

Sure. But so what? Dice have no memory.

Quote: unJon

Sure. But so what? Dice have no memory.
link to original post

Exactly, so why so much discussion and articulation of the HE? Based on your post; it is meaningless. ("So what?")

tuttigym

If I make 400 line bets in 6 hours (1 PL & 2 Comes up at a time) for $25 a piece that'd be $10,000 total bet. The HA would be $10,000 X 1.41% = $141 or just under six $25 bets. With $125 odds or up to $50,000 extra bet, just one extra odds win could make that $141 loss up; or one extra odds loss could nearly double up my loss. But there is no HA on the odds bets so it will have no theoretical difference on my HA of -$141.

Quote: tuttigym

Quote: Mission146

Quote: Bossdog

I can’t figure out how to calculate the true odds of winning on a pass or come bet, ie bet $25 on pass. I know I house edge is 251 to 244 so 1.41% but how is that calculated. Obviously it includes winning ob 7/11 and losing on craps before a point. Help.
link to original post

Are we doing homework? JK. It's fine; I wrote research papers in college for cash, anyway.

The first thing that we note is that we immediately win if the Come Out roll is either 7 or 11. Your example says bet $25, so we might as well.

(2/36 + 6/36) * 25 = 5.55555555556

The second thing we note is that we immediately lose if the Come Out roll is 2, 3 or 12:

-(1/36 + 1/36 + 2/36) * 25 = -2.77777777778

The remaining numbers are 4, 5, 6, 8, 9 and 10. If we roll a Point Number, then the only numbers a Pass Line bet cares about are that Point Number and 7. If you roll the Point Number again, baby gets a new pair of shoes; if you don't, baby is walking home barefoot.

ROLLS:

4 OR 10: 6/36

5 OR 9: 8/36

6 OR 8: 10/36

You can also sum these with the other Come Out probabilities from above to prove 36/36; I do not know how anal the teacher is. JK

At that point, fours and tens are 3/9 to win and 6/9 to lose. Remember, other results just mean we get to stand by a Craps Table longer. We should be very happy to stand there longer; you cannot see 18 Yo's in a row unless you stand by a Craps Table for a significant period of time.

(3/9 * 25) - (6/9 * 25) = -8.33333333333

Above reflects the expected loss from a Come Out Roll of either four or ten. What we have to do now is multiply that by the probability of such a CO. CO is also an abbreviation for Correctional Officer, but college students usually won't have to worry about those. At least...not for anything serious. We should all have a little fun from time to time.

-8.33333333333 * 6/36 = -1.38888888889

The next thing we have to look at is fives and nines. However, if you complete your Stats & Prob courses, as well as other courses, you might have a nice nine to five. That's one way to try to make a living. Dolly Parton has imparted this knowledge unto us; may we carry it with us and use the wisdom passed on by that large-breasted muse, for all time.

(4/10 * 25) - (6/10 * 25) = -5

-5 * 8/36 = -1.11111111111

Finally:

(5/11 * 25) - (6/11 * 25) = -2.27272727273

-2.27272727273 * 10/36 = -0.63131313131

We have already accounted for winning rolls during which we have established a point number. Our only remaining positive Expected Outcome is a Come Out winner, so we must subtract from its Expected Value the Expected Value of all of our barefoot outcomes.

5.55555555556 -2.77777777778 -1.38888888889 -1.11111111111 - 0.63131313131 = -0.35353535353

That means, for every $25 Pass Line bet, we expect to lose $0.353535353535353 cents. I hate change, so this is fine with me.

-0.35353535353/25 = -0.01414141414

Voila! This converts to a House Edge of 1.41414141414141414141414141414141414141414141%. Or, just 1.41%, if you prefer.

Hey, if you play Craps you might go home barefoot, but always give your dealers shoes on their toke bets.
link to original post

This post is very impressive. How long did it take you to complete?

The relevant question here is: Is it REAL?

How many rolls of the dice will it take to complete the above set without an improper duplication of a hand, i.e., two come out 12's or better yet 6 come out 7's?

What are the ODDS that the equation can be performed?

My guess: Winning the lottery would be easier?

tuttigym
link to original post

Is the post real? Yes. You just quoted it. You could probably even print it, if you wanted to.

How long? I don't know. 10 minutes, maybe? I figured it would be faster just to do it again than look for where I'd done it before, which I'm sure I have.

What are the Odds the equation can be performed? I performed the equation with a probability of 100%, as it has already happened. You just quoted it. Prior to completing it, probably 99.99999999% as the only thing that would have prevented me would have been an extremely untimely, and coincidental, death.

Quote: tuttigym

Quote: unJon

Sure. But so what? Dice have no memory.
link to original post

Exactly, so why so much discussion and articulation of the HE? Based on your post; it is meaningless. ("So what?")

tuttigym
link to original post

You should just bet Any 7 every time if you don't believe in House Edge. It's fun and you get to be a contrarian on every roll except the Come Out.

Quote: tuttigym

This post is very impressive. How long did it take you to complete?

The relevant question here is: Is it REAL?

How many rolls of the dice will it take to complete the above set without an improper duplication of a hand, i.e., two come out 12's or better yet 6 come out 7's?

What are the ODDS that the equation can be performed?

My guess: Winning the lottery would be easier?

tuttigym
link to original post

I know you’re a math denier, but I’ll give you an example

An average craps player could easily make 100,000 PL bets over the course of his multi-decade “career”. Over that term, he has a 95% chance of losing between 0.8% and 2% of all money wagered. There will be huge variance from the 1.4% for some sessions/years, but long-term the 1.4% expectation is definitely real. His friends making mostly place bets and hardways will lose several times more over that period, which is also real and essentially guaranteed

Then look at a global company like MGM that takes like 1,000 PL bets per minute. Over the course of just one month, their total hold on all PL bets made is pretty much exactly 1.41%. That is also real

It seems that the only phrase you despise more than “math” is “long-term”. When you think about it, most things only matter over the long-term. Like investing and financial success…it generally doesn’t matter what happens in one day. What matters is how the market performs and how you behave over decades.

Well, 100,000 line bets divided by 400 line bets in my above post equals 250 6-hour sessions or $141 x 250 = $35,250 of HA on $25 line bets. Total bets are 100,000 x $25 = $2.5 million. 0.8% of $2.5 million = $20,000 loss; 2% of $2.5 million = $50,000.
Now to add odds on of $125 (5X odds in this case for the poster), up to $12.5 million in odds; 0.8% x $12.5 million = $100,000, 2% x $12.5 million = $250,000.

So losing $20K to $50K on the line bets is dwarfed by the possible loss of $100K to $250K on odds bets? I thought odds bets were 0 HA, so maybe there's no calculation to make on the odds bets except for the variance which could add up to the 0.8% to 2%.

But it's certainly possible to lose 10% to 20% of total bets in a night on line bets and odds just from bad luck. 400 X $25 = $10,000 bet on the line bets X 10% to 20% = $1,000 to $2,000 loss in 6 hours or 40 to 80 line bets. Add odds of $125 and it'd be another $5,000 to $10,000 lost on odds bets that went down with points not made in one night. I'll be wondering where the upside is because Hot Shooters don't come along often enough. I'll have to go darkside to counter this kind of anticipated losing streak.

Quote: ChumpChange

Well, 100,000 line bets divided by 400 line bets in my above post equals 250 6-hour sessions or $141 x 250 = $35,250 of HA on $25 line bets. Total bets are 100,000 x $25 = $2.5 million. 0.8% of $2.5 million = $20,000 loss; 2% of $2.5 million = $50,000.
Now to add odds on of $125 (5X odds in this case for the poster), up to $12.5 million in odds; 0.8% x $12.5 million = $100,000, 2% x $12.5 million = $250,000.

Speaking of 100k…for the 100,000th time, you don’t add odds bets to your preferred bet level, you reduce your line bet so that the total bet amount (line + odds) is right

If $25 is your comfort level then you should be betting $5 or $10 on the line. Your total average bet will be around $25 and the edge will be 0.37% on your total wagers. No better deal than that except dont pass plus odds. Fairly low standard deviation of 1.3 on average wager

And since we are discussing the long-run, there are no “Hot Shooters” in the long run. Or you could say the “Cold Shooters” will cancel the “Hot Shooters” almost perfectly

I expect a Hot Shooter who can roll 20 rolls or more as one shooter to occur at least once an hour. But it's like keno, sometimes it's more often, sometimes they don't show.

Quote: Mission146

Quote: tuttigym

Quote: Mission146

Quote: Bossdog

I can’t figure out how to calculate the true odds of winning on a pass or come bet, ie bet $25 on pass. I know I house edge is 251 to 244 so 1.41% but how is that calculated. Obviously it includes winning ob 7/11 and losing on craps before a point. Help.
link to original post

Are we doing homework? JK. It's fine; I wrote research papers in college for cash, anyway.

The first thing that we note is that we immediately win if the Come Out roll is either 7 or 11. Your example says bet $25, so we might as well.

(2/36 + 6/36) * 25 = 5.55555555556

The second thing we note is that we immediately lose if the Come Out roll is 2, 3 or 12:

-(1/36 + 1/36 + 2/36) * 25 = -2.77777777778

The remaining numbers are 4, 5, 6, 8, 9 and 10. If we roll a Point Number, then the only numbers a Pass Line bet cares about are that Point Number and 7. If you roll the Point Number again, baby gets a new pair of shoes; if you don't, baby is walking home barefoot.

ROLLS:

4 OR 10: 6/36

5 OR 9: 8/36

6 OR 8: 10/36

You can also sum these with the other Come Out probabilities from above to prove 36/36; I do not know how anal the teacher is. JK

At that point, fours and tens are 3/9 to win and 6/9 to lose. Remember, other results just mean we get to stand by a Craps Table longer. We should be very happy to stand there longer; you cannot see 18 Yo's in a row unless you stand by a Craps Table for a significant period of time.

(3/9 * 25) - (6/9 * 25) = -8.33333333333

Above reflects the expected loss from a Come Out Roll of either four or ten. What we have to do now is multiply that by the probability of such a CO. CO is also an abbreviation for Correctional Officer, but college students usually won't have to worry about those. At least...not for anything serious. We should all have a little fun from time to time.

-8.33333333333 * 6/36 = -1.38888888889

The next thing we have to look at is fives and nines. However, if you complete your Stats & Prob courses, as well as other courses, you might have a nice nine to five. That's one way to try to make a living. Dolly Parton has imparted this knowledge unto us; may we carry it with us and use the wisdom passed on by that large-breasted muse, for all time.

(4/10 * 25) - (6/10 * 25) = -5

-5 * 8/36 = -1.11111111111

Finally:

(5/11 * 25) - (6/11 * 25) = -2.27272727273

-2.27272727273 * 10/36 = -0.63131313131

We have already accounted for winning rolls during which we have established a point number. Our only remaining positive Expected Outcome is a Come Out winner, so we must subtract from its Expected Value the Expected Value of all of our barefoot outcomes.

5.55555555556 -2.77777777778 -1.38888888889 -1.11111111111 - 0.63131313131 = -0.35353535353

That means, for every $25 Pass Line bet, we expect to lose $0.353535353535353 cents. I hate change, so this is fine with me.

-0.35353535353/25 = -0.01414141414

Voila! This converts to a House Edge of 1.41414141414141414141414141414141414141414141%. Or, just 1.41%, if you prefer.

Hey, if you play Craps you might go home barefoot, but always give your dealers shoes on their toke bets.
link to original post

This post is very impressive. How long did it take you to complete?

The relevant question here is: Is it REAL?

How many rolls of the dice will it take to complete the above set without an improper duplication of a hand, i.e., two come out 12's or better yet 6 come out 7's?

What are the ODDS that the equation can be performed?

My guess: Winning the lottery would be easier?

tuttigym
link to original post

Is the post real? Yes. You just quoted it. You could probably even print it, if you wanted to.

How long? I don't know. 10 minutes, maybe? I figured it would be faster just to do it again than look for where I'd done it before, which I'm sure I have.

What are the Odds the equation can be performed? I performed the equation with a probability of 100%, as it has already happened. You just quoted it. Prior to completing it, probably 99.99999999% as the only thing that would have prevented me would have been an extremely untimely, and coincidental, death.
link to original post

I know the POST is real.

10 minutes to produce it is remarkable in that my personal typing skills could duplicate your post in 30+ minutes or more.

You didn't perform anything other than producing a post. The question remains: What are the odds of performing your post at the tables playing the game of craps for real $$$?? A second question: How much of a bankroll would be required to complete the task?

Does anyone suppose the Wizard can answer these questions? Let's see.

tuttigym

Quote: Mission146

Quote: tuttigym

Quote: unJon

Sure. But so what? Dice have no memory.
link to original post

Exactly, so why so much discussion and articulation of the HE? Based on your post; it is meaningless. ("So what?")

tuttigym
link to original post

You should just bet Any 7 every time if you don't believe in House Edge. It's fun and you get to be a contrarian on every roll except the Come Out.
link to original post

Perhaps you should dive in and play the game for a few sessions and produce your own 1.41% HA/HE results. That would "be fun", and you could start your own "adventure" thread too.

tuttigym

Quote: Ace2

Quote: ChumpChange

Well, 100,000 line bets divided by 400 line bets in my above post equals 250 6-hour sessions or $141 x 250 = $35,250 of HA on $25 line bets. Total bets are 100,000 x $25 = $2.5 million. 0.8% of $2.5 million = $20,000 loss; 2% of $2.5 million = $50,000.
Now to add odds on of $125 (5X odds in this case for the poster), up to $12.5 million in odds; 0.8% x $12.5 million = $100,000, 2% x $12.5 million = $250,000.

Speaking of 100k…for the 100,000th time, you don’t add odds bets to your preferred bet level, you reduce your line bet so that the total bet amount (line + odds) is right

If $25 is your comfort level then you should be betting $5 or $10 on the line. Your total average bet will be around $25 and the edge will be 0.37% on your total wagers. No better deal than that except dont pass plus odds. Fairly low standard deviation of 1.3 on average wager

And since we are discussing the long-run, there are no “Hot Shooters” in the long run. Or you could say the “Cold Shooters” will cancel the “Hot Shooters” almost perfectly
link to original post

Now, now Ace, you have to play nice.

tuttigym

Quote: tuttigym

Quote: Mission146

Quote: tuttigym

Quote: unJon

Sure. But so what? Dice have no memory.
link to original post

Exactly, so why so much discussion and articulation of the HE? Based on your post; it is meaningless. ("So what?")

tuttigym
link to original post

You should just bet Any 7 every time if you don't believe in House Edge. It's fun and you get to be a contrarian on every roll except the Come Out.
link to original post

Perhaps you should dive in and play the game for a few sessions and produce your own 1.41% HA/HE results. That would "be fun", and you could start your own "adventure" thread too.

tuttigym
link to original post

I have better adventures to go on than playing Craps for five minutes. I could lay in bed and watch my toenails grow.

Let me be frank with you: You’re the one trying to convince someone of something, not me. I don’t care if you believe the House Edge is real or not.

Of course, you’re not really trying to convince anyone of anything because you’ve presented no position. The only thing that you have done is suggest that the House Edge is not real because it’s not pragmatically reasonable for anyone to physically play long enough to conclusively demonstrate it.

If your argument is that most individual Craps players will not play long enough to have manifested actual results that correlate precisely with a 1.414141414141—% edge, then that is self-evident and I agree with you.

An example of this would be a single decision. You cannot lose 1.41% of a Pass Line bet on a single decision; you can only lose the total amount of the Pass Line bet or profit an amount equal to that.

Most of us don’t dispute things that are clearly evident.

In other words, I don’t know what you’re arguing. Are you arguing that the House Edge is immaterial to anything because you won’t make enough bets to strictly guarantee that your negative results will be precisely in line with the House Edge? I grant that you won’t; I don’t grant that the House Edge is irrelevant. Again, just bet Any Seven all the time if you think, the House Edge being non-existent, that no one bet is worse than any other.

Quote: Mission146

Quote: tuttigym

Quote: Mission146

Quote: tuttigym

Quote: unJon

Sure. But so what? Dice have no memory.
link to original post

Exactly, so why so much discussion and articulation of the HE? Based on your post; it is meaningless. ("So what?")

tuttigym
link to original post

You should just bet Any 7 every time if you don't believe in House Edge. It's fun and you get to be a contrarian on every roll except the Come Out.
link to original post

Perhaps you should dive in and play the game for a few sessions and produce your own 1.41% HA/HE results. That would "be fun", and you could start your own "adventure" thread too.

tuttigym
link to original post

I have better adventures to go on than playing Craps for five minutes. I could lay in bed and watch my toenails grow.

Let me be frank with you: You’re the one trying to convince someone of something, not me. I don’t care if you believe the House Edge is real or not.

Of course, you’re not really trying to convince anyone of anything because you’ve presented no position. The only thing that you have done is suggest that the House Edge is not real because it’s not pragmatically reasonable for anyone to physically play long enough to conclusively demonstrate it.

If your argument is that most individual Craps players will not play long enough to have manifested actual results that correlate precisely with a 1.414141414141—% edge, then that is self-evident and I agree with you.

An example of this would be a single decision. You cannot lose 1.41% of a Pass Line bet on a single decision; you can only lose the total amount of the Pass Line bet or profit an amount equal to that.

Most of us don’t dispute things that are clearly evident.

In other words, I don’t know what you’re arguing. Are you arguing that the House Edge is immaterial to anything because you won’t make enough bets to strictly guarantee that your negative results will be precisely in line with the House Edge? I grant that you won’t; I don’t grant that the House Edge is irrelevant. Again, just bet Any Seven all the time if you think, the House Edge being non-existent, that no one bet is worse than any other.
link to original post

First, thank you for a wonderful and frank post. Second, I have never stated that the house edge is NOT real. The fact that your agreement with me, i.e., performing the "math," is important because it demonstrates how irrelevant that doctrine (for lack of a better term) is to the game of craps. In other words, the pass line bet in and of itself is insignificant in the totality of the game given the various choices a player can make, and it is not essential to winning. In fact, it provides a significant HE once a point is established.

For some unknown reason to me the "establishment" "math" folks have made the HE pass line the center piece of their "one can't win because one cannot beat the house edge" diatribe. It seems that they are saying that the pass line is the only reason players lose at the game of craps. Nothing could be further from the truth. You and I both know that.

I truly appreciate your honesty and recognition of my basic position.

tuttigym

p.s. Do your toenails really grow so fast in five minutes that one can see the change?

Quote: Ace2

Quote: tuttigym

This post is very impressive. How long did it take you to complete?

The relevant question here is: Is it REAL?

How many rolls of the dice will it take to complete the above set without an improper duplication of a hand, i.e., two come out 12's or better yet 6 come out 7's?

What are the ODDS that the equation can be performed?

My guess: Winning the lottery would be easier?

tuttigym
link to original post

I know you’re a math denier, but I’ll give you an example

An average craps player could easily make 100,000 PL bets over the course of his multi-decade “career”. Over that term, he has a 95% chance of losing between 0.8% and 2% of all money wagered. There will be huge variance from the 1.4% for some sessions/years, but long-term the 1.4% expectation is definitely real. His friends making mostly place bets and hardways will lose several times more over that period, which is also real and essentially guaranteed

Then look at a global company like MGM that takes like 1,000 PL bets per minute. Over the course of just one month, their total hold on all PL bets made is pretty much exactly 1.41%. That is also real

It seems that the only phrase you despise more than “math” is “long-term”. When you think about it, most things only matter over the long-term. Like investing and financial success…it generally doesn’t matter what happens in one day. What matters is how the market performs and how you behave over decades.
link to original post

Ace, you are so good at producing fiction that if there was a Pulitzer awarded for such, you could be in the running.

tuttigym

Tuttigym

Consider a fair coin flip. The expectation is for 50% of all flips to be heads. However, you could do 1,000,000 flips and never be at exactly 50% at any time during all those flips. Though there is a 99.99% chance you’d be within 49.8% and 50.2% at the end

Does that mean the 50% isn’t real? Because you can’t demonstrate it with 100% confidence no matter how many flips you make

Quote: Ace2

Tuttigym

Consider a fair coin flip. The expectation is for 50% of all flips to be heads. However, you could do 1,000,000 flips and never be at exactly 50% at any time during all those flips. Though there is a 99.99% chance you’d be within 49.8% and 50.2%

Does that mean the 50% isn’t real? Because you can’t demonstrate it with 100% confidence no matter how many flips you make
link to original post

Again, with the coin flips. Do you consider the math associated with coin flips equivalent to the PL "math"? For me, not even close.

tuttigym

Quote: tuttigym

Quote: Ace2

Tuttigym

Consider a fair coin flip. The expectation is for 50% of all flips to be heads. However, you could do 1,000,000 flips and never be at exactly 50% at any time during all those flips. Though there is a 99.99% chance you’d be within 49.8% and 50.2%

Does that mean the 50% isn’t real? Because you can’t demonstrate it with 100% confidence no matter how many flips you make
link to original post

Again, with the coin flips. Do you consider the math associated with coin flips equivalent to the PL "math"? For me, not even close.

tuttigym
link to original post

Sure. Craps is a series of coin flips with a weighted coin.

Quote: tuttigym

Quote: Ace2

Tuttigym

Consider a fair coin flip. The expectation is for 50% of all flips to be heads. However, you could do 1,000,000 flips and never be at exactly 50% at any time during all those flips. Though there is a 99.99% chance you’d be within 49.8% and 50.2%

Does that mean the 50% isn’t real? Because you can’t demonstrate it with 100% confidence no matter how many flips you make
link to original post

Again, with the coin flips. Do you consider the math associated with coin flips equivalent to the PL "math"? For me, not even close.

tuttigym
link to original post

I was trying to use the simplest possible example, hoping you might grasp coin flips since you clearly don’t grasp craps. Fair coin flips follow a perfect standard normal distribution and are the best way to “dumb down” probability calculations for math-illiterate people. Not referring to anyone specifically

As Unjon said, betting the pass line is exactly like a coin flip. The coin comes up heads 244 out of 495 times. Many coins probably are biased to about this degreee

Quote: tuttigym

Quote: Mission146

Quote: tuttigym

Quote: Mission146

Quote: tuttigym

Quote: unJon

Sure. But so what? Dice have no memory.
link to original post

Exactly, so why so much discussion and articulation of the HE? Based on your post; it is meaningless. ("So what?")

tuttigym
link to original post

You should just bet Any 7 every time if you don't believe in House Edge. It's fun and you get to be a contrarian on every roll except the Come Out.
link to original post

Perhaps you should dive in and play the game for a few sessions and produce your own 1.41% HA/HE results. That would "be fun", and you could start your own "adventure" thread too.

tuttigym
link to original post

I have better adventures to go on than playing Craps for five minutes. I could lay in bed and watch my toenails grow.

Let me be frank with you: You’re the one trying to convince someone of something, not me. I don’t care if you believe the House Edge is real or not.

Of course, you’re not really trying to convince anyone of anything because you’ve presented no position. The only thing that you have done is suggest that the House Edge is not real because it’s not pragmatically reasonable for anyone to physically play long enough to conclusively demonstrate it.

If your argument is that most individual Craps players will not play long enough to have manifested actual results that correlate precisely with a 1.414141414141—% edge, then that is self-evident and I agree with you.

An example of this would be a single decision. You cannot lose 1.41% of a Pass Line bet on a single decision; you can only lose the total amount of the Pass Line bet or profit an amount equal to that.

Most of us don’t dispute things that are clearly evident.

In other words, I don’t know what you’re arguing. Are you arguing that the House Edge is immaterial to anything because you won’t make enough bets to strictly guarantee that your negative results will be precisely in line with the House Edge? I grant that you won’t; I don’t grant that the House Edge is irrelevant. Again, just bet Any Seven all the time if you think, the House Edge being non-existent, that no one bet is worse than any other.
link to original post

First, thank you for a wonderful and frank post. Second, I have never stated that the house edge is NOT real. The fact that your agreement with me, i.e., performing the "math," is important because it demonstrates how irrelevant that doctrine (for lack of a better term) is to the game of craps. In other words, the pass line bet in and of itself is insignificant in the totality of the game given the various choices a player can make, and it is not essential to winning. In fact, it provides a significant HE once a point is established.

For some unknown reason to me the "establishment" "math" folks have made the HE pass line the center piece of their "one can't win because one cannot beat the house edge" diatribe. It seems that they are saying that the pass line is the only reason players lose at the game of craps. Nothing could be further from the truth. You and I both know that.

I truly appreciate your honesty and recognition of my basic position.

tuttigym

p.s. Do your toenails really grow so fast in five minutes that one can see the change?
link to original post

Okay, so you do accept the HE, but only for points that are established?

If you only play the PL, then only the PL matters. If you only play PL + Odds, then that. You can stand and only bet on Yo every single roll, if you wanted, and the only thing that would matter is the roll being eleven or some number other than that.

With that, I don’t know what you mean by totality of the game. The only bets that have any monetary relevance are the ones you are betting. If you’re not playing Craps, then it doesn’t matter at all unless you’re reading or writing about Craps.

Basically, if the PL is, “”Insignificant to the totality of the game,” then so is any other individual bet. Of course, without bets, you’re not actually playing anything. I guess I just don’t understand what you mean by totality of the game. The game is where you have money and what events would cause you to win or lose. Craps has no meaning or significance beyond that.

DP actually has a slightly lower HE per net resolved, but that also makes PL math slightly easier to explain. Plus, most players play PL, so it’s more relevant to them.

Betting on Craps is the main reason people are lifetime losers at Craps. Exceptions, lifetime winners or people playing with an advantage due to some other mechanism.

I say that, absent some other mechanism, people who are ahead lifetime would lose if only they could play long enough. I don’t know precisely how long long enough is.

You dispute my position on that; I do not care. You care only about what has actually happened. If that is all that matters to you, then you really shouldn’t care that I think the House Edge would eventually cause them to become down money lifetime.

And, yes, it would likely take more turns than they can reasonably be expected to perform in a live casino environment to be an ironclad guarantee. Even if not, then it would certainly take more time than they’d ever wish to spend playing Craps.

No, my toenails would not visually grow in that five minutes. At least, not enough for me to tell the difference. Still generally more entertaining than playing Craps. I have a disdain for losing that I didn’t use to have; I lose quite frequently enough even when I have a mathematical advantage.

Quote: Ace2

Quote: tuttigym

Quote: Ace2

Tuttigym

Consider a fair coin flip. The expectation is for 50% of all flips to be heads. However, you could do 1,000,000 flips and never be at exactly 50% at any time during all those flips. Though there is a 99.99% chance you’d be within 49.8% and 50.2%

Does that mean the 50% isn’t real? Because you can’t demonstrate it with 100% confidence no matter how many flips you make
link to original post

Again, with the coin flips. Do you consider the math associated with coin flips equivalent to the PL "math"? For me, not even close.

tuttigym
link to original post

I was trying to use the simplest possible example, hoping you might grasp coin flips since you clearly don’t grasp craps. Fair coin flips follow a perfect standard normal distribution and are the best way to “dumb down” probability calculations for math-illiterate people. Not referring to anyone specifically

As Unjon said, betting the pass line is exactly like a coin flip. The coin comes up heads 244 out of 495 times. Many coins probably are biased to about this degreee
link to original post

Wow, more fiction, folks. The coin flip is a 50/50 proposition so out of 495 times (as above), heads would come up 247.5 times not 244. Or is that the "establishment math" doing the flipping?

If what you and Unjon is correct, then before every come out roll, flip a coin -- heads PL bet; tails DP. That should guarantee successful sessions every time. That is how you must "grasp" your craps participation. Try it and let us all know how it works out for you.

tuttigym

Quote: Mission146

Okay, so you do accept the HE, but only for points that are established?

No. The HE or odds of losing exists on virtually every wager on the cloth in one fashion or another.

Quote: Mission146

If you only play the PL, then only the PL matters. If you only play PL + Odds, then that. You can stand and only bet on Yo every single roll, if you wanted, and the only thing that would matter is the roll being eleven or some number other than that.

I don't play the PL.

Quote: Mission146

With that, I don’t know what you mean by totality of the game. The only bets that have any monetary relevance are the ones you are betting. If you’re not playing Craps, then it doesn’t matter at all unless you’re reading or writing about Craps.

The craps "establishment" play in a windowless room. Their play is embedded in cement. What I mean by that is the standard PL + odds + PB's (3 point Molly) or the opposite equivalent dark side. In order to "win," they need the hot shooter or the very cold table. A choppy table, which is the most frequent occurrences, is the killer. The typical craps player is greedy and undisciplined. Lifetime losers are players that stayed at the table one, two, or three hands too long. Players like Ace or Axel blame the HE for their "long term" failures when quitting sooner while ahead even a little bit could swing losses into wins. I have seen it all too often.

Quote: Mission146

Basically, if the PL is, “”Insignificant to the totality of the game,” then so is any other individual bet. Of course, without bets, you’re not actually playing anything. I guess I just don’t understand what you mean by totality of the game. The game is where you have money and what events would cause you to win or lose. Craps has no meaning or significance beyond that.

Yes, that is correct. "Totality of the game" to me means that craps offers so many options that dwelling on one bet does a disservice to the almost endless options available to a player.

Quote: Mission146

Betting on Craps is the main reason people are lifetime losers at Craps. Exceptions, lifetime winners or people playing with an advantage due to some other mechanism.

I say that, absent some other mechanism, people who are ahead lifetime would lose if only they could play long enough. I don’t know precisely how long long enough is.

You dispute my position on that; I do not care. You care only about what has actually happened. If that is all that matters to you, then you really shouldn’t care that I think the House Edge would eventually cause them to become down money lifetime.

You care. I know because we are having this conversation. I addressed your previous statements above.

Quote: Mission146

And, yes, it would likely take more turns than they can reasonably be expected to perform in a live casino environment to be an ironclad guarantee. Even if not, then it would certainly take more time than they’d ever wish to spend playing Craps.

Thank you.

tuttigym

Quote: tuttigym

This post is very impressive. How long did it take you to complete?

The relevant question here is: Is it REAL?

How many rolls of the dice will it take to complete the above set without an improper duplication of a hand, i.e., two come out 12's or better yet 6 come out 7's?

What are the ODDS that the equation can be performed?

My guess: Winning the lottery would be easier?

tuttigym
link to original post

I don't remember anybody saying that any set of 36 consecutive comeouts must include exactly one 2, two 3s, three 4s, ..., two 11s, and one 12 in order for the probability calculations to be valid.
In fact, if that is a mandatory condition, then you have just stumbled across the 100% Guaranteed Can't Lose Craps System:
1. Track 36 consecutive comeouts.
2. Starting with the next comeout, bet (if you can - for example, I don't think that you can bet that the next roll will be a 6) that the comeout roll will be the same as the one that was 36 comeouts ago.
Proof that this works, under the assumption that every set of 36 comeouts has the same distribution:
Suppose the first comeout in the set of 36 is a 5. Under the assumption, there are 4 comeout rolls of 5 in comeout rolls 1-36; since one of them was in comeout roll 1, there were 3 5s in comeout rolls 2-36. Again according to the assumption, there must be 4 comeout rolls of 5 in comeout rolls 2-37, as this is also a set of 36 consecutive comeout rolls; since there were 3 in rolls 2-36, this requires that comeout roll 37 must also be a 5.
You can make a similar "system" with roulette, if you assume that, on a table with N different numbers, any set of N consecutive spins must include each number exactly once.

Quote: ThatDonGuy

Quote: tuttigym

This post is very impressive. How long did it take you to complete?

The relevant question here is: Is it REAL?

How many rolls of the dice will it take to complete the above set without an improper duplication of a hand, i.e., two come out 12's or better yet 6 come out 7's?

What are the ODDS that the equation can be performed?

My guess: Winning the lottery would be easier?

tuttigym
link to original post

I don't remember anybody saying that any set of 36 consecutive comeouts must include exactly one 2, two 3s, three 4s, ..., two 11s, and one 12 in order for the probability calculations to be valid.
In fact, if that is a mandatory condition, then you have just stumbled across the 100% Guaranteed Can't Lose Craps System:
1. Track 36 consecutive comeouts.
2. Starting with the next comeout, bet (if you can - for example, I don't think that you can bet that the next roll will be a 6) that the comeout roll will be the same as the one that was 36 comeouts ago.
Proof that this works, under the assumption that every set of 36 comeouts has the same distribution:
Suppose the first comeout in the set of 36 is a 5. Under the assumption, there are 4 comeout rolls of 5 in comeout rolls 1-36; since one of them was in comeout roll 1, there were 3 5s in comeout rolls 2-36. Again according to the assumption, there must be 4 comeout rolls of 5 in comeout rolls 2-37, as this is also a set of 36 consecutive comeout rolls; since there were 3 in rolls 2-36, this requires that comeout roll 37 must also be a 5.
You can make a similar "system" with roulette, if you assume that, on a table with N different numbers, any set of N consecutive spins must include each number exactly once.
link to original post

I did not suggest that all come outs be consecutive. However, the 495 hands in the equation have to be, in no particular order, in the same bundle. That is what is represented in the 1.41% HE equation.

tuttigym

Quote: tuttigym

I did not suggest that all come outs be consecutive. However, the 495 hands in the equation have to be, in no particular order, in the same bundle. That is what is represented in the 1.41% HE equation.

tuttigym
link to original post

Back to kindergarten-level probability: coin flips

So you are saying that if, for instance, you flip a fair coin twice and it lands HH or TT, then that means it’s not a 50/50 probability of heads/tails. Every pair of flips must land either TH or HT for a 50/50 coin. Right?

That means you know what the next flip will be with 100% certainty…it’s the opposite of the previous flip.

I case you were wondering, and I know you weren’t, the average number of flips to get at least one head and at least one tail is 3. Does that make it a 1/3 chance of getting a head or tail ?

Quote: tuttigym

I did not suggest that all come outs be consecutive. However, the 495 hands in the equation have to be, in no particular order, in the same bundle. That is what is represented in the 1.41% HE equation.

tuttigym
link to original post

Er, no, as otherwise you have the same "system" - the 496th "hand in the equation" has to be the same as the first, the 497th has to be the same as the second, and so on.

Things like EV and HE are based on the probability of a single event. They do not depend on every possible result happening in any finite set of events. Otherwise, how do you possibly explain how to determine the probability of winning, say, the next Powerball draw?

Quote: tuttigym

No. The HE or odds of losing exists on virtually every wager on the cloth in one fashion or another.

We agree. I maintain that the PL has a lower House Edge than most other bets, those exceptions being DP and ODDS (on either side). I wouldn't use the word, 'Odds,' interchangeably with HE as I take ODDS as synonymous with probability, which is a component of HE.

Quote:

I don't play the PL.

General 'you,' not specific 'you.' Whatever bet, or combination of bets, you are playing is the game. Nothing matters aside from anything pertinent to your bet, or bets.

Quote:

The craps "establishment" play in a windowless room. Their play is embedded in cement. What I mean by that is the standard PL + odds + PB's (3 point Molly) or the opposite equivalent dark side. In order to "win," they need the hot shooter or the very cold table. A choppy table, which is the most frequent occurrences, is the killer. The typical craps player is greedy and undisciplined. Lifetime losers are players that stayed at the table one, two, or three hands too long. Players like Ace or Axel blame the HE for their "long term" failures when quitting sooner while ahead even a little bit could swing losses into wins. I have seen it all too often.

Both you and they can play however you wish. I don't play strictly what you call the, "Establishment Way," in the extremely rare event I play Craps. I make a Crap Check bet on the CO, even though I know the House Edge is terrible. Telling people the HE is terrible and telling them not to play a certain bet are two totally different things. They can play the Pass Line, penny slots, the Big Six wheel or nothing at all---why should I care?

I have never heard of Axel to even play Craps, so I have no idea what you're talking about there.

Quote:

Yes, that is correct. "Totality of the game" to me means that craps offers so many options that dwelling on one bet does a disservice to the almost endless options available to a player.

In this case, I was answering the question the thread was asking. Specifically, Bossdog wanted to know how the House Edge of the PL might be calculated. Therefore, I calculated it. He didn't ask me how the totality of the game could be calculated, or I would have done that.

Therefore, you basically came in and attacked...or 'challenged us'...if I want to be really generous, because we had the audacity to answer a question we were directly asked by someone. That's the way I see it anyway.

Quote:

You care. I know because we are having this conversation. I addressed your previous statements above.

I only care to the extent that I couldn't let your first post stand without some sort of rebuttal. Originally, I cared only to answer the question that the OP directly asked of me, which is what I did.

After that, you came in and your posts took a steaming dump on this thread, which is what your posts tend to do when there is math going on and you find yourself offended by it for reasons beyond my comprehension. You challenged the very notion of PL math, and its relevance, despite the fact that the OP had a question specific to PL math.

In other words, you posted in a conversation that should be of no concern to you, because we all know that, at best, you disregard HE math as unimportant.

That's what I mean when I say I don't care. I care to answer a question that someone asked and I know how to answer.

If you created a thread called, "The House Edge in Craps Doesn't Really Matter," I would open that thread, see that you are the OP, close it and never open it again. That's what I mean when I say I don't care. What you think about anything is immaterial to me.

You can't open a thread in which a person has asked us to do something, I do that, say what I have done is immaterial and then tell me that such is not a challenge to me. That's what I mean when I say I don't care. You challenged me; I did not challenge you; I don't care what you think about any subject that does or ever could possibly exist and have no desire to attempt to change your mind about anything for any reason.

Quote:

Thank you.

tuttigym
link to original post

And a good day to you. Though I imagine we will engage sooner than later if I continue to make the mistake of attempting to answer questions for people.