Craps losses per hour?
Thx, i figured about $12 -$15 an hour? sound about right?
But 100 come bets/hr would simply be: 1.41%*$10*100 ~ $14.10/hr
Quote: bk1248Thx, i figured about $12 -$15 an hour? sound about right?
If right, bear in mind a single session can be as much as 40 times expectations ... losses or otherwise
although playing single odds keeps your variance down to make this less likely
Quote: bk1248.........or I crap out,
You never CRAP OUT!!!!!
You SEVEN OUT.
Quote: FatGeezusYou never CRAP OUT!!!!!
You SEVEN OUT.
Thank you, that's always a pet peeve of mine. I even hear dealers say it too.
EDIT: Ah, looks like MrV got us on a technicality (below). He's right, but I know what you had meant...haha ;)
Quote: FatGeezusYou never CRAP OUT!!!!!
You SEVEN OUT.
Not so.
You "crap out" when you roll a 2, 3 or 12 on the come out roll.
You "seven out" if you set a point but roll a seven before you repeat the point.
Quote:
bk1248
If I bet $10 on pass and $10 on every come bet with max odds until I have bases loaded or I crap out, How much do I generally lose per hour at 10 xs odds and say 100 rolls per hour.
Thx, i figured about $12 -$15 an hour? sound about right?
Why don't you try it and report back how much you lost. You don't win at craps by going by the math of the game. While theoretically it sounds good, just ask all those pass-line betters taking max so-called free odds why they just lost when they are leaving the craps table.
All the free odds does is gets you to bet more chips, it doesn't change the odds of you winning or losing. Bet more lose more, if your on the wrong side of the table! There is nothing in a casino that is free, you are always paying for what you get!
Once the point is established your chances of winning goes down hill, the casinos know that,..that is why they have the so-called free odds, to suck in any sucker they have at the craps tables. Do you really think a casino would give you a bet that they have a 0 house advantage on? You pay for the free odds with your pass-line bet. Then you pay again when you lose those so-called free odds in most likely four rolls of the dice.
If there are 2 to 1 odds on the shooter rolling a 4, how does taking those so-called free odds change that fact, after the point has been established and you just took you so-called free odds? You just locked yourself into a bet that can only end when the shooter 7's out or by some strange luck makes the point of 4.
Quote: superrickAll the free odds does is gets you to bet more chips, it doesn't change the odds of you winning or losing. Bet more lose more, if your on the wrong side of the table! There is nothing in a casino that is free, you are always paying for what you get!.
Actually, betting odds does increase the chance of winning. Over the course of an hour, say 35 line bets, a flat pass bettor needs to win 18 or more of 35 to win. The pass+odds bettor needs to win fewer than 18 of 35 to come out ahead due to the payout ratios on the odds bets. With odds bets, you still need to get lucky to win, but not by as much as if you're just flat betting the line.
Quote: MathExtremistActually, betting odds does increase the chance of winning.
Yes, and you are betting more; i.e. more Action.
You can also keep your Action the same but devote more to the odds bet in a session. Now it becomes indisputably more likely to be a winner. What it boils down to, for it to be the same Action, is to have shorter sessions with less bets.
Quote: MathExtremistActually, betting odds does increase the chance of winning. Over the course of an hour, say 35 line bets, a flat pass bettor needs to win 18 or more of 35 to win. The pass+odds bettor needs to win fewer than 18 of 35 to come out ahead due to the payout ratios on the odds bets. With odds bets, you still need to get lucky to win, but not by as much as if you're just flat betting the line.
I think what he meant was it doesn't increase your chance of winning that particular pass line bet; not that it doesn't increase your chance of winning more money over the course of the session.
Quote:
wudged
I think what he meant was it doesn't increase your chance of winning that particular pass line bet; not that it doesn't increase your chance of winning more money over the course of the session.
You are 100% correct, that is where the misconception is,.. when someone reads about taking the so-called free odds. It doesn't increase your chances of winning anything. What it does do is cut your bankroll down in size with every bet that you make, only because you are now betting more, every time you make that pass-line bet with the so-called free odds.
Most players that buy into the so-called free odds bets never realize that they could make more money if the point was a 4 or 10 just by placing the point! A pass-line bet on the 4 when you are playing with red chips pays you $5 for you line bet and $20 for $10 in odds, so you make $25 on your bets! On the other hand if you place the 4 for $15 you would have been paid $27 for your place bet, that you can take down or turn off at any time you want to!
The so-called free odds bets make no sense what so ever until you use 5 x odds of better. For a low roller taking the so-called free odds can put a end to your play in short order!
Agree.Quote: MathExtremistActually, betting odds does increase the chance of winning.
Over the course of an hour, say 35 line bets, a flat pass bettor needs to win 18 or more of 35 to win.
The pass+odds bettor needs to win fewer than 18 of 35 to come out ahead due to the payout ratios on the odds bets.
With odds bets, you still need to get lucky to win, but not by as much as if you're just flat betting the line.
Some #s. results below were calculated
(I used Excel - and Gambler's Odds 1.3 does the same without any programming except for the game/bet additions)
3 players:
$5 with $10 odds always
(close to same action over N wagers - $5*1/3 + $15*2/3 - to the $12 line bets, right?)
$12 pass line bettor
$1 pass line bettor
sufficient bankroll to make every wager
# line bets (session or lifetime - matters not)
Prob (%) of a NET LOSS - $5 with $10 odds always
Prob (%) of a NET LOSS - $12 flat no odds
Prob (%) of a NET LOSS - $1 flat no odds
35
51.254197%
53.357738%
53.357738%
350
53.722272%
58.368256%
58.368256%
3,500
61.5218%
79.383959%
79.383959%
35,000
82.263452%
99.58597%
99.58597%
Quote: MathExtremistActually, betting odds does increase the chance of winning. Over the course of an hour, say 35 line bets, a flat pass bettor needs to win 18 or more of 35 to win. The pass+odds bettor needs to win fewer than 18 of 35 to come out ahead due to the payout ratios on the odds bets. With odds bets, you still need to get lucky to win, but not by as much as if you're just flat betting the line.
Doesn't taking odds just increase the variance? Not nitpicking here but don't the chances of the wager winning/losing remain the same ex. pass/fail?
Regarding the op. In a scenario with bases loaded and max. odds, assuming odds w.o.c. Just for a hypothetical example I'll throw out there that they win some lose some always off and on, on the come bets. When the shooter gets ready to toss I'll say half the bases are loaded with max. odds. In the natural order of things the co is gonna be 7, 1 in 6 times. Over 18 the 7 is gonna hit +/- 3 times, with bases loaded max. odds [or 3 bases] isn't the op just going to be ruined?
Quote: petroglyphDoesn't taking odds just increase the variance? Not nitpicking here but don't the chances of the wager winning/losing remain the same ex. pass/fail?
For an individual wager, yes. But he was discussing the probability of a winning/losing overall based on the total number of pass line wins does change with placing odds because you are getting > 1 to 1 when you win after the comeout roll. The average amount lost for the total wagers stays the same (the idea of "free odds"), however, as you will have more being wagered after a point number is rolled and you are less than 1 to 1 to hit your point when it's set.
Quote: HexDiceAnd to me, the lower the edge the less you lose = more money.
Since you wrote "And to me" maybe you were stating this as an opinion, but to be technical:
You lose the same amount of money with or without odds unless your odds bets are being rated for comp purposes.
Betting $10 on the pass line with no odds or betting $10 on the pass line with 100X odds both give the house about 14.1 cents on average.
I said "to me" because it's a negative expectation game and you'll never make money.
Quote: HexDiceThe house edge diminishes as you increase the odds. Because the free odds pay at true odds and the pass line is even money (where the house edge lies) as you increase the amount of odds the overall edge drops.
I said "to me" because it's a negative expectation game and you'll never make money.
Percentage wise yes, 100X makes it a near zero-edge game. In terms of absolute hourly loss, it still remains the same (well ignoring that you slightly slow the game down by an odds wager...haha). So personally, I would always bet minimum on the pass line/come bets, and bet whatever amount of odds you feel comfortable with.
Quote: bk1248i didnt say i win i say i lose about 15 an hour to some of the posters.
We've derailed your thread I think after covering your question.
It seems to happen with craps questions a lot...sorry.
Quote: HexDiceMaybe in terms of how you're rated. I don't understand how you say "percentage wise yes" and then say that doesn't effect your hourly loss.
Because when most people consider the house edge percentage wise, they lump the odds bet into the wager.
So if you take 100X odds on a $10 pass line bet after the point is made the combined house edge becomes:
($10*0.0141 + $1000*0.00)/($10+$1000) = 0.0001396 = 0.01396%
If you take no odds after the point is made the house edge is simply:
$10*0.0141/$10 = 0.0141 = 1.41%
This leads to the common misconception that the odds bet must necessarily be better for you. But if your goal is to minimize your expected loss, odds bets do nothing to minimize your expected loss:
Expected loss if you take 100X odds on a $10 pass line bet after the point is made:
$10*0.0141 + $1000*0.00 = $0.141
Expected loss if you take no odds after the point is made:
$10*0.0141 = $0.141
Huh?Quote: tringlomaneBecause when most people consider the house edge percentage wise, they lump the odds bet into the wager.
So if you take 100X odds on a $10 pass line bet after the point is made the combined house edge becomes:
($10*0.0141 + $1000*0.00)/($10+$1000) = 0.0001396 = 0.01396%
care to do this the Wizard's way?
I get
-280/1,339,800
-0.000208986
-0.020899%
aint noQuote: tringlomaneThis leads to the common misconception that the odds bet must necessarily be better for you.
common misconception.
The amount you are short changed on any Odds WIN at Craps is $0.00 exactly.
Every other Craps bet that wins gets a short pay always.
That makes the best bet according to how much one is shorted (shafted) on a win the odds bet at 0%.
sure they do... PER $ wageredQuote: tringlomaneBut if your goal is to minimize your expected loss, odds bets do nothing to minimize your expected loss:
Make your example a fair one and see what the expected loss would be over the SAME amount of resolved action.
In other words, Per $ actual bet resolved they do.
See WinCraps help section on Odds for a great example of the concept.
too much crap in craps
1x odds $10 pass: ((6*$22)-(7*$20))/36 rolls = -$.22/roll
100x odds $10 pass: ((6*$2220)-(7*$1010))/36 rolls = $173.61/roll
That's -1.1% vs 17% per roll. Where did I mess up?
EV = HE * $AvgBetQuote: HexDiceTell me what I'm doing wrong, but I ran the numbers for the EV of placing odds after the 6 was established over 36 rolls and I'm obviously missing something stupid. But it is demonstrating my point.
1x odds $10 pass: ((6*$22)-(7*$20))/36 rolls = -$.22/roll
100x odds $10 pass: ((6*$2220)-(7*$1010))/36 rolls = $173.61/roll
That's -1.1% vs 17% per roll. Where did I mess up?
Per bet resolved
same EV
I get
1x odds $10 pass:
$AvgBet 16 2/3 (16.67) right??
*
he -280/33000
(-0.008484848) =
ev -$0.141414141
per roll just multiply by 165/557
100x odds $10 pass:
$AvgBet 676 2/3 (676.67) right???
*
he -280/1,339,800
(-0.000208986) =
ev -$0.141414141
per roll just multiply by 165/557
right????
https://wizardofodds.com/games/craps/appendix/1/
Quote: 7crapsHuh?
care to do this the Wizard's way?
I get
-280/1,339,800
-0.000208986
-0.020899%
aint no
common misconception.
The amount you are short changed on any Odds WIN at Craps is $0.00 exactly.
Every other Craps bet that wins gets a short pay always.
That makes the best bet according to how much one is shorted (shafted) on a win the odds bet at 0%.
sure they do... PER $ wagered
Make your example a fair one and see what the expected loss would be over the SAME amount of resolved action.
In other words, Per $ actual bet resolved they do.
See WinCraps help section on Odds for a great example of the concept.
too much crap in craps
I didn't do it the Wizard's way because it would make my point more confusing than it apparently already is. I got yelled at on 2+2 for the same argument.
All I am trying to say is this:
Your total expected loss betting the pass line is always this:
(Total amount of pass line wagers) x 1.41%
So I make 100,000 pass line bets at $10 each and take no odds, I would expect to lose: 100,000 x $10 x 1.41% = $14,100
If you make 100,000 pass line bets at $10 each and take 100X odds, I would expect you to lose: 100,000 x $10 x 1.41% = $14,100
Now if you want to say playing a lot of odds is more fun, or odds lowers the house edge on a percentage basis, or you lost less per dollar wagered, that's great. But in the end, on average, the loss incurred by both of us will be the same. That's all I am trying to argue, nothing more, nothing less. Every time I try to argue this though, people seem to have no idea wtf I am talking about. Am I that bad at English? :-\
Quote: HexDiceWell my calculations were after the come out roll, so the house edge is different. Just seems a little too simple to just have a house edge that's a stagnant 1.41% no matter the point or the odds if you're betting the pass. I'd rather not do the research, and I have looked at the wizardofodds page, and it does say the edge per roll goes down with increased odds.
Because handle increases, not because expected loss decreases. The expected loss of a $10 line bet is 14 cents regardless of the odds.
because that smaller edge is over a larger average bet.Quote: HexDiceWell my calculations were after the come out roll, so the house edge is different.
Just seems a little too simple to just have a house edge that's a stagnant 1.41%
no matter the point
or the odds
if you're betting the pass.
I'd rather not do the research, and I have looked at the wizardofodds page,
and it does say the edge per roll goes down with increased odds.
The EV of both players in your example you gave is the same.
either per roll or per bet resolved
This comparison is about apples to oranges. Not a fair one in my book.Quote: tringlomaneSo I make 100,000 pass line bets at $10 each and take no odds, I would expect to lose: 100,000 x $10 x 1.41% = $14,100
If you make 100,000 pass line bets at $10 each and take 100X odds, I would expect you to lose: 100,000 x $10 x 1.41% = $14,100
Here is apples to apples.
no wonder my universe is so upside-in
"So I make 100,000 pass line bets at $10 each and take no odds"
Total resolved bets from this PlayerA =
$1,000,000 and expected LOSS is just $14,100
over an average bet of $10
PlayerB (let us call him Mike S)
"100,000 pass line bets at $10 each and take 100X odds"
Total resolved bets from this PlayerB (on average)=
$67,666,667 - WOW! 67 times more total wagered than PlayerA
(But we knew that. This is for those that read this thread in the years to come)
and expected LOSS is just $14,100 - what? ONLY a $14,100 EV
over an average bet of $676.67
and I am not even impressed.
Both players have different risks. Gambling is about risk taking.
How about a same risk to make it fair.
what would PlayerA's EV be if his average bet was $677 over 100,000 such bets??
(or $10 over 6.7 million bets)
now the same total wagers resolved
Not even close to $14,100
Left to the reader to do the math
again: per $ bet (resolved) is the fair comparison
I ain't at all impressed by showing an exampleQuote: tringlomaneNow if you want to say playing a lot of odds is more fun,
or odds lowers the house edge on a percentage basis,
or you lost less per dollar wagered, that's great.
But in the end, on average, the loss incurred by both of us will be the same.
where one player wagers 67 times more over 100k bets than another.
And as long as bankroll is not an issue,
who has the best chance of showing a NET profit after that many wagers (lifetime)
We all just won!
I agree too.Quote: bk1248If I bet $10 on pass and $10 on every come bet with max odds
until I have bases loaded or I crap out,
How much do I generally lose per hour at 10 xs odds and say 100 rolls per hour.
Thx, i figured about $12 -$15 an hour? sound about right?
No one has yet to mention that 100 rolls is too short for 10X Odds to really hit all of it's possible outcomes.
The distribution skew: 0.4 points to a mean (average)
that is greater than the mode (most likely)
or median (the 50/50 point).
median: -$110
High: $7210
Low: -$4610
The standard deviation is about $1430
so one should not be surprised at one 100 roll session ending up at -$4300 or even $4300 or higher.
Variance rules here with 10X odds
You may be very disappointed as to how often you will actually LOSE say $15 in a 100 roll session.
That value is an average over many many 100 roll sessions.
More than likely one will end up far away from that -$15.
(sim results)
Ending interval of -$30 to $0 (-$15 +/- $15)
prob: 0.01117
or about 1 in 90 sessions
Many think that is too low and should be at least 80%
the truth can hurt sometimes
To lose between
-$110 and -$220
prob: 0.03358
or about 1 in 30 sessions
3 times more likely to end in this interval than around the -$15 value
To lose more than $1500 in 100 rolls
prob: 0.14736
or about 1 in 7 sessions
13 times more likely to lose more than $1500 than lose around the -$15 value
Damn
But there is always good news when most of your money wagered is on the Odds bet in Craps.
A 0% house edge means. when you win, you do not get shorted on any payoff for the money on the odds bet.
It is paid off at true odds
To WIN $1500 or more in 100 rolls
prob: 0.14562
or about 1 in 7 sessions (been there)
again
13 times more likely to WIN $1500 or more
than lose around the -$15 value
Hit this 3 times in a row your next trip
Good Luck
added: results for No odds, Full 2X odds ($25 odds on 6&8)
and 345X odds
$10 Pass and Come bets every roll for 100 rolls
(odds not working on the cor)
######################################
0X odds - $10 pass and come bets every roll
Ending Bankroll
median: -$20
High: 480
Low: -320
StdDev: 97
skew: 0.32
Draw-down (10000 start)
Mean: 9893
Median: 9900
Mode: 9940
Low: 9620
StdDev: 56
one should not be surprised at one 100 roll session ending up at -$300 or even $300 or higher.
Variance rules here with 0X odds
(sim results)
Ending interval of -$30 to $0 (-$15 +/- $15)
prob: 0.16615
or about 1 in 6 sessions
Many think that is too low and should be at least 99%
To lose between
-$110 and -$220
prob: 0.14895
or about 1 in 7 sessions
close to ending in this interval as is the -$15 value
To lose more than $100 in 100 rolls
prob: 0.2181
or about 1 in 5 sessions
more likely to lose more than $100 than lose around the -$15 value
Damn truth again
To WIN $100 or more in 100 rolls
prob: 0.1396
or about 1 in 7 sessions
again
more likely to WIN $100 or more
than lose around the -$15 value
#######################################
Full2X odds (not working on cor)
Ending Bankroll
median: -$35
High: 1930
Low:-1255
StdDev:377
skew: 0.42
Draw-down (10000 start)
Mean: 9650
Median: 9680
Mode: 9830
Low: 8685
StdDev: 195
one should not be surprised at one 100 roll session ending up at -$1130 or even $1130 or higher.
Variance rules here with 2X odds
(sim results)
Ending interval of -$30 to $0 (-$15 +/- $15)
prob: 0.0381
or about 1 in 27 sessions
Many think that is too low and should be at least 95%
the truth can hurt sometimes
To lose between
-$110 and -$220
prob: 0.1234
or about 1 in 8 sessions
at least 3 times more likely to end in this interval than around the -$15 value
To lose more than $380 in 100 rolls
prob: 0.1611
or about 1 in 7 sessions
about 5 times more likely to lose more than $380 than lose around the -$15 value
Damn truth
To WIN $380 or more in 100 rolls
prob: 0.15065
or about 1 in 7 sessions
again
about 5 times more likely to WIN $380 or more
than lose around the -$15 value
#################################################
345X odds (not working on cor)
Ending Bankroll
median: -$60
High: 3070
Low:-2040
StdDev:636
skew: 0.42
Draw-down (10000 start)
Mean: 9427
Median: 9480
Mode: 9780
Low: 7910
StdDev: 328
one should not be surprised at one 100 roll session ending up at -$1908 or even $1908 or higher.
Variance rules here with 345X odds
You may be very disappointed as to how often you will actually LOSE say $15 in a 100 roll session.
That value is an average over many many 100 roll sessions.
More than likely one will end up far away from that -$15.
(sim results)
Ending interval of -$30 to $0 (-$15 +/- $15)
prob: 0.0256
or about 1 in 39 sessions
Many think that is too low and should be at least 90%
the truth can hurt sometimes
To lose between
-$110 and -$220
prob: 0.07255
or about 1 in 14 sessions
almost 3 times more likely to end in this interval than around the -$15 value
To lose more than $650 in 100 rolls
prob: 0.15745
or about 1 in 7 sessions
about 6 times more likely to lose more than $650 than lose around the -$15 value
Damn truth
To WIN $650 or more in 100 rolls
prob: 0.1510
or about 1 in 7 sessions
again
about 6 times more likely to WIN $650 or more
than lose around the -$15 value
Quote: 7crapsI agree too.
No one has yet to mention that 100 rolls is too short for 10X Odds to really hit all of it's possible outcomes.
The distribution skew: 0.4 points to a mean (average)
that is greater than the mode (most likely)
or median (the 50/50 point).
median: -$110
High: $7210
Low: -$4610
The standard deviation is about $1430
so one should not be surprised at one 100 roll session ending up at -$4300 or even $4300 or higher.
Variance rules here with 10X odds
You may be very disappointed as to how often you will actually LOSE say $15 in a 100 roll session.
That value is an average over many many 100 roll sessions.
More than likely one will end up far away from that -$15.
(sim results)
Ending interval of -$30 to $0 (-$15 +/- $15)
prob: 0.01117
or about 1 in 90 sessions
Many think that is too low and should be at least 80%
the truth can hurt sometimes
To lose between
-$110 and -$220
prob: 0.03358
or about 1 in 30 sessions
3 times more likely to end in this interval than around the -$15 value
To lose more than $1500 in 100 rolls
prob: 0.14736
or about 1 in 7 sessions
13 times more likely to lose more than $1500 than lose around the -$15 value
Damn
But there is always good news when most of your money wagered is on the Odds bet in Craps.
A 0% house edge means. when you win, you do not get shorted on any payoff for the money on the odds bet.
It is paid off at true odds
To WIN $1500 or more in 100 rolls
prob: 0.14562
or about 1 in 7 sessions (been there)
again
13 times more likely to WIN $1500 or more
than lose around the -$15 value
Hit this 3 times in a row your next trip
Good Luck
@7craps
Is this regardless of whether or not there's one come bet or constant come betting?
I seem to lose a lot more than $15 per hour. If the law of large numbers catches up with me I'm going to be rich.
That is all? Way too low.Quote: odiousgambitIf right, bear in mind a single session can be as much as 40 times expectations ... losses or otherwise
how about 100 to 300 (3sd)
times expectation with a standard deviation of about $1430
see my last post in this thread using the OP pass and come bets every roll
Quote: odiousgambitalthough playing single odds keeps your variance down to make this less likely
SD of about $220 so a range between +/- 15 to 50 would be expected.
a bit less than 40 on average
Remember, the SD value can also be used to figure on a starting
Bankroll for a required Risk of Ruin.
I use 3 times the SD as to minimize hitting Ruin during any part of the session.
For this 100 roll session a quick sim shows the largest draw-down was -$753
and the chance of running out of a $660 Bankroll about 0.08%
(not likely in one 100 roll session but over many, yep likely for someone)
The OP was pass and come bets every roll.Quote: petroglyph@7craps
Is this regardless of whether or not there's one come bet or constant come betting?
I seem to lose a lot more than $15 per hour. If the law of large numbers catches up with me I'm going to be rich.
The variance is way lower for one come bet max as should be expected.
I am sure I have another post about that . Maybe I can find that link
Hopefully you can just get on a good luck winning streak and win 5 or more sessions in a row.
Quote: MathExtremistBecause handle increases, not because expected loss decreases. The expected loss of a $10 line bet is 14 cents regardless of the odds.
Yep, but craps players seem to be much more concerned about handle increases and not their constant expected loss. Am I able to spend this extra handle on hookers and blow?? lol
Quote: 7crapsThe OP was pass and come bets every roll.
The variance is way lower for one come bet max as should be expected.
I am sure I have another post about that . Maybe I can find that link
Hopefully you can just get on a good luck winning streak and win 5 or more sessions in a row.
@7craps
Thanks for the reply.
You are right you wrote about this on another thread and I shouldn't have been so lazy.
It just seems every time I see somebody getting all the bases covered with the come/odds the devil shows and the whole wad is gone. Even if they won the pass bet, they get killed on all those others.
Of course there's always the chance of a monster roll which is what it would seem to take to make all the comes pay off. When I'm playing, I'm always conscious of the average roll is around 8.5 toss's. So if a shooter throws a crap or two and makes a point that's 3. Cover the bases and that's another 5. The shooter is already out there at the end of an average roll, and all those wagers are locked in?
Is that how you bet? If so, why not just all the way across after the come out?
Quote:7Craps
No one has yet to mention that 100 rolls is too short for 10X Odds to really hit all of it's possible outcomes.
The distribution skew: 0.4 points to a mean (average)
that is greater than the mode (most likely)
or median (the 50/50 point).
median: -$110
High: $7210
Low: -$4610
The standard deviation is about $1430
so one should not be surprised at one 100 roll session ending up at -$4300 or even $4300 or higher.
Variance rules here with 10X odds
This is were most math guys led the suckers down the Yellow Brick Road, they never mention that in a 100 roll session the player could end up losing -$4300 or even $4300 or higher.
The guys that know nothing about the math of the game think that they have the best bet on the table by betting the pass-line and taking full odds. They don’t know that when the math guys are writing about this kind of betting they are writing about hundreds if not thousands of rolls for the math of the game to come out right.
They will never have the bankroll to sustain that type of betting, most players are only at the tables for under 100 rolls of the dice and by then they are out of money!
Yes. Of course it does. Drastically. And the vig on these bets is already pretty lo at 1.41% for Pass/Come. Common sense. No calc's reqd.
Variance no doubt flies all over the place esp w small time gamblers like me taking as much odds as I can afford in my gambling budget. But I don't see how variance would affect a player's win/loss nearly as much as vig, which is tiny on Don't Pass/Don't Come w the highest odds allowed, tho I can't prove it quantatively. That being said I lose at a rate far in excess of that indicated (per unit time) in probability/statistical analyses I've read, and I'm hardly alone. It's the experience of a lot of guys I've talked to. It's very frustrating. I don't get it.
Quote: 7crapsThat is all? Way too low.
how about 100 to 300 (3sd)
times expectation with a standard deviation of about $1430
see my last post in this thread using the OP pass and come bets every roll
SD of about $220 so a range between +/- 15 to 50 would be expected.
a bit less than 40 on average
Remember, the SD value can also be used to figure on a starting
Bankroll for a required Risk of Ruin.
I use 3 times the SD as to minimize hitting Ruin during any part of the session.
For this 100 roll session a quick sim shows the largest draw-down was -$753
and the chance of running out of a $660 Bankroll about 0.08%
(not likely in one 100 roll session but over many, yep likely for someone)
What simulation prog are you using? I'd like to learn a little more abt it esp statistical dist variables, variance, SD and the like.
Thx.
Quote: superrickQuote:7Craps
No one has yet to mention that 100 rolls is too short for 10X Odds to really hit all of it's possible outcomes.
The distribution skew: 0.4 points to a mean (average)
that is greater than the mode (most likely)
or median (the 50/50 point).
median: -$110
High: $7210
Low: -$4610
The standard deviation is about $1430
so one should not be surprised at one 100 roll session ending up at -$4300 or even $4300 or higher.
Variance rules here with 10X odds
This is were most math guys led the suckers down the Yellow Brick Road, they never mention that in a 100 roll session the player could end up losing -$4300 or even $4300 or higher.
The guys that know nothing about the math of the game think that they have the best bet on the table by betting the pass-line and taking full odds. They don’t know that when the math guys are writing about this kind of betting they are writing about hundreds if not thousands of rolls for the math of the game to come out right.
They will never have the bankroll to sustain that type of betting, most players are only at the tables for under 100 rolls of the dice and by then they are out of money!
Yeah, but a guy's bankroll may be considered as infinite in practical terms as one's whole lifetime craps betting may be considered as one very long game, non gambling intervals being mathematically entirely irrelevant. Guy loses, comes back again 2 mo later. Unlimited bankroll, in practical terms. I don't think these 'Banker's Ruin' theories make much sense in a craps context.
Quote:
APEppink
Yeah, but a guy's bankroll may be considered as infinite in practical terms as one's whole lifetime craps betting may be considered as one very long game, non gambling intervals being mathematically entirely irrelevant. Guy loses, comes back again 2 mo later. Unlimited bankroll, in practical terms. I don't think these 'Banker's Ruin' theories make much sense in a craps context.
So with that way of thinking , you must think that the players that come to these boards are here to learn how to lose,..is that so?
Most players do not have an unlimited bankroll, at some point they go broke, or just figure out that they can’t beat the casinos!
You may not view it this way, but everybody plays craps in the short run, never in the long run, unlike what all the math guys want everybody to believe. You will not have the bankroll that can sustain the amount of time you would need for the math of the game to even out,.. to what the math of the game says it should be.
Quote:
7Craps
No one has yet to mention that 100 rolls is too short for 10X Odds to really hit all of it's possible outcomes.
The distribution skew: 0.4 points to a mean (average)
that is greater than the mode (most likely)
or median (the 50/50 point).
median: -$110
High: $7210
Low: -$4610
The standard deviation is about $1430
so one should not be surprised at one 100 roll session ending up at -$4300 or even $4300 or higher.
Variance rules here with 10X odds
The above example should show you just how negative the game of craps can be for someone using a pass-line bet with ten times odds!
Do you have $4300 to lose every time you go to the craps tables? Just how long could you stand at a craps table and keep betting your 10 X odds? Do you have the money to keep coming back and chasing your losses, well that is what you would have to do if you wanted to win, and I hope that we all know that chasing losses is just plain stupid!
Quote: 7CrapsNo one has yet to mention that 100 rolls is too short for 10X Odds
Saying "too short" is probably right for getting a handle from raw data, but that is putting it mildly; 100 million rolls is more like it. Otherwise a short session is what makes the most sense with negative expectation.
for these here, WinCraps Pro and ClassicQuote: APEppinkWhat simulation prog are you using?
I'd like to learn a little more abt it esp statistical dist variables, variance, SD and the like.
Thx.
The Help section is very helpful
(WinCraps is free to own)
I also sometimes use for Craps
GAMBLSIM by Steve Fry
both his version and a modified version I did
I also use my own code in Excel and the R program (R is free too)
Good Luck
Quote: 7crapsfor these here, WinCraps Pro and Classic
The Help section is very helpful
(WinCraps is free to own)
I also sometimes use for Craps
GAMBLSIM by Steve Fry
both his version and a modified version I did
I also use my own code in Excel and the R program (R is free too)
Good Luck
Thx.
