Craps - General Questions
I'm a floor supervisor at a local casino and was wondering about a strange play. A player was betting both the pass and don't pass at the same time. My question is if he was betting the do side for $10, I would give him an average bet of $10. Now that he is betting both sides, and probably not risking any funds, what would the average bet be? I know if this patron bet both the red and black on roulette, he should get an average bet of both bets on red and black, since the house advantage of 5.26% make both bets a loser over the long run.
Show me a player making opposite, or near opposite, bets and I'll show you a player up to something. He is probably trying to take advantage of a promotion or comps. If I ran a casino, I would give credit only for money being risked. One could argue he is risking $10, because a 12 will cause the pass to win and the don't pass to push. However, that will happen 1 in 36 pass line bets only. If I ran a casino, I would give him an average bet of $0.
I was playing craps at www.gamehouse.com and bet $20 on the horn and won $60 on a roll of 11. If the horn bet is spread out between 2,3,11,12, shouldn't I have won $75 ($5X15)?
No, you were paid correctly. The 11 does pay 15:1 on the $5 of your bet. However you lost the other $15 on the 2, 3, and 12. So $75-$15=$60. Instead of taking the $15 from your bet, they take it from the winnings.
What are your casino choices in Atlantic City for craps and Spanish 21? When you mention that it is wise to take full odds on a bet while playing craps, do you mean to match your bet with an equal odds bet, or to make the highest allowed odds bet along with your bet (ex: at a table with 10x max odds place a $1 bet with a $1 free odds bet, or a $1 bet with a $10 free odds bet). I'm a little confused on that. I love your site, and honestly see it as "a diamond in the rough" among gambling advice web sites. Personally, I like to know what the mathematical odds are when it comes to wagering my hard earned money! Thanks in advance for answering my questions!
Thanks for the compliment. The Spanish 21 rules are the same across Atlantic City. I only know of two that have the game, the Tropicana and the Claridge, but there could be others by now. If I'm not mistaken, the best craps game is at the Sands, which offers 5X odds. When I say to take the maximum odds I mean bet the maximum allowed on the odds. For example, $50 after a $10 line bet. Keep in mind that you won't win more money by taking the odds, you just get to bet more without losing more in the long run.
What is the casino average hold for craps?
I don't know the hold for any game. For the benefit of other readers, the hold percentage is the ratio of casino profit to chips purchased at the table. Since the same chips will circulate back and forth between the players and dealer for an unknown period of time, the mathematician has no way of calculating the hold or hold percentage.
When reading literature about the formula casinos use to determine comps, the only formula I ever see used as an example is a formula for blackjack. Assuming the casino determines your average bet by your spread, what formula is typically used by casinos to determine expected craps losses, which in turn, determines available comps.
I asked my friend Larry Drummond, a craps dealer and former webmaster of Next Shooter for help on this question. Larry can be a bit abrasive but is a good source of hard to find information on craps. Here is what he said, "Comps for craps vary from Casino to Casino and from Boxman to Boxman. A player should get to know the Boxman. The Boxman sets the players average bet and tracks the TIME that a player is at the table. It is easier for the Boxman to track action for COMPS. if the player is consistent in their wagering pattern. Now, I ask you ... if a player goes $52 or $54 across after a Point is established with a $5 flat bet on the Pass Line. Is that a $57 or $59 average? ... Or a $5 average with a whole bunch of other INDIVIDUAL bets? The answer is... IT DEPENDS ON HOW WELL YOU KNOW THE BOXMAN and HOW MANY TIMES TO YOU ATTEND THIS PARTICULAR CASINO."
Larry added in another e-mail the following, "In addition to the information I already sent to you ... ODDS on Pass Line and Come bets are often NOT included in the AVERAGE for comps. Same with LAYING ODDS on the Don’t Side... as in the long run this should be a WASH. But ... If a smart boxman wants someone who is spending big bucks on ANY 7, the worst bet on the table... he would probably average the ODDS and the LAYS to keep the sucker coming back to the casino ... you can re-word this to make it a little more palatable for your site ... In addition ... A good boxman will COMP to the MAX if he sees that the PLAYER is "betting for the boys."
Will you be doing a risk of ruin analysis for Craps? The only analysis I have found online to date seemed to be flawed.
I wasn’t planning on it. There are so many betting patterns in craps that one analysis would only fit a small percentage of craps players.
Mr. Wizard, you’re site is great. I think you may be the only honest expert on the internet. My question is this. I know how odds are calculated in craps but I cannot shake the feeling that once a player gets up multiple bets, either through placing them or betting the pass line and making successive come bets, that the odds shift dramatically into the houses favor. It only takes one seven to wipe out all the bets at one time. In order to win, you must hit each number and after it is hit, a seven would wipe out the remaining numbers.
Thank you for the kind words. I still say that the house edge is not dependent on the number of come bets you make. Yes, it is depressing to establish one come point after another and then lose everything on a seven. However there are other times when the shooter takes ages before rolling a seven and you win lots of come bets along the way.
What is your opinion of Card Craps as played by many of the casinos in the San Diego area?
In California dice alone can not be used to determine the outcome of a game. To get around this law many casinos use a hybrid of cards and dice, or cards only. My crap section now addresses some of the ways this is done.
I realize that decisions per hour in games like blackjack and craps can depend heavily on factors like the number of other players at the table, the hand shuffle vs. machine shuffle, shooter and dealer speed. Still, I was curious if you could give me a rough approximation of how many decisions per hour an individual can expect at a mostly-full craps table and a blackjack table with both a hand shuffle and machine shuffle. This would help me estimate my expected loss per hour and weigh it against the comps I am being offered.
The following tables show the number of hands/tosses per hour in blackjack, craps, and roulette. The source of the tables is Casino Operations Management by Jim Kilby.
Hands per Hour in Blackjack
| Players | Hands per hour |
| 1 | 209 |
| 2 | 139 |
| 3 | 105 |
| 4 | 84 |
| 5 | 70 |
| 6 | 60 |
| 7 | 52 |
Rolls per Hour in Craps
| Players | Rolls per hour |
| 1 | 249 |
| 3 | 216 |
| 5 | 144 |
| 7 | 135 |
| 9 | 123 |
| 11 | 102 |
In craps 29.6% of total rolls are come out rolls, on average.
Spins per Hour in Roulette
| Players | Spins per hour |
| 1 | 112 |
| 2 | 76 |
| 3 | 60 |
| 4 | 55 |
| 5 | 48 |
| 6 | 35 |
Do you have any good rules/setups for playing Craps at home for actual cash. I understand that to keep things legal, I can’t take a ’house cut’, but assuming a buy-in is enforced (like playing poker at home) is there a good system to play privately for money without becoming "the house" and paying winners out of my own pocket?
You could do a tournament. Every player will buy in for the same amount of non-cashable chips. Establish somebody to be the banker, paying off bets as in normal craps. Whoever has the most chips after some benchmark, for example x 7-outs, wins the pool. Since you will have an even chance with everybody else, I think it would be okay to ask for tips for the use of your house.
Just wanted to know if you knew where the majority of casinos here in Vegas buy their craps tables. And if these companies sold their tables to the public?
Two suppliers of gaming tables I am aware of are TCS John Huxley and Midwest Game Supply. They probably do sell to the public. The price of a craps table at Midwest Game Supply is $3,950. If a used one will suffice, the Gambler's General Store sells used tables.
What casinos in Las Vegas have a small table, called a tub, for craps?
According to the the Bone Man at NextShooter.com, here is where and when you can find the tubs:
One Tub at Wild, Wild West (probably open only evenings, weekdays, and on weekends).
One Tub at Ellis Island (probably open only evenings, weekdays, and on weekends).
One Tub at Circus Circus in West Casino section, hardly ever open unless on busy holiday.
2010 update: I hear the Ellis Island replaced the tub with a full craps table.
What percentage of rolls in craps are come out rolls?
The expected number of total rolls is 1671/196 = 8.5255. Interestingly, the expected number of rolls for a point is exactly 6. That leaves 2.5255 come out rolls. So the percentage of come out rolls is 2.5255/8.5255 = 29.6%.
I am wondering which will come up more rolling a pair of dice — an odd or even total?
The answer is 50/50. This will be true for any number of dice rolled, not just two.
A bit off-topic, but I've always thought an odd/even set of bets would be a good way to replace the dreaded big 6/8 bets in craps. To give the house an advantage, here are my proposed pay tables and analysis.
Odd Bet
| Event | Pays | Combinations | Probability | Return |
|---|---|---|---|---|
| 3 or 11 | 1.5 | 4 | 0.111111 | 0.166667 |
| 5 or 9 | 1 | 8 | 0.222222 | 0.222222 |
| 7 | 0.5 | 6 | 0.166667 | 0.083333 |
| Even | -1 | 18 | 0.500000 | -0.500000 |
| Total | 36 | 1.000000 | -0.027778 |
Even Bet
| Event | Pays | Combinations | Probability | Return |
|---|---|---|---|---|
| 2 or 12 | 3 | 2 | 0.055556 | 0.166667 |
| 4 or 10 | 1 | 6 | 0.166667 | 0.166667 |
| 6 or 8 | 0.5 | 10 | 0.277778 | 0.138889 |
| Odd | -1 | 18 | 0.500000 | -0.500000 |
| Total | 36 | 1.000000 | -0.027778 |
Please note that I claim all rights with this publication.
This question is raised and discussed in my forum at Wizard of Vegas.
What is the etiquette of tipping the shooter in craps?
There is absolutely no expectation of tipping the shooter ever! I would go as far as to ask you not to, lest it become a "thing," and leeches start hanging around the table, only betting on their turn, and shaking other players down for tips. This whole culture of tipping in casinos is getting completely out of hand.
This question is raised and discussed in my forum at Wizard of Vegas.
Assume I'm playing craps at a table with 100x odds. I'm debating whether to make a place bet on the 6 or 8 or a put bet. How much odds would I have to put on a put bet to have a better value than the place bet.
Good question. The house edge on a place bet on the 6 or 8 is 1.52%. At 5x odds, the overall house edge is exactly the same on a put bet on the 6 or 8 at 1.52%. At 6x odds, it drops to 1.30%. So, it takes 6x odds to become a better value.
What is the Iron Cross strategy in craps and what do you think of it?
The Iron Cross is a way of betting the field and place bets to win on any roll of the dice except a 7. The field already covers the 2, 3, 4, 9, 10, 11, and 12. The player will add to that place bets on the 5, 6, and 8 to cover the rest of the numbers, besides 7. The following table shows what the math looks like with a $5 field bet, $5 place bet on the 5, and $6 place bets on the 6 and 8.
Iron Cross
| Dice Total | Win | Combinations | Probability | Return |
|---|---|---|---|---|
| 2 | 10 | 1 | 0.027778 | 0.277778 |
| 3 | 5 | 2 | 0.055556 | 0.277778 |
| 4 | 5 | 3 | 0.083333 | 0.416667 |
| 5 | 2 | 4 | 0.111111 | 0.222222 |
| 6 | 2 | 5 | 0.138889 | 0.277778 |
| 7 | -22 | 6 | 0.166667 | -3.666667 |
| 8 | 2 | 5 | 0.138889 | 0.277778 |
| 9 | 5 | 4 | 0.111111 | 0.555556 |
| 10 | 5 | 3 | 0.083333 | 0.416667 |
| 11 | 5 | 2 | 0.055556 | 0.277778 |
| 12 | 15 | 1 | 0.027778 | 0.416667 |
| 36 | 1.000000 | -0.250000 |
The lower right cell of the table shows an expected loss of $0.25. The total amount bet is $22. This makes the overall house edge $0.25/$22 = 1/88 = 1.14%.
At this point you may be wondering how this house edge can be lower than the house edge of each individual bet. The answer is because the house edge of 1.52% placing the 6 and 8 and 4.00% placing the 5 is based on per bet resolved. If define the house edge on place bets on a per roll basis, then the house edge placing the 6 or 8 is 0.46% and on the 5 is 1.11%.
We can get at the 1.14% house edge by taking a weighted average of all bets made, as follows:
($5*2.78% + $5*1.11% + $12*0.46%)/22 = $0.25/$22 = 1.14%.
Be wary of casinos that pay only 2 to 1 for 12 on the field bet. Insist on getting the full 3 to 1. The short pay doubles the house edge on that bet from 2.78% to 5.56%.
As to my opinion, compared to most games, 1.14% is a pretty good bet. However, you could do much better in craps. For example, with 3-4-5x odds, making the pass and come bets, with full odds, you can get the house edge down to 0.37%. Doing the opposite, betting the don't pass and don't come, plus laying full odds, results in a house edge of 0.27%.What is the standard deviation in craps, assuming a pass line bet and taking 3-4-5x odds? How about a don't pass bet and laying 3-4-5x odds?
The standard deviation, relative to the pass bet, with full 3-4-5x odds is 4.915632.
The standard deviation, relative to the don't pass bet, laying full 3-4-5x odds is 4.912807.
What is your analysis of the following way in craps to parlay a $5 bet into $1,200? Start with a $5 bet on the 4. If that wins, you parlay the winnings on the 5. If that wins, parlay those winnings on the 6. You keep going, betting on the 8, 9, and then 10. You may assume the player adds $1 after a wins on the 4 and 8, to keep the bets in round numbers.
The probability of winning on the 4 is 3/(3+6) = 3/9 = 1/3. A place bet on the 4 pays 9 to 5, so if that bet wins you'll have a total of $9 + $5 = $14.
Next, the player adds $1 to his bet for a total of $15 on the 5. The probability of winning on the 5 is 4/(4+6) = 4/10 = 2/5. A place bet on the 5 pays 7 to 5, so if that bet wins you'll have a total of $21 + $15 = $36. The probability of getting at least this far is (1/3)*(2/5) = 13.33%.
Next, the player bets $36 on the 6. The probability of winning on the 6 is 5/(5+6) = 5/11. A place bet on the 6 pays 7 to 6, so if that bet wins you'll have a total of $42 + $36 = $78. The probability of getting at least this far is (1/3)*(2/5)*(5/11) = 2/33 = 6.06%
Next, the player bets $78 on the 8. The probability of winning on the 8 is 5/(5+6) = 5/11. A place bet on the 8 pays 7 to 6, so if that bet wins you'll have a total of $91 + $78 = $169. The probability of getting at least this far is (1/3)*(2/5)*(5/11)^2 = 10/363 = 2.75%
Next, the player adds $1 from his pocket to the $169 and bets $170 on the 9. The probability of winning on the 9 is 4/(4+6) = 2/5. A place bet on the 9 pays 7 to 5, so if that bet wins you'll have a total of $238 + $170 = $408. The probability of getting at least this far is (1/3)*(2/5)^2*(5/11)^2 = 4/363 = 1.10%
Finally, we're ready to bet on the 10. With the lower house edge on the buy bet, let's assume the player bets that. You didn't specify whether the player must prepay the commission or only pay it on wins. Let's look at prepaying the commission first. Under that rule, the bet amount should be evenly divisible by $21. Let's assume the player bets $380 on the 10, prepays a 5% commission of $19 and pockets the other $9 from his $408.
The probability of winning on the 4 is 3/(3+6) = 3/9 = 1/3. A winning $380 bet will pay $760 in winnings, for a total of $760+$380 = $1,140. The probability of getting at least this far is (1/3)^2*(2/5)^2*(5/11)^2 = 4/1089 = 0.37% = 1 in 272.25.
Remembmer the player bet $5+$1+$1 on the way, but pocketed $9 after the win on a 9, for a net win of $1,142. If we define the house edge as the expected loss to the original $5 bet, then it would be $1.06/$5.00 = 21.16%.
Next, let's see what happens if the commission is paid on wins only on the 10. There, buy bets on the 10 should be evenly divisible by $20. Let's assume the player pockets $8 and bets the other $400.
A winning $400 bet will pay $780 in winnings, for a total of $780+$400 = $1,180.
Remembmer the player bet $5+$1+$1 on the way, but pocketed $8 after the win on a 9, for a net win of $1,181. If we define the house edge as the expected loss to the original $5 bet, then it would be $0.92/$5.00 = 18.44%.
So, we can't quite get to $1,200 unless the player pulls more money out of his pocket after a win on 9, or somewhere else along the way. I can't endorse this strategy in terms of value, but it does seem like it would have a high fun and excitement factor.
On Roll to Win craps games the player may make both lay and place-to-lose bets. Here are the odds offered on the place-to-lose bets:
- 4 and 10: 5 to 11
- 5 and 9: 5 to 8
- 6 and 8: 4 to 5
The lay bets pay fair odds, except the player must pay a 5% commission, based on the win amount, if he wins.
My question is which type of bet offers the better odds?
The following table shows the house edge both ways by the number bet on. You can see the house edge is lower on the lay bets for all points except the 6 and 8.
House Edge on Place to Lose and Lay Bets
| Number | Place to Lose | Lay |
|---|---|---|
| 4 or 10 | 3.03% | 1.67% |
| 5 or 9 | 2.50% | 2.00% |
| 6 or 8 | 1.82% | 2.27% |
I hear the MGM properties in Vegas now charge the vig on buy bets after a win only. How does this change the house edge?
I believe the already charged the vig after a win only on the 4 and 10. So, there is no effect there. Here is the house edge on the 4 and 10 all three ways you might bet them:
- Place bet (pays 9 to 5) — 6.67%
- Buy bet (commission always paid) — 4.76%
- Buy bet (commission paid on wins only) — 1.76%
On the 5 and 9, this is good news, lowing the house edge from 4.00% on a place bet to 2.00%.
- Place bet (pays 7 to 5) — 4.00%
- Buy bet (commission always paid) — 4.76%
- Buy bet (commission paid on wins only) — 2.00%
On the 6 and 8, it doesn't matter, as the house edge on place bets is still lower.
- Place bet (pays 7 to 6) — 1.52%
- Buy bet (commission always paid) — 4.76%
- Buy bet (commission paid on wins only) — 2.27%
This question is asked and discussed in my forum at Wizard of Vegas.
If I make a buy bet for $20 on the 4 and 10 and place $30 on the 5, 6, 8, and 9, what is my house edge? Please assume the commission on the 4 and 10 is paid on wins only. Please calculate it whether I:
- Leave the bets up for one roll only
- Leave the bets up until some significant event happens (any roll between 4 and 10)
- Leave the bets up until they are all resolved.
The first table shows my analysis for leaving the bets up for one roll only. The return column is calculated as the win*probability/(total bet). The lower right cell shows a house edge of 0.69%.
One Roll Analysis
| Roll | Bet | Net Win | Combinations | Probability | Return |
|---|---|---|---|---|---|
| 2 | 0 | 0 | 1 | 0.027778 | 0.000000 |
| 3 | 0 | 0 | 2 | 0.055556 | 0.000000 |
| 4 | 20 | 39 | 3 | 0.083333 | 0.020313 |
| 5 | 30 | 42 | 4 | 0.111111 | 0.029167 |
| 6 | 30 | 35 | 5 | 0.138889 | 0.030382 |
| 7 | 0 | -160 | 6 | 0.166667 | -0.166667 |
| 8 | 30 | 35 | 5 | 0.138889 | 0.030382 |
| 9 | 30 | 42 | 4 | 0.111111 | 0.029167 |
| 10 | 20 | 39 | 3 | 0.083333 | 0.020313 |
| 11 | 0 | 0 | 2 | 0.055556 | 0.000000 |
| 12 | 0 | 0 | 1 | 0.027778 | 0.000000 |
| 160 | 36 | 1.000000 | -0.006944 |
The second table shows my analysis for leaving the bets up until a bet is resolved. In other words, rolling again after a total of 2, 3, 11, or 12. The return column is calculated as the win*probability/(total bet). The lower right cell shows a house edge of 0.83%.
One Significant Roll Analysis
| Roll | Bet | Net Win | Combinations | Probability | Return |
|---|---|---|---|---|---|
| 4 | 20 | 39 | 3 | 0.100000 | 0.024375 |
| 5 | 30 | 42 | 4 | 0.133333 | 0.035000 |
| 6 | 30 | 35 | 5 | 0.166667 | 0.036458 |
| 7 | 0 | -160 | 6 | 0.200000 | -0.200000 |
| 8 | 30 | 35 | 5 | 0.166667 | 0.036458 |
| 9 | 30 | 42 | 4 | 0.133333 | 0.035000 |
| 10 | 20 | 39 | 3 | 0.100000 | 0.024375 |
| Total | 160 | 30 | 1.000000 | -0.008333 |
The third table shows my analysis for leaving the bets up until all are resolved. The return column is calculated as the win*probability/(total bet). The lower right cell shows a house edge of 2.44%.
Roll Until All Bets Resolved Analysis
| Win | 4,10 Rolled |
5,9 Rolled |
6,8 Rolled |
Combinations | Probability | Return |
|---|---|---|---|---|---|---|
| -160 | 1 | 0 | 0 | 2,677,114,440 | 0.200000 | -0.200000 |
| -101 | 0 | 1 | 0 | 594,914,320 | 0.044444 | -0.028056 |
| -88 | 0 | 0 | 1 | 823,727,520 | 0.061538 | -0.033846 |
| -95 | 2 | 0 | 0 | 1,070,845,776 | 0.080000 | -0.047500 |
| -42 | 0 | 2 | 0 | 74,364,290 | 0.005556 | -0.001458 |
| -16 | 0 | 0 | 2 | 149,768,640 | 0.011189 | -0.001119 |
| -30 | 1 | 1 | 0 | 267,711,444 | 0.020000 | -0.003750 |
| -29 | 1 | 0 | 1 | 421,812,160 | 0.031512 | -0.005712 |
| -36 | 0 | 1 | 1 | 562,464,448 | 0.042020 | -0.009455 |
| -23 | 1 | 1 | 1 | 800,192,448 | 0.059780 | -0.008593 |
| 36 | 2 | 1 | 0 | 751,055,104 | 0.056109 | 0.012625 |
| 30 | 2 | 0 | 1 | 93,017,540 | 0.006949 | 0.001303 |
| 23 | 1 | 2 | 0 | 127,949,276 | 0.009559 | 0.001374 |
| 43 | 0 | 2 | 1 | 136,097,920 | 0.010168 | 0.002733 |
| 49 | 1 | 0 | 2 | 276,379,776 | 0.020648 | 0.006323 |
| 29 | 0 | 1 | 2 | 259,917,112 | 0.019418 | 0.003519 |
| 42 | 2 | 1 | 1 | 383,915,862 | 0.028681 | 0.007529 |
| 95 | 1 | 2 | 1 | 280,463,688 | 0.020953 | 0.012441 |
| 108 | 1 | 1 | 2 | 430,248,448 | 0.032143 | 0.021696 |
| 101 | 2 | 2 | 0 | 626,008,276 | 0.046767 | 0.029522 |
| 102 | 2 | 0 | 2 | 48,772,745 | 0.003644 | 0.002323 |
| 88 | 0 | 2 | 2 | 101,392,694 | 0.007575 | 0.004166 |
| 114 | 2 | 2 | 1 | 243,130,194 | 0.018164 | 0.012942 |
| 167 | 2 | 1 | 2 | 263,665,646 | 0.019698 | 0.020560 |
| 160 | 1 | 2 | 2 | 409,147,802 | 0.030566 | 0.030566 |
| 173 | 2 | 2 | 2 | 679,339,612 | 0.050752 | 0.054875 |
| 232 | 0 | 0 | 0 | 832,156,379 | 0.062168 | 0.090144 |
| Total | 13,385,573,560 | 1.000000 | -0.024848 |
Recall from your statistics call the probability of an event x NOT happening is exp(-x). It is then easy to say the probability it has happened at least once is 1-exp(-x). The following list shows the probability for any length of time x the given points have been rolled. Then, integration over all periods of time x from 0 to infinity. I prefer the integral calculator at www.integral-calculator.com/. Finally, remember to weight these probabilities by similar events. For example, the probability of rolling a 4 is the same as rolling a 10.
- 4 or 10 -- (1-exp(-3x/36))*exp(-3x/36)*exp(-4x/36)^2*exp(-5x/36)^2*exp(-x/6)/6
- 5 or 9 -- (1-exp(-x/9))*exp(-5x/36)^2*exp(-3x/36)^2*exp(-x/9)exp(-x/6)/6
- 6 or 8 -- (1-exp(-5x/36))*exp(-4x/36)^2*exp(-3x/36)^2*exp(-5x/36)exp(-x/6)/6
- 4 and 10 -- (1-exp(-3x/36))^2*exp(-4x/36)^2*exp(-5x/36)^2*exp(-x/6)/6
- 5 and 9 -- (1-exp(-4x/36))^2*exp(-5x/36)^2*exp(-3x/36)^2*exp(-x/6)/6
- 6 and 8 -- (1-exp(-5x/36))^2*exp(-4x/36)^2*exp(-3x/36)^2*exp(-x/6)/6
- 4 and 5 -- (1-exp(-3x/36))*(1-exp(-4x/36))*exp(-5x/36)^2*exp(-4x/36)*exp(-3x/36)*exp(-x/6)/6
- 4 and 6 -- (1-exp(-3x/36))*(1-exp(-5x/36))*exp(-4x/36)^2*exp(-5x/36)*exp(-3x/36)*exp(-x/6)/6
- 5 and 6 -- (1-exp(-4x/36))*(1-exp(-5x/36))*exp(-3x/36)^2*exp(-5x/36)*exp(-4x/36)*exp(-x/6)/6
- 4,5,6 -- (1-exp(-3x/36))^1*exp(-3x/36)^1*exp(-4x/36)^1*(1-exp(-4x/36))^1*(1-exp(-5x/36))^1*exp(-5x/36)^1*exp(-x/6)/6
- 4,5,10 -- (1-exp(-3x/36))^2*(1-exp(-4x/36))*exp(-5x/36)^2*exp(-4x/36)*exp(-x/6)/6
- 4,6,10 -- (1-exp(-3x/36))^2*(1-exp(-5x/36))*exp(-4x/36)^2*exp(-5x/36)*exp(-x/6)/6
- 4,5,9 -- (1-exp(-4x/36))^2*(1-exp(-3x/36))*exp(-5x/36)^2*exp(-3x/36)*exp(-x/6)/6
- 5,6,9 -- (1-exp(-4x/36))^2*(1-exp(-5x/36))*exp(-3x/36)^2*exp(-5x/36)*exp(-x/6)/6
- 4,6,8 -- (1-exp(-3x/36))^1*exp(-3x/36)*exp(-4x/36)^2*(1-exp(-5x/36))^2*exp(-x/6)/6
- 5,6,8 -- (1-exp(-3x/36))^0*exp(-3x/36)^2*exp(-4x/36)^1*(1-exp(-4x/36))*(1-exp(-5x/36))^2*exp(-5x/36)^0*exp(-x/6)/6
- 4,5,6,10 -- (1-exp(-3x/36))^2*exp(-4x/36)^1*(1-exp(-4x/36))^1*(1-exp(-5x/36))^1*exp(-5x/36)^1*exp(-x/6)/6
- 4,5,6,9 -- (1-exp(-3x/36))^1*exp(-3x/36)^1*exp(-4x/36)^0*(1-exp(-4x/36))^2*(1-exp(-5x/36))^1*exp(-5x/36)^1*exp(-x/6)/6
- 4,5,6,8 -- (1-exp(-3x/36))^1*exp(-3x/36)^1*exp(-4x/36)^1*(1-exp(-4x/36))^1*(1-exp(-5x/36))^2*exp(-5x/36)^0*exp(-x/6)/6
- 4,5,9,10 -- (1-exp(-3x/36))^2*exp(-3x/36)^0*(1-exp(-4x/36))^2*(1-exp(-5x/36))^0*exp(-5x/36)^2*exp(-x/6)/6
- 4,6,8,10 -- (1-exp(-3x/36))^2*exp(-3x/36)^0*(exp(-4x/36))^2*(1-exp(-5x/36))^2*exp(-5x/36)^0*exp(-x/6)/6
- 5,6,8,9 -- (1-exp(-3x/36))^0*exp(-3x/36)^2*(1-exp(-4x/36))^2*(1-exp(-5x/36))^2*exp(-5x/36)^0*exp(-x/6)*exp(-x/6)/6
- 4,5,6,9,10 -- (1-exp(-3x/36))^2*exp(-3x/36)^0*(1-exp(-4x/36))^2*(1-exp(-5x/36))^1*exp(-5x/36)^1*exp(-x/6)/6
- 4,5,6,8,10 -- (1-exp(-3x/36))^2*(1-exp(-4x/36))^1*exp(-4x/36)*(1-exp(-5x/36))^2*exp(-x/6)/6
- 4,5,6,8,9 -- (1-exp(-3x/36))^1*exp(-3x/36)^1*(1-exp(-4x/36))^2*(1-exp(-5x/36))^2*exp(-x/6)/6
- 4,5,6,8,9,10 -- (1-exp(-3x/36))^2*(1-exp(-4x/36))^2*(1-exp(-5x/36))^2*exp(-x/6)/6
What is your opinion of the Doey Don't strategy in craps?
For those not familiar with it, the Doey Don't works as follows:
- On a come-out roll, make both a pass and don't pass bet, for the same amount.
- If a point is rolled, bet the odds on it.
The thinking is the don't pass bet is a hedge against a 7-out on the come-out roll. Players who play the doey don't would probably say the pass and don't offset each other, letting the player enjoy the odds bet with no house advantage.
The flaw is if a 12 is rolled on the come-out roll. The pass will lose and the don't pass will push, resulting in a loss of one unit. The probability of a 12 is 1/36, making the expected loss with this strategy 1/36 = 2.78% of one unit. Meanwhile, the expected loss making just a pass bet and taking the odds 7/495 = 1.41% of one unit.
However, the Doey Don't has less volatility. Assuming 3-4-5x odds, here is the standard deviation both ways:
- Pass + full odds: 4.915632
- Doey Don't: 4.085789
The bottom line is I don't recommend the Doey Don't, because the expected loss is 1.36% of a unit more.
It's ironic you test online craps games for fairness, when your own craps game is so seriously flawed. I lost the pass bet by 72 7-out's before I made a point!
To test your theory, I played through 50 pass line bets where the result was either winning by making a point or a 7-out. This took 248 rolls.
In the following table, I document my results.
Craps Data
| Point | Wins | Losses | Total | Prob. Win | Exp. Wins |
|---|---|---|---|---|---|
| 4 | 3 | 5 | 8 | 33.33% | 2.666667 |
| 5 | 3 | 3 | 6 | 40.00% | 2.400000 |
| 6 | 5 | 5 | 10 | 45.45% | 4.545455 |
| 8 | 6 | 4 | 10 | 45.45% | 4.545455 |
| 9 | 4 | 3 | 7 | 40.00% | 2.800000 |
| 10 | 1 | 8 | 9 | 33.33% | 3.000000 |
| Total | 22 | 28 | 50 | 19.957576 |
The right cell shows an expected number of wins of 19.96, given the points rolled. My actual wins was 22. The probability of exactly 22 wins is 7.66%. The probability of less than 22 is 64.73%. The probability of 23 or more is 27.61%. So, this test shows my results were close to expectations.
For all the details, please see this video response to your challenge.
What's your opinion of the Marching Soldier strategy in craps?
For the benefit of other readers, let me explain what the Marching Soldier is. Briefly, it's a method of parlaying $5 into $1,200. You do have to add more money to it sometimes, so that wins are evenly divisible by $1. Here are the specifics of how it works on a table where you must pre-pay the commission on the 4 and 10:
- Make $5 place bet on the 4.
- If step 1 wins, then take $14 win* plus $1 from your stack and make $15 place bet on the 5.
- If step 2 wins, then make $36 place bet on the 6.
- If step 3 wins, then make $78 place bet on the 8.
- If step 4 wins, then bet $169 win plus $1 from your stack on the 9.
- If step 5 wins, then you will have $408. Take $12 from your stack to pay the $20 commission on a $400 buy bet on the 10.
- If step 6 wins, then take down the $1200.
- If any bet loses, then you're done.
*: By "win" I include the original bet amount.
You may also do this in the reverse order, starting with 10 and ending with 4.
If you make it all the way to the end, you'll have a net win of $1,181, after deducting the $5 in initial and $14 in subsequent wagers from your stack.
The probability of success is 0.3673%, or 1 in 272.
If you play at a table where the commission on the 4 and 10 is paid after a win only, then instead of adding $12 from your pocket after a win on the 9, I would put $8 into it. Then you'll have an even $400 on the 10, for which you'll get back $1,180 if you win. This more generous rule results in an expected net win of $1,188.
If we define the house edge as the expected loss to the expected wager, it is 19.76% if the commission must be paid in advance and 17.03% if it is paid after a win only.
This question is asked and discussed in my forum at Wizard of Vegas.