- Calculating The Probability Of Winning Craps
- Probability Of Winning Craps
- Probability Of Winning A Field Bet In Craps
You can use probability to figure out the odds of winning and losing in the popular casino dice game of craps. In the game of craps, on your first roll (called the come out roll), three outcomes are possible: Natural: Rolling a total of 7 or 11 — automatically wins. Craps: Rolling a total of 2, 3, or 12 — automatically loses. Interestingly, in a game of Crapless Craps, the ability to lose a Pass Line bet to craps on a come out roll is removed – hence the name 'crapless' craps. This improves your odds of winning by removing the house edge and increasing your expected value. Finally, (d) show that the probability that a player wins a game of craps is exactly 244/495. I'm sure I can figure out (b) and (c) after I figure out how to do part (a). I know that the chance of rolling a 4 would be 1/12, but I'm pretty sure there has to be more to the probability than that.
These 5 simple tips on how to win at craps will improve your odds of winning while playing craps. There are 11 numbers possible with a pair of dice; some easier to roll than others. Those unfamiliar with the game of craps are usually put off by the large betting fields and seemingly endless rules and betting combinations. You could expect on average 18.94 points made on a 6 or 8, 13.33 on a 5 or 9, and 8.33 on a 4 or 10. Are the craps probability numbers with the odds taken 100% reliable. Also is the gaming industry your full time profession, and do you visit Atlantic City often? Also, how do you simulate billions and billions of hands, spins, and rolls.
The 3 pass line wins can be any combination of:
Come out win, Come out win, Come out win
Come out win, Come out win, Point win
Come out win, Point win, Come out win
Come out win, Point win, Point win
Point win, Come out win, Come out win
Point win, Come out win, Point win
Point win, Point win, Come out win
Point win, Point win, Point win
i believe its about:
1-((.507)^3+3((.493)(.507^2))+3((.493^2)(.507)))
(244/495)^3 = 0.119771609 exactly 3
you want 3 or more
formula for the sum of a geometric series is a/1-r
where a is the first term
r = the ratio
a = (244/495)^3
r = 244/495
so (244/495)^3 / (1-(244/495) = 0.236202973
But now you want this to end. Multiply the above result by 1-(244/495)
0.119771609
((244/495)^3 / (1-(244/495)) * (1-(244/495))
added my table
| at least in a row then lose | Prob | 1 in |
|---|---|---|
| 2 | 0.242979288 | 4.1 |
| 3 | 0.119771609 | 8.3 |
| 4 | 0.059038934 | 16.9 |
| 5 | 0.02910202 | 34.4 |
| 6 | 0.014345238 | 69.7 |
| 7 | 0.007071188 | 141.4 |
| 8 | 0.003485596 | 286.9 |
| 9 | 0.001718152 | 582.0 |
| 10 | 0.000846928 | 1,180.7 |
| 11 | 0.000417475 | 2,395.4 |
| 12 | 0.000205786 | 4,859.4 |
| 13 | 0.000101438 | 9,858.3 |
| 14 | 5.00017E-05 | 19,999.3 |
| 15 | 2.46473E-05 | 40,572.4 |
| 16 | 1.21494E-05 | 82,308.7 |
| 17 | 5.98878E-06 | 166,978.8 |
| 18 | 2.95205E-06 | 338,748.0 |
| 19 | 1.45515E-06 | 687,214.1 |
| 20 | 7.17286E-07 | 1,394,143.4 |
Calculating The Probability Of Winning Craps
pass line win probability = 244/495
(244/495)^3 = 0.119771609 exactly 3
Mustangsally: Maybe I'm missing something, but I get different results from what you show. Suppose (for simplicity) that it was a coin flip with p=0.5 instead of 244/495. To get at least one win in a row would be P=0.5. To get at least two in a row is P=0.5^2, etc. To get the answer for exactly n in a row, you need to multiply the 'at least' by the probability of losing on the n+1 try.
For the pass line problem, I think the 0.119771609 figure is for 3 or more, not for exactly 3.
Isn't that correct, or what did I miss?
For the pass line problem, I think the 0.119771609 figure is for 3 or more, not for exactly 3.
Isn't that correct, or what did I miss?
Sally did it differently by starting with 3 pass line wins in a row in 3 trials.
The OP asked a unique question.
Most ask the probability of winning 3 pass line bets in a row. And for 3 trials it is simply p^3
Sally's math shows 3 in a row in 3 trials and the OPs Q arrives at the same value. It should.
IF the OP had asked what is the probability of winning 3 pass line bets in a row then losing, we would have p^3 * q or 0.06073
Let us see if OP is happy and replies.
added
average number of trials to see a run of 3 or more pass line wins: 16.466
4 or more: 33.404
5 or more: 67.765
6 or more: 137.475
Multiple streaks of pass line winners.
15 trials about 30 minutes of play at 100 rolls per hour
30 trials about 1 hour of play
Example: 30 pass line trials
Probability Of Winning Craps
about a 90% chance of at least 3 pass line wins in a row at least one time
(244/495)^3 = 0.119771609 exactly 3
you want 3 or more
formula for the sum of a geometric series is a/1-r
where a is the first term
r = the ratio
a = (244/495)^3
r = 244/495
so (244/495)^3 / (1-(244/495) = 0.236202973
But now you want this to end. Multiply the above result by 1-(244/495)
0.119771609
((244/495)^3 / (1-(244/495)) * (1-(244/495))
added my table
| at least in a row then lose | Prob | 1 in |
|---|---|---|
| 2 | 0.242979288 | 4.1 |
| 3 | 0.119771609 | 8.3 |
| 4 | 0.059038934 | 16.9 |
| 5 | 0.02910202 | 34.4 |
| 6 | 0.014345238 | 69.7 |
| 7 | 0.007071188 | 141.4 |
| 8 | 0.003485596 | 286.9 |
| 9 | 0.001718152 | 582.0 |
| 10 | 0.000846928 | 1,180.7 |
| 11 | 0.000417475 | 2,395.4 |
| 12 | 0.000205786 | 4,859.4 |
| 13 | 0.000101438 | 9,858.3 |
| 14 | 5.00017E-05 | 19,999.3 |
| 15 | 2.46473E-05 | 40,572.4 |
| 16 | 1.21494E-05 | 82,308.7 |
| 17 | 5.98878E-06 | 166,978.8 |
| 18 | 2.95205E-06 | 338,748.0 |
| 19 | 1.45515E-06 | 687,214.1 |
| 20 | 7.17286E-07 | 1,394,143.4 |
Calculating The Probability Of Winning Craps
pass line win probability = 244/495
(244/495)^3 = 0.119771609 exactly 3
Mustangsally: Maybe I'm missing something, but I get different results from what you show. Suppose (for simplicity) that it was a coin flip with p=0.5 instead of 244/495. To get at least one win in a row would be P=0.5. To get at least two in a row is P=0.5^2, etc. To get the answer for exactly n in a row, you need to multiply the 'at least' by the probability of losing on the n+1 try.
For the pass line problem, I think the 0.119771609 figure is for 3 or more, not for exactly 3.
Isn't that correct, or what did I miss?
For the pass line problem, I think the 0.119771609 figure is for 3 or more, not for exactly 3.
Isn't that correct, or what did I miss?
Sally did it differently by starting with 3 pass line wins in a row in 3 trials.
The OP asked a unique question.
Most ask the probability of winning 3 pass line bets in a row. And for 3 trials it is simply p^3
OP wanted to add the probability of 3 *or more* and *followed by a loss*.
Sally's math shows 3 in a row in 3 trials and the OPs Q arrives at the same value. It should.
IF the OP had asked what is the probability of winning 3 pass line bets in a row then losing, we would have p^3 * q or 0.06073
Let us see if OP is happy and replies.
added
average number of trials to see a run of 3 or more pass line wins: 16.466
4 or more: 33.404
5 or more: 67.765
6 or more: 137.475
Multiple streaks of pass line winners.
15 trials about 30 minutes of play at 100 rolls per hour
30 trials about 1 hour of play
Example: 30 pass line trials
Probability Of Winning Craps
about a 90% chance of at least 3 pass line wins in a row at least one time
about a 58% chance of at least 3 pass line wins in a row at least two times
about a 23% chance of at least 3 pass line wins in a row at least three times
here is the losing streaks (miss) for the pass line per N trials
added
average number of trials to see a run of 3 or more pass line wins: 16.466
4 or more: 33.404
5 or more: 67.765
6 or more: 137.475
You guys are great. Thanks for the detailed responses.
Probability Of Winning A Field Bet In Craps
How did you calculate the average number of trials to see a run of 3, 4, 5, 6 or more pass line wins?
How do you define a trial? Would each shooter be a new trial? Or does a new trial begin after any losing pass line bet, in which case a single shooter could have multiple trials that end and start over with a losing pass line bet from throwing 2, 3, or 12 on a come out roll?
If a bettor were to power press the pass line bet with a $100 wager:
| at least in a row then lose | Prob | 1 in | Bet | Win | Lose |
|---|---|---|---|---|---|
| 1 | 0.492929293 | 2.0 | $100.00 | $100.00 | $100.00 |
| 2 | 0.242979288 | 4.1 | $200.00 | $300.00 | $100.00 |
| 3 | 0.119771609 | 8.3 | $400.00 | $700.00 | $100.00 |
| 4 | 0.059038934 | 16.9 | $800.00 | $1,500.00 | $100.00 |
| 5 | 0.02910202 | 34.4 | $1,600.00 | $3,100.00 | $100.00 |
| 6 | 0.014345238 | 69.7 | $3,200.00 | $6,300.00 | $100.00 |
Does this mean that on average you would be betting about $1600 to win $700 on a press of 3 pass line wins for a net loss of $900?
$3,300 to win $1,500 on 4 pass line wins losing $1,800?
$6,700 to win $3,100 on 5 pass line wins losing $3,600?
$13,700 to win $6,300 on 6 pass line wins losing $7,400?
What's the best way to interpret this expected value for power pressing a pass line bet?
