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Probabilities in Craps |
Why are the odds of a hard four different from the odds of a hard six? Isn't there just one way out of thirty-six possible combinations to hit doubles (double 1,2,3...)? – James from Santa Cruz
Yes, the probability of each double is 1/36. However, you have to compare that to the probability of rolling a losing combination. For a hard four, there are 8 losing rolls (two each of 1-6, 2-5, 3-4, and 1-3), so the probability of winning is 1/9. For a hard six, there are ten losing rolls (two each of 1-6, 2-5, 3-4, 1-5 and 2-4), so the probability of winning is 1/11. The hard six pays more because the probability of winning is less. November 4, 2007I rolled four hard 4's without rolling a 7 or an easy 4. Any idea what the odds on doing that is? Can it be
calculated?
The probability of winning the hard 4 bet is
1/9. So the probability of winning four times in a row is
(1/9)4 = 1 in 6561.
Aug. 25, 2003
On average, after establishing a point in craps how
often will the player make the point?
Given that a point was made 5/12 of the time it
will be a 6 or 8, 4/12 a 5 or 9, and 3/12 a 4 or 10. The
probability of making a 6 or 8 is 5/11, a 5 or 9 is 4/10,
and a 4 or 10 is 3/9. So the probability of making a
point, given that a point was established is,
(5/12)*(5/11)+(4/12)*(4/10)+(3/12)*(3/9) = 40.61%.
June 27, 2003
Based on approximately 150 rolls per hour in dice, how
many decisions with regard to the point will be made. I was
told by someone that there is a decision every 3.6 rolls. Is
this correct? - Jeff from Las Vegas, US
The following are the possible outcomes of the
pass/come bet and their associated probabilities:
- Player wins on come out roll: 22.22%
- Player loses on come out roll: 11.11%
- Player wins on a point: 27.07%
- Player loses on a point: 39.60%
So the player will win on a point about 1 in 3.7
rolls. Nov. 28,
2002
First let me say I think your web site is absolutely
outstanding. Thanks. I watched a new craps game being played
at Grand Casino, Biloxi, MS. called "Four The Money". To win
the shooter must throw the dice 4 times without a 7 coming
up. What are the odds of throwing the dice:
4 times without throwing a 7?
3 times without throwing a 7?
2 times without throwing a 7?
1 times without throwing a 7?
How does the math work for this? Thanks - Stan Abadie from
Harahan, Louisiana
You're welcome, thanks for the kind words. The
probability of throwing the dice n times without a 7, and
then throwing a 7, is (5/6)n*(1/6). The
probability of throwing n non-sevens, without specifying
the next throw would be (5/6)n. So the
probability of throwing the dice at least four times
without a seven would be
(5/6)4=625/1296=0.4823.
Mar. 6, 2002
On average, during the course of 100 points being
established in craps: (1) How many of those would be 4/10,
5/9 or 6/8, (2) During the 100 how many times would each
point (4/10, 5/9, 6/8) vs. a 7 being rolled. — Jon Moriarty
from Danville, New Hampshire
Of those 100 points established, on average 45.45 would be on a 6 or 8, 40 would be on a 5 or 9, and 33.33 would be on a 4 or 10. You could expected 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. Jan. 14, 2001
Do you have any idea what the "record" is for the most
sevens thrown with a pair of fair dice in craps is? I had
someone tell me it was 84, but the odds against that many
sevens in a row being thrown is so long I'm skeptical. It
seems it's more possible that 84 consecutive passes have
come out, but even that's a million to one shot
(figuratively--literally, it's much worse). I tried to look
on the Web but have no idea where I would find something
like that. ñ Mike Manuel from ?
All I know is that a downtown Las Vegas casino,
I think the Californian, boasts of being the home of the
longest single shooter in craps history. I think his roll
lasted about three hours. How many times he rolled I do
not know. The probability of rolling 84 sevens in a row
is 1 in
50,021,738,714,629,100,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
(rounded). The probability of winning 84 pass line bets
in a row is 1 in
534,241,247,406,472,000,000,000,000(rounded). I doubt
that either has ever happened.
Dec. 31, 2000
How often can you roll a pair of dice 28 times without
getting a 7? How do you figure this? Congratulation on your
site, it's great. - Arturo Gomez from Mexico City, Mexico
Thanks for the compliment. I take it you mean
what is the probability of rolling a pair of dice 28
times without getting a 7. The probability of not rolling
a 7 on any one roll is 5/6. The probability of not
rolling a 7 in 28 rolls is (5/6)28 = 0.006066,
or about 1 in 165. Oct. 5,
2000
Q: What are the odds (and frequency) in 100,000 rolls
of the dice (craps/dont pass line) of losing a DP bet
2x's.3x's 4x's 5x's 6x's 7x's 8x's or 9x's in a row. -
Derick from Minneapolis, USA
A: My craps
appendix shows how to work out the odds for any one
bet. There you will see the probability of losing the
don't pass bet is 2928/5940. The probability of losing n
bets in a row is (2928/5940)n. The frequency
in 100,000 of losing n is a row is 100,000 *
(2928/5940)n, assuming that each roll is the
beginning of a new sequence. In other words counting
overlapping DP streaks. Note that this also counts ties
as breaking a series. Aug. 20,
2000
Q: What is the average number of rolls (throws?) till
a shooter "sevens out"? I know that a 7 will appear every 6
rolls, but with come-out 7-11s and craps, plus the
possibility of shooters making multiple points, I think the
average number of rolls may be higher than expected. Is
there any mathematical reference material on this? -
Grshooter from Kansas City, Missouri
First, if the probability of an event is p then the expected number of trials for it to occur is 1/p. Let's call x the expected number of rolls per shooter. The probability that any given round will end in one roll (with a 2, 3, 7, 11, or 12) is 1/3. If the player rolls a 4 or 10 on the come out roll the expected number of additional rolls is 4, because the probability of rolling a 4 or 7 is (6+3)/36 = ¼. . Likewise If the player rolls a 5 or 9 on the come out roll the expected number of additional rolls is 3.6 and for a 6 or 8 is 36/11. Assuming a point was thrown the probability of it being a 4 or 10 is 3/12, a 5 or 9 is 4/12, and a 6 or 8 is 5/12. So the expected number of throws per round is 1+(2/3)*((3/12)*4 + (4/12)*3.6 + (5/12)*(36/11)) = 3.375758. Next, the probability that the player will seven out is (2/3)*((3/12)*(2/3) + (4/12)*(3/5) + (5/12)*(6/11)) = 0.39596. The probability that player will not seven out is 1 - 0.39596 = 0.60404. So...
x = 3.375758 + 0.60404*x
0.39596*x = 3.375758
x = 8.52551 November 28, 2004
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