I am working on a problem to use probabilities for a game with 2 dice. I have researched many methods (Bayesian, etc.) but I am not sure what the right path would be to answer these questions. Here is the use case:

Rules of the Game

  1. The player rolls a pair of fair 6-sided dice. Then, takes the sum of both dice.

    • If the sum is 7 or 11 -> Player Wins!
    • If sum is 2, 3, or 12 -> Player Lost!
    • If the sum is anything else -> Record Sum as “X” and continue to step 2.

  2. The player rolls again.

    • If the sum=X ->Player Wins!
    • If the sum is 7->Player Lost!
    • If the sum is anything else, repeat step 2.

Something strange occurs: One of the dice is not FAIR! It always comes up in the range of 2-5 with equal probability but never 1 or 6!


  1. Let's assume we are on our first roll and we do not win or lose. Instead, we get the sum X, which has roll probability of p. Given that you have already made it to this point, what is the chance of winning going forward? 

  • 2
    $\begingroup$ This game is called Craps $\endgroup$
    – Henry
    Sep 22, 2021 at 11:31
  • $\begingroup$ You've edited out the questions that are addressed by number in the accepted answer, which makes it very confusing to read. $\endgroup$ Sep 22, 2021 at 21:40

1 Answer 1

  1. If $X$ and $Y$ are independent random variables, then $P(X \cap Y) = P(X) \times P(Y)$. You can use this to compute the frequency of rolls. Let $X$ be the fair die and $Y$ the biased die. Let $Z = X+Y$. I'll start you off

$$P(Z=2) = P( (X=1) \cap (Y=1)) = P(X=1) \times P(Y=1) = \dfrac{1}{6} \times 0 = 0$$

$$P(Z=3) = P( (X=1) \cap (Y=2)) + P((X=2) \cap (Y=1)) = \dfrac{1}{6}\times\dfrac{1}{4} + \dfrac{1}{6}\times0 = \dfrac{1}{24}$$

and so on...

  1. This would be $P(Z=11) + P(Z=7)$. You will need to compute them as I have done above. Note that you can make 7 and 11 in two ways though, which is why $P(Z=3)$ is the sum of two probabilities.

  2. Given the answer to 2, what do you think?

  3. This is more interesting. Suppose you roll $x$ and $P(Z=x)= p$. You can now win in a couple of ways.

a) You roll an $x$ on your second roll. If $Z_i$ is the $i^{th}$ roll, then the probability you win on your next roll is $p$ since rolls are independent.

However assume you roll neither an $x$ nor a $7$. We know the probability this happens (hint: its 1 - something). Let's call it $\theta$ for now. The probability you win on the third roll is $\theta \times p$. The probability of a win on the fourth roll is $\theta^2 \times p$, and so on. The probability of winning going forward involves these quantities. You should be able to write out the probability of winning and spot a geometric series, and so the sum (since probabilities are less than 1) can be written down analytically.

  1. You knot the probability of winning and the probability of losing from the above questions, so this is a straight forward computation.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.