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Here is the situation:

You roll $x$ fair $6$-sided dice. On a roll of $6$, you re-roll the dice and add the result to the previous roll. If you roll a $6$ again, you continue re-rolling until you don't get a $6$.

I'm trying to figure out the odds of getting at least one result of (or over) $Y$ on these dice.

ViHdzP
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Gabriel Lorquet
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  • Unless I misunderstand the premise, this is just like rolling $x$ fair 5 sided dice. – Peter Woolfitt Jan 01 '15 at 11:29
  • @PeterWoolfitt: It is equivalent but not immediately because you have to check that the probability of ending the rolling process is $1$, or equivalently that the probability of rolling on for ever is $0$. It is entirely possible but with $0$ probability. – user21820 Jan 01 '15 at 11:42
  • @PeterWoolfitt: You missed the adding of the previous roll, there's actually a chance of getting 36723945302753249239645962349329234 from one die in this game. – Henrik supports the community Jan 01 '15 at 11:45
  • @Henrik: Oh I misread too. – user21820 Jan 01 '15 at 11:46
  • @Henrik I think that the question leaves that point ambiguous. – Peter Woolfitt Jan 01 '15 at 11:55
  • @PeterWoolfitt: After a re-read, I must agree that's it's not clearwhat s meant. And reading like you do, it is comleteæy equivalent to rolling a set of fair 5 sided dices (with the ending questions). – Henrik supports the community Jan 01 '15 at 12:09
  • http://math.stackexchange.com/questions/1644794/exploding-a-k-a-open-ended-dice-pool http://math.stackexchange.com/questions/391792/probability-of-rolling-n-successes-on-an-open-ended-exploding-dice-roll http://math.stackexchange.com/questions/1119872/on-average-how-many-times-must-i-roll-a-dice-until-i-get-a-6 – Mox Oct 11 '16 at 21:32

1 Answers1

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The dice are independent so we only need to solve for one die. Since the probability of re-rolling it forever is $\lim_{n\to\infty} (\frac{1}{6})^n = 0$, the die will with probability $1$ end up showing a non-$6$ value.

Let $p_k$ be the probability that the value is $k$, for any $k \in [1..5]$. Then $p_k = \frac{1}{6} + \frac{1}{6} p_k$ because there is a $\frac{1}{6}$ probability of getting $k$ on the first roll and $\frac{1}{6}$ probability of getting a $6$ on the first roll, after which the die (having forgotten that it has been rolled before) behaves exactly as before, in which case it has probability $p_k$ of finally showing a $k$. This easily gives $p_k$.

If on re-rolling we do not add the previous rolls of the same die, then we are done. If not...

Now we need to add the values gotten by adding all the previous rolls. Note that it is just $6$ times the number of re-rolls, and the probability of re-rolling $k$ times is $\frac{5}{6}\frac{1}{6^k}$. Thus the probability of getting a total of $6a+b$ is $\frac{5}{6}\frac{1}{6^a} p_b$.

Now it is easy to find the probability of getting for one die a total score of more than $y$. Let $q$ be that probability. Then the probability of getting at least one such total among $x$ dice is $1-(1-q)^x$.

user21820
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