Suppose you continue to roll a fair 6-sided die until you either roll a one, roll two 2's in a row, or roll three 3's in a row. What is the probability for each of these game endings?

Edit: My thinking is as follows. Since die rolls are independent of one another, $\mathbb{P}$(roll a 1) = $\frac 16$, $\mathbb{P}$(roll two 2's) = $\frac 16$ $\cdot$ $\frac 16$, $\mathbb{P}$(roll three 3's) = $\frac 16$ $\cdot$ $\frac 16$ $\cdot$ $\frac 16$. The game can only end in the three possible ways listed above, and thus form a partition with these three probabilities. Then the proportion of each individual probability that comprises the partition is the probability that the game will conclude with that ending. For example, the probability for the game to end by rolling a 1 would be $\frac {\frac 16}{\frac 16 + \frac {1}{36} + \frac {1}{216}}$. Is this a correct approach?

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    $\begingroup$ What have you tried till now ? Pl let us know, and also where you are stuck. $\endgroup$ – true blue anil Apr 7 at 8:47
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    $\begingroup$ What have you tried so far? $\endgroup$ – Moko19 Apr 7 at 8:47
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    $\begingroup$ For reasons that are not obvious to me, the three probabilities seem to be multiples of $\frac1{351}$ $\endgroup$ – Henry Apr 7 at 9:26
  • $\begingroup$ I have edited my post, if you wouldn't mind checking $\endgroup$ – Riley Lenway Apr 7 at 9:56
  • $\begingroup$ As far as I can tell $\frac {\frac 16}{\frac 16 + \frac {1}{36} + \frac {1}{216}}=\frac{36}{43}\approx 0.8372$ is slightly too low $\endgroup$ – Henry Apr 7 at 10:03

Let's define $P_n$ as the probability of stopping after exactly $n$ rolls, and $Q_n$ as the probability of stopping by the time we reach $n$ rolls.

We can express the probability of stopping on the $n$th roll after a $1$ as $A_n=\frac{1}{6}(1-Q_{n-1})$ (and $A_1=\frac{1}{6}$, after a pair of 2s as $B_n=\frac{1}{36}(1-Q_{n-2})$ (and $B_1=0, B_2=\frac{1}{36}$), and after a trio of 3s as $C_n=\frac{1}{216}(1-Q_{n-3})$ (and $C_2=C_1=0, C_3=\frac{1}{216}$).

We can now build the rest our recursive definitions: $$P_n=A_n+B_n+C_n$$ $$Q_n=\sum_{j=1}^{n}P_j$$

The chances of ending after a 1 are $\sum_{j=1}^\infty A_j$, after a pair of 2s are $\sum_{j=1}^\infty B_j$, and after a trio of 3s are $\sum_{j=1}^\infty C_j$.

Using a computer for computation, I got that these are $\frac{36}{43},\frac{6}{43}$ and $\frac{1}{43}$ respectively. I could conjecture a reason for this, but it would be only a guess

  • $\begingroup$ $B_n$ is only $1/36$ of those whose $n-2$th roll is not 2. Otherwise it contributes to $B_{n-1}$ $\endgroup$ – Empy2 Apr 8 at 10:47

You are in one of four states:
B: Just rolled a 2
C: Just rolled one 3
D: just rolled two 3s
A: everything else.

When in state A, let the chances of finishing with 1,2s or 3s be A1, A2 and A3. Similar with the other states. Note A1+A2+A3=1.

Depending on the next roll, $$A1=\frac16+\frac16B1+\frac16C1+\frac36A1$$ Each of the twelve probabilities has a similar equation, leading to a $12×12$ matrix equation.
I think the denominators turn out to be $351$ because of the determinant of that matrix.


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