Roll, roll, roll the dice

Let $D_1, D_2$ be the random variables representing the values rolled for the two dice. We observe that the expected value of $D_1 + D_2$ , regardless of whether doubles are rolled, is $E[D_1 + D_2] = E[D_1]+E[D_2] = 2E[D_1] = 2(1+2+3+4+5+6)/6 = 7$ . (Note also $E[D_1 + D_2 \mid D_1 \ne D_2 ] = 2(3+4+2(5)+2(6)+3(7)+2(8)+2(9)+10+11)/30 = 7$ .) Furthermore, note that for any roll of the pair of dice, the probability of observing $D_1 = D_2$ is $\Pr[D_1 = D_2] = 6/36 = 1/6$ . Hence the probability that Pinar rolls zero doubles is $5/6$ , exactly one double is $(1/6)(5/6) = 5/36$ , exactly two doubles is $(1/6)(1/6)(5/6) = 5/216$ , and exactly three doubles is $(1/6)^3 = 1/216$ , and these are the only possible outcomes. Furthermore, because each roll of the pair of dice is independent and identically distributed, the expected value of the token's position for zero doubles is $7$ , exactly one double is $7 + 7 = 14$ (since she rolls the dice twice), exactly two doubles is $3(7) = 21$ , and exactly three doubles is $36$ . Hence the unconditional expected value is $\frac{5}{6} \cdot 7 + \frac{5}{36} \cdot 14 + \frac{5}{216} \cdot 21 + \frac{1}{216} \cdot 36 = \frac{607}{72},$ so $a + b = 607 + 72 = 679$ .

We first find the expected sum on a pair of dice for each roll. Let $(m,n)$ denote the result of each roll, with $m$ and $n$ being the numbers on the first and second die respectively. Observe that we can pair up all possible rolls such that each roll $(m,n)$ is paired with a different roll $(7-m,7-n)$ , with the mean sum of the two rolls being $7$ . Therefore, the expected sum on a pair of dice for each roll must be $7$ .

We now tweak the rules a little and suppose that Pinar does not move the token to $36$ if she rolls doubles a third time, but just move the tokens in the usual manner on the third move regardless of the result. We shall calculate the expected value of the square that the token would end on. On the first throw, the token is expected to move $7$ squares to the right. The second move will occur with $\frac{1}{6}$ probability, and if it occurs, the token is also expected to move $7$ squares to the right. The same is true for third move, but it occurs with $\frac{1}{36}$ probability. Therefore, the expected value of the final position of the token will be $7(1+\frac{1}{6}+\frac{1}{36}) = \frac{301}{36}$ .

But we know that Pinar will just move the token to $36$ if she rolls doubles a third time. Observe that if we know that Pinar threw a double, then we can still do the pairing of possible rolls as we did earlier and we can still conclude that the expected sum on the pair of dice must be $7$ . So the expected position of the token if Pinar threw three doubles (in the game with the tweaked rule) is $7 \times 3 = 21$ . But in the actual game, Pinar would have moved the token to $36$ , so the expected "undercount" of the position of the token is $36 - 21 = 15$ . We only undercount the position if Pinar threw three doubles, and this happens with $\frac{1}{216}$ probability. So we shall add $15 \times \frac{1}{216} = \frac{5}{72}$ to $\frac{301}{36}$ to obtain $\frac{607}{72}$ as the expected value of the final position of the token.

The answer we want is thus $607 + 72 = 679$ .

1st dice roll: The probability for each of the normal outcomes from a single pair of dice is the same except for the even numbers, which are reduced by 1/36 (the doubles outcome). For example - The probability of 7 is still 6/36 = 1/6, but the probability of 6 is 5/36 -1/36 = 4/36 = 1/9

2nd dice roll: Multiply the probability of each individual double on the first roll with the probability of the second dice roll with the same parameters as the first. Record for each outcome. For example: - Probability of getting a 5 after two dice rolls = $1/36 \times 1/18$ = 1/648

3rd Dice Roll: Multiply the probability of the various combinations of doubles by the probability of the third dice role (same parameters as the first) for all values up to 36 which is the probability of rolling three doubles of any kind. For example: - The probability of getting 12 after 3 dice rolls = $(3/1296 \times 2/36) + (2/1296 \times 4/36) + (1/1296 \times 4/36) = 1/259$ - The probability of getting 36 after dice rolls = $1/6 \times 1/6 \times 1/6 = 1/216$

Expected value: Add the probabilities up for each specific outcome, then multiply the result by the outcome. Add up all of these results together to come up with the expected value of 8 31/72 or 607/72

607 + 72 = 679

We calculate the probability of getting doubles $0,$ $1,$ $2,$ or $3$ times. We summarize in the following table:

$\begin{array}{|r|cccc|} \hline \mbox{Doubles} & 0 & 1 & 2 & 3\\ \hline \mbox{Chance} & \frac{5}{6} & \frac{5}{36} & \frac{5}{216} & \frac{1}{216}\\ \hline \end{array}$

When doubles are rolled, the expected value of the roll is $\frac{2 + 4 + 6 + 8 + 10 + 12}{6} = 7$ and when doubles are not rolled the expected value is the same. Thus, when Pinar rolls $0$ doubles, the expected position of the token is $0.$ When she rolls $1$ it is $7 + 7 = 14,$ when she rolls $2$ it is $7 + 7 + 7 = 21,$ and when she rolls $3$ it is $36.$ Thus, the expected position of the token is $7\left(\frac{180}{216}\right) + 14\left(\frac{30}{216}\right) + 21\left(\frac{5}{216}\right) + 36\left(\frac{1}{216}\right) = \frac{1785 + 36}{216}= \frac{607}{72}.$

So $a + b = 679.$

The probability that the first roll is not a double is $\frac{5}{6}$ . The possible values for the sum of the die are $3, 5, 7, 9, 11$ , so the expected value is $(\frac{5}{6})(\frac{3+5+7+9+11}{5}) = \frac{35}{6}$ .

Next, we consider when the first roll is a double, but the second is not. The probability of this occurring is $(\frac{1}{6})(\frac{5}{6})$ , or $\frac{5}{36}$ . The possible values for the sum of the die on the second roll are still $3, 5, 7, 9, 11$ , but now we must add the amount that was moved with the previous roll, which could be $2, 4, 6, 8, 10, 12$ , so the expected value is $(\frac{5}{36})(\frac{(2+3)+(2+5)+... (4+3)+(4+5)+... (6+3)+(6+5)+...(8+3)+(8+5)+...}{30}) = \frac{35}{18}$ .

Next, we consider when the first and second rolls are doubles, but the third is not. The probability of this occurring is $(\frac{1}{6})^2(\frac{5}{6})$ , or $\frac{5}{216}$ . The possible values for the sum of the die on the third roll are still $3, 5, 7, 9, 11$ , but now we must add the amount that was moved with the previous rolls, which could now be $4, 8, 12, 16, 20, 24$ (two doubles), so the expected value is $(\frac{5}{216})(\frac{(4+3)+(4+5)+... (8+3)+(8+5)+... (12+3)+(12+5)+...(16+3)+(16+5)+...}{30}) = \frac{35}{72}$ .

Lastly, we consider when all three rolls are doubles. The probability of this occurring is $(\frac{1}{6})^3$ , or $\frac{1}{216}$ , so the expected value is $\frac{5}{216}(36)=\frac{1}{6}$ .

We add these all up, getting $\frac{35}{6}+ \frac{35}{18}+\frac{35}{72}+\frac{1}{6}=\frac{607}{72}$ , so the answer is $607+72=679$