2018 AIME II Problems/Problem 13

Revision as of 19:59, 3 July 2018 by Smurty (talk | contribs) (Solution 4)

Problem

Misha rolls a standard, fair six-sided die until she rolls 1-2-3 in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

Let $P_{odd}=\frac{m}{n}$, with the subscript indicating an odd number of rolls. Then $P_{even}=1-\frac{m}{n}$.

The ratio of $\frac{P_{odd}}{P_{even}}$ is just $\frac{P_{odd}}{1-P_{odd}}$.

We see that $P_{odd}$ is the sum of $P_3$,$P_5$,$P_7$,... , while $P_{even}$ is the sum of $P_4$, $P_6$, $P_8$,... .

$P_3$, the probability of getting rolls of 1-2-3 in exactly 3 rolls, is obviously $\frac{1}{216}$.

We set this probability of $P_3$ aside, meaning we totally remove the chance of getting 1-2-3 in 3 rolls. Now the ratio of $P_4+P_6+P_8+...$ to $P_5+P_7+P_9+...$ should be equal to the ratio of $\frac{P_{odd}}{P_{even}}$, because in this case the 1st roll no longer matters, so we can disregard the very existence of it in counting how many times of rolls, and thus, 4 rolls, 6 rolls, 8 rolls... would become an odd number of rolls (while 5 rolls, 7 rolls, 9 rolls... would become even number of rolls).

Notice $P_4+P_6+P_8+...=P_{even}$, and also $P_5+P_7+P_9+...=P_{odd}-P_3=P_{odd}-\frac{1}{216}$

So we have $\frac{P_{even}}{P_{odd}-\frac{1}{216}}=\frac{P_{odd}}{P_{even}}$.

Finally, we get $P_{odd}=\frac{m}{n}=\frac{216}{431}$. Therefore, $m+n = \boxed{647}$.

Solution 2

Call the probability you win on a certain toss $f_n$, where $n$ is the toss number. Obviously, since the sequence has length 3, $f_1=0$ and $f_2=0$. Additionally, $f_3=\left(\frac{1}{6}\right)^3$. We can call this value $x$, to keep our further equations looking clean. We can now write our general form for $f$ as $f_n=x(1-\sum_{i=1}^{n-3}f_i)$. This factors the probability of the last 3 rolls being 1-2-3, and the important probability that the sequence has not been rolled in the past (because then the game would already be over). Note that $\sum_{i=1}^{\infty}f_i=1$ since you'll win at some point. An intermediate step here is figuring out $f_n-f_{n+1}$. This is equal to $x(1-\sum_{i=1}^{n-3}f_i)-x(1-\sum_{i=1}^{n-2}f_i)=x(\sum_{i=1}^{n-2}f_i-\sum_{i=1}^{n-3}f_i)=xf_{n-2}$. Adding up all the differences, i.e. $\sum_{i=2}^{\infty}(f_{2n-1}-f_{2n})$ will give us the amount by which the odds probability exceeds the even probability. Since they sum to 1, that means the odds probability will be half of the difference above one-half. Subbing in our earlier result from the intermediate step, the odd probability is equal to $\frac{1}{2}+\frac{1}{2}x\sum_{i=2}^{\infty}f_{2n-3}$. Another way to find the odd probability is simply summing it up, which turns out to be $\sum_{i=1}^{\infty}f_{2n-1}$. Note the infinite sums in both expressions are equal; let's call it $P$. Equating them gives $\frac{1}{2}+\frac{1}{2}xP=P$, or $P=\frac{1}{2-x}$. Finally, substituting $x=\frac{1}{216}$, we find that $P=\frac{216}{431}$, giving us a final answer of $216 + 431 = \boxed{647}$. --DanDan0101

Solution 3

Let $S(n)$ be the number of strings of length $n$ containing the digits $1$ through $6$ that do not contain the string $123$. Then we have $S(n) = 6 \cdot S(n-1) - S(n-3)$ because we can add any digit to end of a string with length $n-1$ but we have to subtract all the strings that end in $123$. We rewrite this as \begin{align*} S(n) &= 6 \cdot S(n-1) - S(n-3) \\ &= 6 \cdot (6 \cdot S(n-2) - S(n-4)) - (6 \cdot (S(n-4) - S(n-6)) \\ &= 36 \cdot S(n-2) - 12 \cdot S(n-4) + S(n-6) \end{align*} We wish to compute $P=\sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}}$ since the last three rolls are $123$ for the game to end. Summing over the recursion, we obtain \[\sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}} =36 \cdot \sum_{n=0}^\infty \frac{S(2n-2)}{6^{2n+3}} - 12 \cdot  \sum_{n=0}^\infty \frac{S(2n-4)}{6^{2n+3}}+ \sum_{n=0}^\infty \frac{S(2n-6)}{6^{2n+3}}\]Now shift the indices so that the inside term is the same: \begin{align*} \sum_{n=3}^\infty  \frac{S(2n)}{6^{2n+3}} &= \frac{36}{6^2} \cdot \sum_{n=2}^\infty \frac{S(2n)}{6^{2n+3}} - \frac{12}{6^4} \cdot \sum_{n=1}^\infty \frac{S(2n)}{6^{2n+3}} + \frac{1}{6^6} \cdot \sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}}  \\  \left(P - \frac{S(0)}{6^3} -  \frac{S(2)}{6^5} -\frac{S(4)}{6^7}  \right) &= \frac{36}{6^2} \cdot \left( P -   \frac{S(0)}{6^3} -  \frac{S(2)}{6^5}\right) - \frac{12}{6^4} \cdot \left(  P - \frac{S(0)}{6^3} \right) + \frac{1}{6^6} \cdot P  \end{align*}Note that $S(0) = 1, S(2) = 26$ and $S(4) = 6^4 - 2 \cdot 6 = 1284$. Therefore, \begin{align*} \left(P - \frac{1}{6^3} -  \frac{36}{6^5} -\frac{1284}{6^7}  \right) = \frac{36}{6^2} \cdot \left( P -   \frac{1}{6^3} -  \frac{36}{6^5}\right) - \frac{12}{6^4} \cdot \left(  P - \frac{1}{6^3} \right) + \frac{1}{6^6} \cdot P   \end{align*}Solving for $P$, we obtain $P = \frac{216}{431} \implies m+n = \boxed{647}$. -Vfire


2018 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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Solution 4

Let $A=\frac{1}{6} \begin{bmatrix} 5 & 1 & 0 & 0 \\ 4 & 1 & 1 & 0 \\ 4 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$. $A$ is a transition matrix for the prefix of 1-2-3 matched so far. The state corresponding to a complete match has no outgoing probability mass. The probability that we roll the dice exactly $k$ times is $(A^k)_{1,4}$. Thus the probability that we roll the dice an odd number of times is $1-\left(\sum_{k=0}^\infty A^{2k}\right)_{1,4} = 1-(I - A^2)^{-1}_{1,4} = \frac{216}{431}$. Thus the answer is $216+431=\boxed{647}$.


2018 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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