Difference between revisions of "2018 AIME II Problems/Problem 13"
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Obviously, since the sequence has length 3, <math>f_1=0</math> and <math>f_2=0</math>. | Obviously, since the sequence has length 3, <math>f_1=0</math> and <math>f_2=0</math>. | ||
Additionally, <math>f_3=\left(\frac{1}{6}\right)^3</math>. We can call this value <math>x</math>, to keep our further equations looking clean. | Additionally, <math>f_3=\left(\frac{1}{6}\right)^3</math>. We can call this value <math>x</math>, to keep our further equations looking clean. | ||
| − | We can now write our general form for <math>f</math> as <math>f_n=x(1-\sum_{i=1}^{n-3}f_i)</math>. This factors the probability of the last 3 rolls being 1-3 | + | We can now write our general form for <math>f</math> as <math>f_n=x(1-\sum_{i=1}^{n-3}f_i)</math>. 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 <math>\sum_{i=1}^{\infty}f_i=1</math> since you'll win at some point. | Note that <math>\sum_{i=1}^{\infty}f_i=1</math> since you'll win at some point. | ||
An intermediate step here is figuring out <math>f_n-f_{n+1}</math>. This is equal to <math>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}</math>. | An intermediate step here is figuring out <math>f_n-f_{n+1}</math>. This is equal to <math>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}</math>. | ||
Revision as of 12:45, 25 March 2018
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 
where 
and 
are relatively prime positive integers. Find 
.
Solution
Call the probability you win on a certain toss 
, where is the toss number.
Obviously, since the sequence has length 3, 
and 
.
Additionally, 
. We can call this value 
, to keep our further equations looking clean.
We can now write our general form for 
as 
. 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 
since you'll win at some point.
An intermediate step here is figuring out 
. This is equal to 
.
Adding up all the differences, i.e. 
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 
.
Another way to find the odd probability is simply summing it up, which turns out to be 
. Note the infinite sums in both expressions are equal; let's call it 
. Equating them gives 
, or 
.
Finally, substituting 
, we find that 
.
The sum of the numerator and denominator is thus 
and that is the answer.
--DanDan0101
| 2018 AIME II (Problems • Answer Key • Resources) | ||
| 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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