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Difference between revisions of "2016 AIME I Problems/Problem 13"
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==Solution==
 
==Solution==
 
Clearly Freddy's <math>x</math>-coordinate is irrelevant, so we let <math>E(y)</math> be the expected value of the number of jumps it will take him to reach the river from a given <math>y</math>-coordinate. Observe that <math>E(24)=0</math>, and <cmath>E(y)=1+\frac{E(y+1)+E(y-1)+2E(y)}{4}</cmath> for all <math>y</math> such that <math>1\le y\le 23</math>. Also note that <math>E(0)=1+\frac{2E(0)+E(1)}{3}</math>. This gives <math>E(0)=E(1)+3</math>. Plugging this into the equation for <math>E(1)</math> gives that <cmath>E(1)=1+\frac{E(2)+3E(1)+3}{4},</cmath> or <math>E(1)=E(2)+7</math>. Iteratively plugging this in gives that <math>E(n)=E(n+1)+4n+3</math>. Thus <math>E(23)=E(24)+95</math>, <math>E(22)=E(23)+91=E(24)+186</math>, and <math>E(21)=E(22)+87=E(24)+273=\boxed{273}</math>.
 
Clearly Freddy's <math>x</math>-coordinate is irrelevant, so we let <math>E(y)</math> be the expected value of the number of jumps it will take him to reach the river from a given <math>y</math>-coordinate. Observe that <math>E(24)=0</math>, and <cmath>E(y)=1+\frac{E(y+1)+E(y-1)+2E(y)}{4}</cmath> for all <math>y</math> such that <math>1\le y\le 23</math>. Also note that <math>E(0)=1+\frac{2E(0)+E(1)}{3}</math>. This gives <math>E(0)=E(1)+3</math>. Plugging this into the equation for <math>E(1)</math> gives that <cmath>E(1)=1+\frac{E(2)+3E(1)+3}{4},</cmath> or <math>E(1)=E(2)+7</math>. Iteratively plugging this in gives that <math>E(n)=E(n+1)+4n+3</math>. Thus <math>E(23)=E(24)+95</math>, <math>E(22)=E(23)+91=E(24)+186</math>, and <math>E(21)=E(22)+87=E(24)+273=\boxed{273}</math>.
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==Video Solution==
 +
https://youtu.be/T5F4e8NxfG0
 +
 
==Video Solution==
 
==Video Solution==
 
For those who want more explanation, here is a video explaining the solution: https://www.youtube.com/watch?v=jQCGkOGMlFQ
 
For those who want more explanation, here is a video explaining the solution: https://www.youtube.com/watch?v=jQCGkOGMlFQ

Latest revision as of 19:41, 14 May 2022

Problem

Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line $y = 24$. A fence is located at the horizontal line $y = 0$. On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a point where $y=0$, with equal likelihoods he chooses one of three directions where he either jumps parallel to the fence or jumps away from the fence, but he never chooses the direction that would have him cross over the fence to where $y < 0$. Freddy starts his search at the point $(0, 21)$ and will stop once he reaches a point on the river. Find the expected number of jumps it will take Freddy to reach the river.

Solution

Clearly Freddy's $x$-coordinate is irrelevant, so we let $E(y)$ be the expected value of the number of jumps it will take him to reach the river from a given $y$-coordinate. Observe that $E(24)=0$, and \[E(y)=1+\frac{E(y+1)+E(y-1)+2E(y)}{4}\] for all $y$ such that $1\le y\le 23$. Also note that $E(0)=1+\frac{2E(0)+E(1)}{3}$. This gives $E(0)=E(1)+3$. Plugging this into the equation for $E(1)$ gives that \[E(1)=1+\frac{E(2)+3E(1)+3}{4},\] or $E(1)=E(2)+7$. Iteratively plugging this in gives that $E(n)=E(n+1)+4n+3$. Thus $E(23)=E(24)+95$, $E(22)=E(23)+91=E(24)+186$, and $E(21)=E(22)+87=E(24)+273=\boxed{273}$.

Video Solution

https://youtu.be/T5F4e8NxfG0

Video Solution

For those who want more explanation, here is a video explaining the solution: https://www.youtube.com/watch?v=jQCGkOGMlFQ

See also

2016 AIME I (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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