A flea

Probability Level 4

A flea is currently on number 11 of a number line (with length 20) marked 0, 1, 2, ..., 19, 20. Every second the flea jumps randomly towards either of the two extremes, 0 and 20, making a movement of 1 at a time.

What is the expected number of seconds until the flea ends on either one of the two extremes?


Bonus : What is the probability that the flea ends first on 0 (which is a distance of 11 away from its starting position)?


The answer is 600.

Filippo Olivetti by Filippo Olivetti , 20, Italy

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Filippo Olivetti
Jul 20, 2017

I've previously numbered all the points of the segment according to their position. So A A is an extreme, point 1 1 is right than A and so on until point 20 20 which is the rightmost point as the above picture shows.

Let T i T_i , with i ∈ 1 , 2 , . . . , 20 i \in {1,2,...,20} , be the expected value of the number of seconds which the flea takes in order to end on point A A starting on point i i . Now, let's analyze a particular T i T_i , for instance T 8 T_{8} . If the flea is at point 8 8 , then it can end on point 7 7 or point 9 9 with equal probability, so: T 8 = 1 2 ( T 7 + 1 ) + 1 2 ( T 9 + 1 ) T_{8} = \frac{1}{2} (T_7+1) + \frac{1}{2} (T_9+1) where + 1 +1 is given by the jump, which, from hypothesis, takes 1 second. Repeating the same argument for all the point we get the system of equation: { T 1 = 1 2 ( 1 ) + 1 2 ( T 2 + 1 ) . . . T i = 1 2 ( T i − 1 + 1 ) + 1 2 ( T i + 1 + 1 ) . . . T 20 = T 19 + 1 \begin{cases} T_1 = \frac{1}{2} (1) + \frac{1}{2} (T_2+1)\\ ... \\ T_i = \frac{1}{2} (T_{i-1}+1) + \frac{1}{2} (T_{i+1}+1)\\ ...\\ T_{20} = T_{19}+1 \\ \end{cases} Substituing the last equation into the previous equations: → { T 1 = 1 2 ( T 2 ) + 3 2 T 2 = T 1 + 37 . . . T i = T i − 1 + 2 ( 20 − i ) + 1 . . . T 19 = T 18 + 3 T 20 = T 19 + 1 \rightarrow \begin{cases} T_1 = \frac{1}{2} (T_2) + \frac{3}{2} \\ T_2 = T_1 + 37 \\ ... \\ T_i =T_{i-1} + 2(20-i)+1\\ ...\\ T_{19} = T_{18}+3 \\ T_{20} = T_{19}+1 \\ \end{cases}

Hence, T 9 = 280 T_9 =280 and T 11 = 320 T_{11}=320 . The answer is 600 \boxed{600} (i made a calculus mistake so the answer is wrong: someone can check this new answer?)

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks. I've updated the answer to 600.

Brilliant Mathematics Staff - 3 years, 10 months ago

The correct formula is T i = i ( 40 − i ) T_i = i(40 - i) .

First, the equation T 1 = 1 2 T 2 + 3 2 T_1 = \frac{1}{2} T_2 + \frac{3}{2} should be T 1 = 1 2 T 2 + 1 T_1 = \frac{1}{2} T_2 + 1 .

From the formulas T 1 = 1 2 T 2 + 1 T_1 = \frac{1}{2} T_2 + 1 and T 2 = T 1 + 37 T_2 = T_1 + 37 , T 1 = 39 T_1 = 39 and T 2 = 76 T_2 = 76 . Then T i = 2 ( 20 − i ) + 1 + T i − 1 = [ 2 ( 20 − i ) + 1 ] + [ 2 ( 21 − i ) + 1 ] + T i − 2 = [ 2 ( 20 − i ) + 1 ] + [ 2 ( 21 − i ) + 1 ] + [ 2 ( 22 − i ) + 1 ] + T i − 3 = ⋯ = [ 2 ( 20 − i ) + 1 ] + [ 2 ( 21 − i ) + 1 ] + [ 2 ( 22 − i ) + 1 ] + ⋯ + ( 2 ⋅ 17 + 1 ) + T 2 = 2 [ ( 20 − i ) + ( 21 − i ) + ⋯ + 17 ] + ( 4 − i ) + ( i − 2 ) + 76 = 2 ⋅ ( 20 − i ) + 17 2 ⋅ ( i − 2 ) + ( i − 2 ) + 76 = 40 i − i 2 . \begin{aligned} T_i &= 2(20 - i) + 1 + T_{i - 1} \\ &= [2(20 - i) + 1] + [2(21 - i) + 1] + T_{i - 2} \\ &= [2(20 - i) + 1] + [2(21 - i) + 1] + [2(22 - i) + 1] + T_{i - 3} \\ &= \dotsb \\ &= [2(20 - i) + 1] + [2(21 - i) + 1] + [2(22 - i) + 1] + \dots + (2 \cdot 17 + 1) + T_2 \\ &= 2[(20 - i) + (21 - i) + \dots + 17] + (4 - i) + (i - 2) + 76 \\ &= 2 \cdot \frac{(20 - i) + 17}{2} \cdot (i - 2) + (i - 2) + 76 \\ &= 40i - i^2. \end{aligned}

Jon Haussmann - 3 years, 10 months ago

What does the expected no of seconds exactly mean? I mean it is equally possible never to reach extremes if the fly keeps jumping back and forth or to reach it in 9 seconds if it keeps moving in one direction only if moving in both directions are equally probable.

Akash Pratap Singh - 3 years, 9 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...