Wandering Sammy

Sammy is lost and starts to wander aimlessly. Each minute, he walks one meter forward with probability 1 2 \frac{1}{2} , stays where he is with probability 1 3 \frac{1}{3} , and walks one meter backward with probability 1 6 \frac{1}{6} .

After one hour, what is the expected value for the forward distance (in meters) that Sammy has traveled?


The answer is 20.

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7 solutions

Eli Ross Staff
Sep 1, 2015

Let X i X_i be the forward distance in meters that Sammy travels during the i i th minute and note that

E ( X i ) = 1 2 ( 1 ) + 1 3 ( 0 ) + 1 6 ( 1 ) = 1 3 E(X_i) = \frac{1}{2}\cdot (1) + \frac{1}{3}\cdot (0) + \frac{1}{6} \cdot (-1) = \frac{1}{3}

for all i . i. Then, the forward distance Sammy has traveled after one hour is simply X 1 + X 2 + + X 60 . X_1+X_2+\ldots+X_{60}. By linearity of expectation,

E [ i = 1 60 X i ] = i = 1 60 E [ X i ] = 60 1 3 = 20. E\left[\sum_{i=1}^{60}X_i\right] = \sum_{i=1}^{60} E\left[X_i\right] = 60 \cdot \frac{1}{3} = 20.

Why in the description of the event going forward 1/6 is multiplied by -1 ? Thanks

guido barta - 5 years, 8 months ago

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That's the probability of walking one meter backward.

Kenneth Choo - 5 years, 2 months ago
Aditya Saroj
Sep 3, 2015

Half of the time Sammy moves forward, so at the rate of each minute he will move 30m forward. He moves backward one-sixth of the time. So, 1/6 X 60 = 10. So distance travelled = 30 - 10 = 20m (NOTE: We can ignore the probability of staying where he is, as it does not affect the distance. It's in the left out 20 minutes as 1/3 X 60 = 20)

O9 o o O

Oportunidad

Alan Jara - 4 years, 5 months ago

If Sammy spent half his time walking forwards, after 60 minutes he would have spend 60/2 = 30 minutes walking forwards. If he walks 1 metre per minute, in 30 minutes he would have walked 30 metres.

If Sammy spent a sixth of his time walking backwards, after 60 minutes he would have spend 60/6 = 10 minutes walking backwards. If he walks 1 metre per minute, in 10 minutes he would have walked back 10 metres.

If he walked 30 metres forward and 10 metres backwards, he would have walked 30 - 10 = 20 metres forwards in total.

Arthur Conmy
Jul 8, 2017

Let E [ X n ] E[X_n] denote the expected distance travelled by Sammy in n n minutes.

Then E [ X n + 1 ] = 1 2 ( E [ X n ] + 1 ) + 1 3 E [ X n ] + 1 6 ( E [ X n ] 1 ) E[X_{n+1}]=\frac{1}{2}(E[X_n]+1) + \frac{1}{3}E[X_n] + \frac{1}{6}(E[X_n]-1) , based off the probabilities of moving 1 1 meter forward, 0 0 meters forward and 1 1 meter backwards (equivalently 1 -1 meters forward).

This simplifies to E [ X n + 1 ] = E [ X n ] + 1 3 E[X_{n+1}]=E[X_n]+\frac{1}{3} , or E [ X n ] = n 3 + E [ X 0 ] E[X_{n}]=\frac{n}{3}+E[X_0] .

Since E [ X 0 ] = 0 E[X_0]=0 , as Sammy can't have moved before the first second, E [ X 60 ] = 60 3 = 20 E[X_{60}]=\frac{60}{3}=\boxed{20}

Abner Estrada
Oct 7, 2017

60 mins ( 1/2 - 1/3 + 1/6) = 20 m

Alaa Ahmed Fawzi
Sep 4, 2015

We can ignore the probability of staying where he is, as it does not affect the distance. It's in the left out 20 minutes as 1/3 X 60 = 20

Elijah Frank
Dec 27, 2020

60 1/2+60 1/3+60 1/6 = 60/3 = 20 meters or the míddle value (I think is geometric mean) and multiply 60 1/3 = 20.

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