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A grasshopper is jumping on a number line and starts at its home at zero (i.e., the “origin”). Its first jump will be length 1/2, its second jump will be length 1/4, its third jump will be length 1/8 and so on. (To make this explicit, its N-th jump will be length (1/2)^N.)

However, before the jumping begins, it drinks a little too much grasshopper juice and loses all sense of direction. For each jump, it will randomly hop left or right along the number line with equal probability.

After infinitely many jumps, the grasshopper’s head is once again clear and it wants to return home to the origin. On average, what is the expected distance it travels to return to the origin? (Note that no matter which side of the origin the grasshopper is on, “distance” is defined as being zero or positive, but cannot be negative.)

Try to solve this puzzle in your head.


This puzzle was previously posted at: https://fivethirtyeight.com/features/can-you-help-the-grasshopper-get-home/

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  • $\begingroup$ The follow up question in the source seems interesting. If the grasshopper jumping distance is (2/3)^N $\endgroup$
    – justhalf
    Jul 11, 2023 at 2:48

3 Answers 3

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Two observations:

1. The first jump determines which side of the origin the grasshopper is, because the total length of all jumps is 1, and the first jump is 1/2. So after the first jump, no jump goes past zero/home.
2. After the first jump, each jump is equally likely to be towards home as away from home, so the expected value of this jump's contribution to the distance from home is zero.

Therefore the expected distance is

1/2.

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The answer is:

1/2

The sum of all jumps after the first is equal to the length of the first jump; no jump after the first will carry the grasshopper past its home. Therefore, each jump after the first does not change the expected distance from home.

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    $\begingroup$ There seems to be something missing in this answer, as it doesn't mention the equal probability of jumping left and right (if this probability is not equal, the answer is not 1/2) $\endgroup$
    – justhalf
    Jul 10, 2023 at 4:19
  • $\begingroup$ Actually, it only the first jump has to be fair. On later jumps, it only matters that the chance of a positive jump is independent of position. $\endgroup$ Jul 16, 2023 at 17:24
  • $\begingroup$ ah, the expectation being taken on later jump is canceled from the first jump being left or right, got it. Jaap's answer canceled the expectation by each jump itself, but this is only possible since the probability is 0.5. Your answer is more general, and thus better, then. Thanks for the explanation! I made a small edit to change my downvote to upvote. $\endgroup$
    – justhalf
    Jul 17, 2023 at 2:44
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~1/3 as it is told to us that the grasshopper can jump either left or right with equal probability. By that logic, one can correctly assume that the grasshopper will alternate between left and right on each jump. Assuming he jumps to the right first(a length of 1/2), and then back to the left (placing him at a distance of now 1/4 from the origin), he will next, by my assumption, jump to the right again (placing him 3/8 from the origin). If you continue to add and subtract by the formula given to us, (1/2)^N, you will eventually come to a value close to 1/3, and if you do this forever and type this alternating series into something like Desmos with N going to infinity, you will get an answer of 1/3 (or 0.3333 repeating).

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    $\begingroup$ That two outcomes each occur with 50% probability absolutely does not mean that the outcomes alternate in sequence. Flipping a coin and getting heads does not mean the next flip will be tails. $\endgroup$ Jul 10, 2023 at 14:00
  • $\begingroup$ Also, even if the grasshopper is specified to be jumping alternately, the answer would be exactly 1/3, not "~1/3" or approximately 1/3. $\endgroup$
    – justhalf
    Jul 11, 2023 at 2:48
  • $\begingroup$ Thank you for your answer but I intended the direction of the grasshopper jumps to be random. I also intended that each random jump direction to be independent of all the other jumps. Because of your answer I decided to edit my question to include the word "randomly". $\endgroup$ Jul 11, 2023 at 8:54

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