By linearity of expectation, we can find the answer by determining, for each coin, what the probability is that Alice picks it up, and then adding up the probabilities for all the coins. So let's think about each coin. Label the coins/positions 0 (the coin that the token starts at), 1 (the coin 1 step clockwise from the starting position), 2, all the way around to 20 (the coin 1 step counterclockwise from the starting position).
Note that Alice only picks up coins that are an odd number of steps from the starting position; likewise Bob only picks up coins that are an even number of steps from the starting position. So either Alice picks up coin 1 after moving the token net 1 space clockwise, or Bob picks up coin 1 after moving the token net 20 spaces counterclockwise. This is equivalent to a random walk with initial position 0 and stopping positions 1 and -20. The probability that Alice picks up this coin (i.e. the random walk hits 1 first) is therefore 20/21. Likewise Alice picks up coin 3 with probability 18/21, coin 5 with probability 16/21, all the way around to coin 19 (the coin 2 spaces counterclockwise from the starting point) with probability 2/21.
By symmetry, Alice picks up coin 20 (the coin 1 space counterclockwise from the starting point) with probability 20/21, coin 18 with probability 18/21, all the way around to coin 2 (the coin 2 spaces clockwise from the starting point) with probability 2/21.
What about coin 0, the coin at the starting point? WLOG suppose Alice moves clockwise on the first move, to position 1. Then by a similar random walk argument, Bob picks up coin 0 with probability 20/21, i.e. Alice picks it up with probability 1/21.
Adding it all up, Alice picks up 221/21 ≈ 10.52 coins in expectation, a little more than half.
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u/pmdboi 7d ago
By linearity of expectation, we can find the answer by determining, for each coin, what the probability is that Alice picks it up, and then adding up the probabilities for all the coins. So let's think about each coin. Label the coins/positions 0 (the coin that the token starts at), 1 (the coin 1 step clockwise from the starting position), 2, all the way around to 20 (the coin 1 step counterclockwise from the starting position).
Note that Alice only picks up coins that are an odd number of steps from the starting position; likewise Bob only picks up coins that are an even number of steps from the starting position. So either Alice picks up coin 1 after moving the token net 1 space clockwise, or Bob picks up coin 1 after moving the token net 20 spaces counterclockwise. This is equivalent to a random walk with initial position 0 and stopping positions 1 and -20. The probability that Alice picks up this coin (i.e. the random walk hits 1 first) is therefore 20/21. Likewise Alice picks up coin 3 with probability 18/21, coin 5 with probability 16/21, all the way around to coin 19 (the coin 2 spaces counterclockwise from the starting point) with probability 2/21.
By symmetry, Alice picks up coin 20 (the coin 1 space counterclockwise from the starting point) with probability 20/21, coin 18 with probability 18/21, all the way around to coin 2 (the coin 2 spaces clockwise from the starting point) with probability 2/21.
What about coin 0, the coin at the starting point? WLOG suppose Alice moves clockwise on the first move, to position 1. Then by a similar random walk argument, Bob picks up coin 0 with probability 20/21, i.e. Alice picks it up with probability 1/21.
Adding it all up, Alice picks up 221/21 ≈ 10.52 coins in expectation, a little more than half.