Patrick, Paul, Peter and Phillip are drawing toothpicks. One of the boys is holding four toothpicks in his hand without revealing which one is the shortest. One of the other three boys draws a toothpick without revealing it. Then the next one draws a toothpick among the remaining three, again without revealing his pick, then the third one draws the toothpick among the remaining two, and finally the boy holding the toothpicks keeps the remaining one in his hand for himself. After each boy has a toothpick, their picks are simultaneously revealed. The one who has the shortest toothpick is “selected” for an action.
Peter wants to be “selected” the most. What should be Peter’s optimal strategy in order his odds to draw the shortest toothpick are the highest possible? I.e., should he draw as the first one, the second, or the third, or it’s better for him to hold the toothpicks at all?
Peter’s optimal strategy is to hold the toothpicks. It gives him the highest probability to draw the short toothpick.
Suppose Patrick is drawing as the first one. In this case the probability for him to draw the short toothpick is, obviously, 1/4.
Suppose Paul is drawing as the second one. After Patrick’s pick there are only two combinations of toothpicks which can be left for Paul. These are:
a) three long toothpicks;
b) two long toothpicks and one short toothpick.
Since Patrick has been drawing from among four toothpicks, the probability for combination a is 1/4 and probability for combination b is 3/4 (1-1/4). In case combination a is true there is a zero probability for Paul to draw the short toothpick (because it has been drawn by Patrick already). In case combination b is true there is a 1/3 probability for Paul to draw the short toothpick. Based on this, the overall probability Paul draws the short toothpick is a combined probability of both combinations:
1/4 x 0 + 3/4 x 1/3 = 1/4 (Paul).
Suppose Phillip is drawing as the third one. After Paul’s pick there are only two combinations of toothpicks which can be left for Phillip. These are:
c) two long toothpicks;
d) one long toothpick and one short toothpick.
Since Paul has been drawing from among three toothpicks, the probability for combination c is 1/3 and probability for combination d is 2/3 (1 – 1/3). In case combination c is true there is a zero probability for Phillip to draw the short toothpick (because it has been drawn by Paul already). In case combination d is true there is a 1/2 probability for Phillip to draw the short toothpick. Based on this, the overall probability Phillip draws the short toothpick is a combined probability of both combinations:
1/3 x 0 + 2/3 x 1/2 = 1/3 (Phillip).
Peter is the one who is holding the toothpicks. After Phillip’s pick there are only two combinations of toothpicks which can be left in Peter’s hand:
e) a long toothpick;
f) the short toothpick.
Since Phillip has been drawing from among two toothpicks, the probability for combination e is 1/2 and probability for combination f is also 1/2 (1 – 1/3). In case combination e is true there is a zero probability for Peter to get the short toothpick. In case combination f is true the probability Peter gets the short toothpick equals 1. Based on this, the overall probability Peter draws the short toothpick is a combined probability of both combinations:
1/2 x 0 + 1/2 x 1 = 1/2 (Peter).