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Patrick's optimal strategy is to draw either as the first or as the second one. Peter's optimal strategy is to hold the toothpicks.
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).
As it can be seen it doesn't matter too much for Patrick whether to be the first or to be the second. Paradoxically, the probability for him to get the short toothpick is the same in both cases - 1/4. In contrast probability for Peter to get the short toothpick is the highest in case he is the one holding the toothpicks in his hand.
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