Another one from Cut the Knot : Probability Riddles by Alexander Bogomolny. This is riddle 3.9, my paraphrase.
Riddle
A standard pack of 52 playing cards contains as we know four aces. If the pack is thoroughly shuffled and we then work our way through it, the first ace we encounter may be at any of the positions . Which of these has the highest probability of being the position where the first ace is found?
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My Slightly Simplified Version of the Solution
Let be the probability that the first ace is at position . We count the number of possible ways the aces may be arranged if the first ace is at position .
If the first ace is at position , the remaining must be each be located at one of the positions from to , which can be done in ways, all equally likely to occur, given the pack is in random order.
Now we calculate the total number of ways the fours aces can be arranged altogether, irrespective of where the first ace is. This is .
The probability is therefore . This plainly decreases with and so its maximum is when , and (as we would naturally expect).