A game consists of 5 cups turned upside down, under one of the cups is a prize. 5 friend's pick a cup in turn and lifts it up, if they get the prize, they win , but if not, the cup is removed and the next friend picks. What position is it best to pick?
The first friend to choose has 5 cups to pick from and only 1 that will win them the prize, so their probability of winning is 1/5.The 2nd friend relies on the first friend to not win, this with a probability of 4/5, and also for them to pick the winning cup, a probability of 1/4. As both of these events have to occur we need to multiply the probabilities to get 4/5 x 1/4 = 1/5.Similarly for the 3rd player, they require both 2 previous players to 'lose' and them to win, with probability 4/5 x 3/4 x 2/3 = 1/5This argument applies for both other players.Therefore, all friends have probability 1/5 of winning the prize, and therefore it does not matter what position you pick a cup in.Extensions: What if there are n friends and n cups? What is there are 5cups, 5 friends but now 2 prizes?
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