Patti (whipartist) wrote,
Patti
whipartist

My answer to the probability puzzle



This reduces to a five-card problem-- four aces and the two of clubs.

Using our five-card deck, 20% of the time, the first card in the deck will be the two of clubs. In this case, we really care about the third card off the deck, because the second card will be the first ace. Obviously, the third card in the deck will never be the 2c when the first card is. 25% of this time, the third card will be the ace of spades. .20 * .25 = .05.

Running total: 5% ace of spades.

20% of the time, the first card in the deck will be the ace of spades. In that case, 25% of the time the next card will be the 2c, and the second card will never be the As when the first one is. Again, .2 * .25 = .05.

Running total: 5% ace of spades, 5% 2 of clubs.

The remaining 60% of the time, the first card in the deck will be one of the other aces. When that happens, the 2c will be the next card 25% of the time, and the As will be the next card 25%
of the time. .60 * .25 = .15 for the As, and .15 for the 2c.

Running total: 20% ace of spades, 20% 2 of clubs. That leaves 60% of the time for the next card to be neither of those two.

These numbers are for the five-card case. When you add the other 47 cards to the deck the absolute probabilities change, but the relative probabilities don't. In fact, the absolute probability of the next card being (any card) is 1/N, where N is the number of cards in the deck. In a 52-card deck, the next card after the first ace will be the ace of spades 1 time in 52, and it will be the two of clubs 1 time in 52.
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