Re: A card game probability



On Jan 26, 11:42 am, "Faton Berisha" <fberi...@xxxxxxxxxx> wrote:
On Jan 25, 9:25 pm, "jsepp...@xxxxxxxxx" <jsepp...@xxxxxxxxx> wrote:

Take a card deck with 52 cards. Pick cards one by one and compute the
cards by ace, two, three,..., jack, queen, king, ace,..., king, ace,
..., king, ace, ..., king. What is the probability that at least once
you turn a card of the same value as you say aloud?

The probability that you turn the first card of the same value as one
of the n cards turned before exactly in your (n+1)-st trial is

p_n = 3n/(52-n) prod_{j=1}^{n-1}(1-3j/(52-j)).

Hence, the probability that you turn such a card in n trials is

sum_{j=1}^{n-1} p_j.

Can you explain what this has to do with the problem posed? What is the
value of n supposed to be? What answer do you get from this formula? I
would expect the correct answer to be approximately 1 - (12/13)^52.
This would be the answer if all the probabilities were independent
(which of course they aren't).

.