Re: A card game probability

On Jan 26, 4:15 pm, matt271829-n...@xxxxxxxxxxx wrote:
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).

OK, lets formulate the problem as I understood it:

You draw cards from a 52 cards deck until you draw two cards of a same
value. We consider that each card has a positive integer value in the
range 1,...,13.

Obviously, you cannot do it on your first draw (''trial"), hence p_0=0.

The probability of drawing such a card in your second trial is p_1=3
times 1 / (52-1).

Now, probability that you don't do it in your second and do it in the
third trial (i.e. you turn a card whose value is equal to one of the
two cards already flipped) is

p_2 = 3 times 2/ (52-2)(1 - 3 times 1/ (52-1)),

and so forth, with p_n being the probability for drawing the first such
card in the n-th trial.

The sum p_1+p_2+...+p_n gives the probability to draw at least one such
card in n trials.

Now, when I read your message, and again the original question, it
seems to me too that what the poster meant was to find the probability
of drawing at least once a card with a value (n-1) mod 13 + 1, where n
is the number of the trial. It is, of course, a completely different
problem.

.

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