Re: A card game probability

matt271829-news@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).
[...]

I think your estimate of 1 - (12/13)^52 gives the exact probability
for a variant problem, where, after each card is picked and its
numerical value checked, the card is replaced in the deck and then
the deck is reshuffled.

We do this 52 times, comparing with "ace, 2, 3, 4, ... King, ace, 2 ..
.... Queen , King". There's a 1/13 chance of value match for any of
the 52 trials, so Prob(no match 52 times consecutively) = (12/13)^52
and so Prob( 1 or more matches) = 1 - (12/13)^52 ~= 0.9844...
That's correct for the variant problem, with cards being put
back in the deck, and 52 trials.

I wrote a program that uses pseudo-random numbers to get an estimate
of the probability of at least one match in the original, more
complicated problem, where the cards aren't replaced and we simply
check the face values of the 52 cards in turn.

Now, the program has tested about 2500 million decks.
The computer-estimated probability of 1 or more matches is 0.9838 .
So the approximation of 1- (12/13)^52 = 0.9844.. is probably
pretty close.

Since there are 4 suits, we could ask actually 4 problems,
one for a pack with spades only, one with spades and hearts,
one with spades, hearts and diamonds, and one with all:
13, 26, 39 and 52 cards resp.

The case of 1 suit: -> derangement problem
"" "" 2 suits: ?
"" "" 3 suits: ?
"" "" 4 suits: about 0.9838

David Bernier
.

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