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A practical question. Given two decks of cards (52 cards) if i pick 15 cards randomly from one deck and then 5 cards from another deck, what is the probability that exactly 3 cards, between two picked sets, will be same ? What is the probability that at least 3 cards will be the same?

Thank you so much !!

Apendix: added on 31.12.13

let me see if i got this right. if i modify the above example so that i am returning my picked cards back to the deck (sampling with replacement) then once i pick 15 cards i count the number of different cards which gives me the probability for picking a matched card from the second deck. thherefore i have :

$p(pick)=15/52 $

$q(not\_pick) = 1-15/52$

and then :

my probability for picking a matching pair is $\binom {5} {2}p(pick)q(not\_pick) \times \binom {5} {2} p(pick)q(not\_pick) \times \binom {5} {2} p(pick)q(not\_pick)$

or did i messed up again??

and thank you so much for all your help :)

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  • $\begingroup$ Hint : hypergeometric distribution $\endgroup$ – Peter Dec 30 '13 at 21:38
  • $\begingroup$ When you say the same do you mean same suit or same card? $\endgroup$ – okarin Dec 30 '13 at 22:22
  • $\begingroup$ If it is same card, there are only two decks so... $\endgroup$ – okarin Dec 30 '13 at 22:22
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We interpret "the same" as meaning that among our $20$ cards, there are exactly $3$ identical pairs, as in $\spadesuit$ Q, $\spadesuit$ Q, $\diamondsuit$ $4$, $\diamondsuit$ $4$, and $\heartsuit$ $7$, $\heartsuit$ $7$.

Let our sample space consist of all ordered pairs $(A,B)$, where $A$ is a hand of $15$ from the first deck, and $B$ is a hand of $5$ from the second deck. The sample space has $\binom{52}{15}\binom{52}{5}$ equally likely elements.

Call an element of the sample space good if there are exactly $3$ matching cards in the two hands. We count the number of good elements.

Which cards match? They can be chosen in $\binom{52}{3}$ ways. For each such way, the hand from the first deck can be completed to $15$ in $\binom{49}{12}$ ways. And for each such completion, the hand from the second deck can be completed to $5$ in $\binom{37}{2}$ ways.

So our sample space has $\binom{52}{3}\binom{49}{12}\binom{37}{2}$ good elements. For the probability, divide.

For "at least $3$ matches," calculate in a similar way the probability of $4$ matches, and of $5$ matches, and add.

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$\textbf{Hint:}$ There are two decks. You are trying to find three cards that are exactly the same.

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