# Probability of drawing a match between two decks

• August 21st 2013, 07:20 AM
Probability of drawing a match between two decks
I'm now flipping through some old exam questions and found this:

Q. Design an experiment of having two people drawing N times from two identical decks consist of N cards. Define the event "match" if both people draw the same card during the same draw, for example, both people drawing the card "King of Heart" during the 7th draw would be a match. What is the probability of having a match after all N draws without replacement?

Solution:

Define A to be the event of a match.

Definte $A_k$ to be event of a match during the k-th draw; note that the events of $A_k's$ are not independent.

Then $A = \bigcup _{k=1}^N A_k$

We then have $P(A)=P( \bigcup _{k=1}^N A_k )=$ $\sum ^N_{k-1}P(A_k)- \sum _{i

Now it says that $P(A_k) = \frac { \binom {N}{1} (N-1)! (1)(N-1)! }{N!N!} = \frac {1}{N}$

$P(A_i \cap A_j) = \frac { \binom {N}{2} (N-2)! (1)(1)(N-2)! }{N!N!} = \frac {(N-2)!}{N!}$

$P(A_i \cap A_j \cap A_k) = \frac { \binom {N}{3} (N-3)! (1)(1)(1)(N-3)! }{N!N!} = \frac {(N-3)!}{N!}$

How does one formulate $P(A_k), P(A_i \cap A_j), P(A_i \cap A_j \cap A_k)$? I don't really understand why we get all the factorals up there...

Thank you!
• August 21st 2013, 01:23 PM
Soroban
Re: Probability of drawing a match between two decks

What level is this course work?
The solution is quite intricate.
(I can't follow the given solution.)

Quote:

Design an experiment of having two people drawing cards from two identical decks of $N$ cards each.
Define the event "match" if both people draw the same card during the same draw.
For example, both people drawing the $K\heartsuit$ during the 7th draw would be a match.
What is the probability of having a match after all N draws without replacement?

The first player's deck (deck $A$) can be any order.

Suppose the second player's deck (deck $B$) has no matches with deck $A$.
Then deck $B$ is a derangement of deck $A$,
. . a permutation in which no cards match.

For $N = 52$, the probability of no matches is approximately $\frac{1}{e}$

Therefore:- $P(\text{at least one match}) \;\approx\;1-\frac{1}{e} \: =\:0.632120559...$

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