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I was wondering what could be the probability of getting at least a pair of cards , when every time you draw 6 cards at random from a fresh deck of cards.

I calculated it as: 3/51 * 48/50 * 44/49 * 40/48 * 36/47 * 15(6C2) = 0.48

Let me know if it is correct?

2 Answers 2

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What's the probability of not getting a pair of cards after drawing n cards?

  • n = 1: p = 1
  • n = 2: p = 1 * (48/51)
  • n = 3: p = 1 * (48/51) * (44/50)

etc...

You want 1 - p.


You could also verify that the result is roughly correct by running many simulations and using this to estimate the probability:

import random  got_pair = 0 n = 6 trials = 10000  deck = [value for suit in 'CDHS' for value in 'A23456789TJQK'] for i in range(trials):     random.shuffle(deck)     if len(set(deck[:n])) < n:         got_pair += 1  print(float(got_pair) / trials) 

Result

0.6542 

Note that the result is only an approximation, but it can be a useful aid to check that you didn't make an error with the mathematics.

  • 0
    I guess that edit with the program code is somewhat less relevant now that the question has been migrated....2012-07-25
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We are counting the probability of at least one pair, meaning that we are allowing "three of a kind" or "two pairs."

How many no pair hands are there? We must choose $6$ different denominations from the $13$ denominations available. There are $\binom{13}{6}$ ways to do this.

For every choice of denominations, there are $4^6$ ways to choose the actual cards.

So the number of no pair hands is $\binom{13}{6}4^{6}$.

To find the probability of a no pair hand, divide by $\binom{52}{6}$. So the probability of getting at least one pair is $1-\frac{\binom{13}{6}4^6}{\binom{52}{6}}.$

Remark: To count the number of hands that have exactly one pair, do this. The denomination can be chosen in $\binom{13}{1}$ ways. For each choice, the actual card can be chosen in $\binom{4}{2}$ ways. Now choose the denominations we will have singles in. This can be done in $\binom{12}{4}$ ways. And now the actual singletons can be chosen in $4^4$ ways, for a total of $\binom{13}{1}\binom{4}{2}\binom{12}{4}4^4.$ For the probability we have precisely one pair, divide by $\binom{52}{6}$.