Probability of Getting a pair of cards 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?
 A: 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.
A: 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 $\binom{52}{6}$.
