# How do you calculate the odds getting a single pair in Texas Hold 'Em?

Given that I am dealt one card, what are the odds that I will then make a pair either from the next card dealt to me or from the river of 5 cards played out?

I'm thinking something like: given I have one card already, I figure I have a 3/51 chance in getting its pair (ignoring cards being dealt to other players). But I come unstuck when trying to then figure out the next 5 cards in the river.

Would they be cumulative - so 3/51 + (3/50 + 3/49 + 3/48 + 3/47 + 3/46)?

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As ncmathsadist says, it is easier to calculate the probability of not getting a pair, then subtract from $1$. It depends upon whether you want to calculate the chance of pairing the first card, or the chance of getting a pair when dealt $7$ cards (your two hole cards plus the five of the board). To not pair the first card, the chance on the second is $\frac{48}{51}$ as you have to avoid $3$ cards of what is left. Assuming you missed on the second, the chance on the third is $\frac{47}{50}$, so the chance of pairing the first in two tries is $1-\frac{48\cdot 47}{51\cdot 50}$. If you are calculating the chance of any pair, missing on the second card is again $\frac{48}{51}$, but missing on the third is $\frac{44}{50}$ as there are now $6$ cards that can pair you. So getting any pair in three cards is $1-\frac{48\cdot 44}{51\cdot 50}$. The pattern should be clear enough.

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Hint. It is easier to compute the probability of failing to get a pair.

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How so? Please assume that I'm not an expert on probability. –  HorusKol Jul 1 '11 at 2:26
If you want to make sure you do not get a pair, throw out all the cards of the same value as your first card. Then select the next four cards from this last batch. Then the number of ways of getting a pair equals the total ways of selecting 5 cards minus the number of ways of not getting a pair. Can you take it from there? –  gary Jul 1 '11 at 2:47

You are given the card, say with value $1$ for definitness. Now, we use the fact that:

P(getting a pair)+P(not getting a pair)=1 , since the two are mutually-exclusive.

Then let's look at the two cases:

i)Not getting a pair with the second card:

Then the second card is not a $1$. So the second card can be chosen out of the 48 cards that are not $1$'s in 48 ways. But there is a total of 51 ways of choosing the second card.

ii)Not getting a pair in the 5 cards you are given:

Then you can choose the 4 remaining cards out of the 48 cards that are different from 1. So choose any 4 out of 48. The total number of choices you can make out of 4 cards is just the number of choices of 4 cards out of 51.

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