Poker probability for two of a kind

[cabernet]

What is the probability that at least two of a kind will be dealt in a hand of 5 cards using a standard deck of 52 cards? I tried to work this problem by using the idea of complement and first figuring out the probability that no cards would match. 1 - (52 *48*44*42*36)/ (52*51*50*49*48) which results in a probability of about 49%. I would like to have confirmation that this answer is correct or another way of solving the problem.---a puzzled math teacher

 

[mattmns]

http://mathworld.wolfram.com/Poker.html

Says that the actual probability of 2 of a kind is 42.3%

Keep in mind that 1 - P(no cards match) is not the probability of getting a 2 of a kind. Instead it is the probability of getting a 2 of a kind, a 3 of a kind, a 4 of a kind, 2 pair, or a full house. And if you add the counts (provided in the link above) of each of these together and divide by the total number of hands, you will get .4929171669 which is exactly your number above.

So to get P(2 of a kind) you would need to subtract the probability of a 3 of a kind, 4 of a kind, 2 pair, and full house from your current number.

 

[Physics_wiz]

Anyone know how to find this answer using combinations?

I thought you find the total number of ways to pick 5 from 52 (2598960) and then find the number of ways to pick a 5 cards with no repetition or 13 choose 5 (1287)

Then do (2598960-1287)/2598960 but this is not the right answer. Anyone know what I did wrong?

 

[mattmns]

There are 4 suits, so you would need 4 times 13 choose 5, but then you have all the mixtures of each one of these, which makes it more complicated than what you have. For example, someone can have a 2, 3, 4, 5, 6 in 2 (or 3 or 4) different suits, and that is not included in what you had.

In the link I posted above they show the combination.

 

[robert Ihnot]

Maybe the surest way to do this is to look at the possibility of no pair in the hand. This means we take one card from 5 sets of 4 cards and we do this in 13 ways:

(4^5*13C5)/(52C5)=50.7%. So then at least one pair would be 49.3%.

 

[HallsofIvy]

Of course, none of that includes straight or flush.

 

[robert Ihnot]

Halls of Ivy: Of course, none of that includes straight or flush.

That is no problem since if you have a pair, you never hold a straight or a flush!

Cabernat: What is the probability that at least two of a kind will be dealt?

However, you could interpreting the original problem to include the possibility of holding a straight or flush. For a flush we have 4 suits and 5 choices out of 13: 4*(13C5)/52C5=.00198. For a straight, we have four choices for each of five cards and 40 such cases, taking the ace as both the lowest and highest card: 40x4^5/52C5 = .0158 Then there is the small overlap cosisting of straight flushes and royal ones, there being 40 such cases: 40/52C5 = .0000154. This adds about .0178=1.78% to the previous figure of 49.3% giving about 51.08%.