Two players are dealt five cards each from a standard well shuffled 52-card deck. a) What is the probability that both hands contain same pair? b) What is the probability that both hands contain same specific pair? Other three cards dont matter.

Im kind of stuck with this now. I know the basics of probability and combinations. But this one seems to be too hard for me now.

I used the formula

Pr(2 players have same pair in 5 cards) = P(A has 1 pair)*P(B has same pair)

+ P(A has 2 pair)*P(B has at least 1 of the 2 paIrs).

I got an answer of 0.48%. For a specific pair, I got 0.037%.

While the probability of a player getting at least one pair is close to 50%, the chance the other player gets both of the two remaining cards in the rank is quite low, therefore the very low probability.

I found this to be somewhat tricky. Can anyone confirm?

I'm getting the same results: a. 4/833 and b. 4/10829
Hopefully I haven't overcounted again

C(13,1)*C(4,2)*C(2,2)*(48! / 3!3!42!) / (52! / 5!5!42! ) for a

Eventually got the same results. A bit different method thou I think, but must be right. Thank you for your help.

Does that mean it counts if they both have the same trips or quads?

To a few decimal places, yes, but I don't think that's exactly the same result as statman's. What you did double-counts two-pair.

Here is what I did.

The probability that A has exactly one pair is P(A1)= 13*C(4,2)C(12.2)* 16/C(52,5). Given A has 1 pair, the probability B has that pair is Pr(B1|A1)= C(2,2)C(45,3)/C(47,5).

The probability A has 2 pair is P(A2) =C(13,2)*6^2*44/C(47,5). Given A has two pair, the probability B has at least one of the two is Pr(B2|A2) =[ 2*1*1*C(45,3)-43]/C(47,5).

On multiplying to get the unconditional probabilities and adding, I got 0.004787.

He didn't say it was holdem.

ETA: Or Omaha.

Lol, funny thing is I knew he wasn't talking about hold'em, but for a moment I forgot there's only 4 of each rank. My bad.

So yes, it's probably what statman said. If we take kapw's 4/833 and subtract P(share a pair with two-pair), we should get statman's 0.004787.

Gtg I'll check later.

When I plug all that in, my calculator says .00178

13*6*66*16*C(45,3) / C(52,5) / C(47,5)
+
78*36*44*(2*C(45,3)-43) / C(47,5)^2
=
.0017811263

Edit: I think I found one thing you mistyped: the 2nd one should also be divided by C(52,5)*C(47,5).

But that only makes the result smaller.

Also: 13*C(4,2)C(12.2)* 16/C(52,5) has to change to 13*6*C(12,3)*4^3

I think you guys are missing the case where the first hand dealt has two pairs in it. This will give the second hand better odds of being able to get a matching pair -- all these calculations are assuming they only have one shot.

Note: didn't actually read the posted answers too clearly this could have been address already.

Figured I'd try it myself, and I typed my calculations, but kapw already pointed out statman's other typos.

Using statman's approach, with the correction kapw showed, I got .0047872618 in agreement with statman.

But now I have to see why subtracting the two-pair cases from 4/833 doesn't yield that answer.

I gave up for now but I'm very curious to see what else is missing (when I subtract I'm still over statman's figure)

Kapw's typo finds are right on. I actually used what he showed; here, hopefully, is a correct summary of what I did:

The probability that A has exactly one pair is P(A1)= 13*C(4,2)C(12,3)* 64/C(52,5). Given A has 1 pair, the probability B has that pair is Pr(B1|A1)= C(2,2)C(45,3)/C(47,5).

The probability A has 2 pair is P(A2) = C(13,2)*6^2*44/C(52,5). Given A has two pair, the probability B has at least one of the two is Pr(B2|A2) =[2*1*1*C(45,3)-43]/C(47,5).

On multiplying to get the unconditional probabilities and adding, I got 0.004787.

The above does account for players getting two pairs.

In a way I'm a bit relieved to see the issues some are having because as i said, I found this to be somewhat tricky. If we could have Bruce back, he would have solved it in minutes and probably give two or three ways of doing it.

I think I see what's happening.
For my own clarity, I'm gonna convert this to my style, and I won't count order of players like you did.

13*3!!*C(48,6)*C(6,3) / C(52,10) / C(9,4)

The key is the C(6,3). There are that many ways to group the unimportant cards into two groups of 3. (Which, btw, counts order of groups only because the groups are distinguishable by the pair suits. I did not count order of groups when grouping the pair suits, which is why the expression for that was 3!! instead of C(4,2). )

But what if one player is to have two-pair? Then there are no longer C(6,3) valid ways to group those extra 6 cards. There are 2*4 = 8.

And if both players have two-pair, there are C(4,2)*2 = 12 ways to group the 6 cards.

So the two-pair cases aren't just counted twice. I think the cases where 1 player has it are counted 2*20/8 = 5 times, and by the same reasoning, the cases where both players have it are counted 2*20/12 = 10/3 times.

If the above is true, we have to subtract the first cases 4 times and the 2nd cases 7/3 times.

Let A:= P(just one player has 2p | they share a pair)
Let B:= P(both players)

What I'd like to see be true is: 4/833 - 4A - (7/3)B = .0047872618

Well I'm leaving now so I'll calculate A & B tomorrow.

I think it's easier to do it another way. Start by thinking about the ten cards in both hands combined. They must contain all four cards of a rank in order to be divided into two five card hands that share a pair.

There are C(52,10) = 15,820,024,220 ways to deal ten cards.

There are C(48,6) = 12,271,512 ways to deal all four Aces, plus six other cards. So the chance of having all four Aces among the 10 cards is 12,271,512 / 15,820,024,220 = 6 / 7,735.

Now pick five cards from among the 10. There are C(10,5) = 252 ways to do it. There are C(4,2)*C(6,3) = 120 of them that include two Aces in each hand.

So the chance of two five card hands each containing a pair of Aces is 120*6 / (252*7,735) = 4 / 10,829.

Given that happened, what's the chance that the hands also share a pair of Kings? This is the same problem, except we deal two three card hands from a 48 card deck. There are C(48,6) = 12,271,512 ways to deal the six cards, C(44,2) = 946 of them contain all four Kings. There are C(6,3) = 20 ways to split the six cards into two hands of three, C(4,2)*C(2,1) = 12 of them result in two Kings in each hand. 946*12 / (12,271,512*20) = 11,352 / 245,430,240 = 1 / 21,620.

The probability that two five card hands each contain exactly one pair of the same rank is 13 * (4 / 10,829) * (1 - 6 / 21,620) = 21,614 / 4,502,365.

The probability that two five card hands each contain two pairs of the same ranks is (4 / 10,829) * (12 + 11 + 10 + . . . + 1) / 21,620 = 6 / 4,502,365.

I also prefer this way (though I wouldn't call it easier). It's similar to statman's, but without the conditional probabilities. However, I think somewhere you have overlapping combos, because after adding your two results at the end, your answer is 4/833.

I don't know why I defined A & B as conditional probabilities. Those should be intersections.

Furthermore, the "A" cases aren't double-counted to begin with. (Plus even if they were, there would be a better way to go about subtracting them. The plan I outlined before was going to avoid overlap in the subtractions, but better is to allow overlap and alternate between subtracting and adding.)

All that needs to be subtracted from 4/833 is a factor of P(both players have the same 2pair).

= 78*3!!*C(4,2)*44*43 / C(52,10) / C(9,4) = 6 / 4502365 like Aaron said

or you can say: 5^2 *C(13,2)*3!!*C(4,2) / C(52,8) / C(7,3)
just like you could have got 4/833 by saying: C(5,2)^2 * 13 / C(52,4)

It looks to me like 4/833 - 11*(6/4502365 ) = statman's answer

Why that's an 11, I don't know yet.

Lol the magic 11, I spent quite a lot of time to figure out how it works out. I think you don't need it.
In statman's calcs you need to add Pr(B has a common pair | A has full house)*Pr(A has FH) =

13*C(4,3)*12*C(4,2)*1*C(45,3)/(C(52,5)*C(47,5))
Although it looks like a very "rare" event it changes the results
Adding everything else in statman's calculation gives : 21614 / 4502365

which is the same as 4/833 - 6/4502365

Looks like Kapw has resolved the issue.

Yes! You got it kapw, case closed. Neither of statman's probabilities include the full house. And it makes much more sense for the subtraction factor to be a 1 than an 11, because the two-pair was double-counted afaict not counted 12 times lol.

Another way to get the FH probability: C(5,2)*13*12*3!!*4*2! / C(52,7) / C(7,2)

I wouldn't have noticed that the FH probability is exactly 10x the two-pair probability.

I built up from the 91 atomic possibilities rather than trying to parse the combination possibilities.

If two hands share a pair, it can be either one of the 13 ranks, or one of the 78 combinations of ranks.

It's easy to compute the probability that two hands share a pair of A's, 4 / 10,829.

It's easy to compute the probability that two hands share a pair of A's and a pair of K's, 4 / (10,829 * 21,620).

To compute the probability that two hands share a pair, sum:

4 / 10,829 for the hands with AA as the only or higher shared pair
4 / 10,829 * (1 - 1 / 21,620) for the hands with KK as the only or higher shared pair (subtracting the AA/KK already counted)
4 / 10,829 * (1 - 2 / 21,620) for the hands with QQ as the only or higher pair

and so on. You get:

4 / 10,829 * (13 - 78 / 21,620) = kapw7's 4/833 - 6/4,502,365

This problem illustrates how easy it is to either double count or leave out possibilities as shown by my initial posting.

If possible, in doing these types of problems, it is a good idea to do it more than one way. Of course when you do that you may overcount one way and undercount the other. If you can't resolve and the results are very close, take the average .