Determining the odds of certain cards being dealt

Factual Questions
1

If I deal three cards face up, how do I determine the odds that exactly two out of three of them are predetermined cards?

On a poker message board I belong to, someone asked what the chances of this scenario happening is (game is Texas Hold 'em):

Each player is dealt two cards, they might be:

player 1 : Q, 10
player 2 : Q, 10
player 3 : Q, 10

Three community cards are dealt, they are:

Q, !0, 5

What the poster wants to know is, what are the odds of everyone being dealt the same cards that end up making the highest possible two pair once the community cards are dealt? This is the way I’m going about it (I’m horrible at math):

The first player’s first card can be anything greater than two, so the odds of this happening are 48/52. The first players second card must be higher than two and any other card than the one he was just dealt, so his odds are 44/51.

The second player’s first card, must be the same as one of the first player’s cards, so the odds are 6/50. The odds that the next card are the same as the first player’s other card is 3/49.

The third player’s first card, 4/48, the second, 2/47.

This is where I’m stuck. Another problem is once the next part is figured out, how do you determine the odds that the one community card that doesn’t pair up with the players cards, is of lower value, since we don’t necessarily know what exact card the players were dealt (they don’t have to be Q, 10)?

2

If you’re familiar with binomial coefficients, it’s a pretty straighforward formula. I’ll write [sub]n[/sub]C[sub]k[/sub] as (n, k).

The probability of that scenario is (4, 1)[sup]6[/sup]/(49, 2)[sup]3[/sup], or 4[sup]6[/sup]/1176[sup]3[/sup].

3

Did you miss the part where I said I’m horrible at math? I’m not familiar with bipedal coiffures, or whatever you said. It doesn’t seem to me to be that straight forward though. Does your formula take in to account that the odd community card is of lower value than the other two?

4

Nah, I missed the requirement that the third card be lower. That’s a significantly more complicated problem, and requires computation tools that I don’t have here.

5

Okay, I think I figured it out. I just pretended that the community cards were dealt first. It shouldn’t matter, should it?

The first community can be anything. The second must be any other (48/51). The third different than the others (44/50).

Player ones first card must be one of the two highest cards (6/50). His second, the higher of the next two (3/49).

Player two’s first card must be the same as one of player two’s cards (4/48). Player two’s second card must be the same as player one’s other card (2/47)

Player three’s first card must be the same as one of player ones’s cards (2/46). Player three’s second card must be the same player one’s other card (1/45).

If I multiply the fractions, I should have the correct answer, right?

6

It looks right.

7

:smiley: I loved this line. Laughed right out loud, I did.

8

Don’t take this bet in Vegas. If I calculated correctly, the odds of this happening are 47,965,542 to 1.

9

How do I figure out this problem:

If you are dealt three cards, what are the chances that one is an ace and another a king?

10

Does the hand specifically have to be be AKx or if the hand came AAK or AKK would that also qualify?

I mean, not that I know the answer or how to figure it out, but it will have a bearing on the solution. I think.

11

Yeah, AAK and AKK also qualify.

12

Three is a small enough number of cards that we can look at that one case by case.

There are three possibilities for the first card: It’s an ace (1/13), it’s a king (1/13), or it’s something else (11/13).

First case first: If it’s an ace, then the second card is either a king (4/51) or something else (47/51). If it’s something else, then the third card is a king (4/50) or something else. So if the first card is an ace, then there’s a (4/51) + (47/51)*(4/50) chance of getting what you want, or 0.152157.

Likewise, if the first card is a king, there’s also a 0.152157 chance of getting what you want.

For the third case, you need the second and third card to be an ace and a king. So the second card must be one of them (8/51), and the second must be the other (4/50). So if your third card is something else, you only have a 0.012549 chance of getting what you want.

Putting this all together, we have a 1/13 chance of case 1, in which case the chance is 0.152157. We have a 1/13 chance of case 2, in which case the chance is 0.152157. And we have a 11/13 chance of case 3, in which case the chance is 0.012549. So our total chance is (1/13)*0.152157 + (1/13)*0.152157 + (11/13)*0.012549, or a total chance of 0.03403, or about three and a half percent.

13

That’s easy–it’s (13, 1)[sup]2[/sup]/(52, 3), or 169/1082900.

:smiley:

14

That’s only .0156% of the time ultrafilter. Chronos’ answer looks more righterer.

15

Chronos’s analysis assumes that (in the first case for example) that the third card has to be a non ace or king. In other words his analysis is for the probability of getting an A-K-x where x cannot be another A or K.

Going on the basis of how the problem was described where the x can be either another Ace or King, then the probability would be:

84(52+51+50) = 0.036923
525150

16

Does the fact that you would burn a card have any effect on the odds?

17

No, burning doesn’t affect the odds at all.

18

That wouldn’t change the odds from 3.7% to 0.156%, would it?

19

On a related noted here’s a good rule of thumb for holdem I just learned:

Count your post flop outs, multiply them by 2, and add 1… this is roughly the chance of hitting an out.

For example:

You need a queen or a 10 to come on the turn or the river and you have neither in your hand. Therefore, you have 8 outs (four queens, and four 10’s), 8 times 2 plus 1 equals a 17% chance of seeing a queen or a ten.

20

No, I don’t assume that. The only case that could come up would be in the first subcase of the first case I examine: First card is an ace or a king, second card is the other one of those two. At that point in my analysis, though, I didn’t calculate at all based on what the third card might be. It might be another ace, but I don’t care: Once my first two cards are AK, I know that I’ve gotten what I want with a 100% chance. If I had wanted to exclude getting another high card, I would have had to multiply case 1 (and case 2) by an additional factor of (44/50).