[Buddha]

I was wondering what the odds are for someone getting the exact same 2 cards in back to back hands. I was dealing the other day at a game, and a guy had two black nines in back to back hands from the same deck. I know the odds have got to be pretty bad for that to happen, and not sure of the math to figure it out. Anyone able to help?

[miandsh2000]

some discussion had started here - down around #18

[_GUN_]

Originally Posted by OutOfCrown

Ever gotten pocket Aces back to back? Odds of that are about 48000:1 against (slightly worse than that if you want the exact number).

http://www.chiptalk.net/forum/online...ight=math+mind

[Dazzler]

There are 52*51 combinations of starting hands = 2,652 hands - divided by 2 as 6c7h is the same as 7h6c = 1,326 unique hands

1326 * 1326 = 1,758,276:1

[bjjensen]

Originally Posted by Buddha

I was wondering what the odds are for someone getting the exact same 2 cards in back to back hands. I was dealing the other day at a game, and a guy had two black nines in back to back hands from the same deck. I know the odds have got to be pretty bad for that to happen, and not sure of the math to figure it out. Anyone able to help?

The odds are approximately 1 in a million .

Actually, they're much better than that-- a little less than 1/1000. By my calculations (which may be inaccurate, since it's a weekend ), that will happen about 7 in every 10,000 hands. Here's my math:

The first hand doesn't matter since you're not trying to get two specific cards on the first hand-- just to duplicate it on the second hand. After the first hand, you now have determined what you need to duplicate (in this case, the two black 9's). The probability of getting one of the two cards you need dealt to you is 2/52. If that is accomplished, you only need one more card, and so the probability of getting it is 1/51. Multiply those probabilities together, and you get 2/52*1/51= 0.000754, or 7.54/10,000ths.

You can also arrive at this via the mathematics of combinations: 52!/2!(50!) yields the same result.

Of course, if you wanted to be dealt the same two cards in sequence, say the 9 of spades then the 9 of clubs (again, provided you don't care what the first two cards are, just that you are re-dealt the same two in sequence), then your probabilities drop to 0.000377.

Want two specific cards two times in a row? That's where you're about 1 in 2 million. Want two specific cards dealt in the same sequence two times in a row? You're about 1 in 10 million.

I hope this was helpful. Class dismissed.

bjjensen

[Dazzler]

Yes bjjensen's right. If you aren't fussed about which hand is replicated then the odds are 1326:1

If you stipulate the hand before the first is dealt then the odds are 1,758,276:1

[bjjensen]

Originally Posted by Dazzler

There are 52*51 combinations of starting hands = 2,652 hands - divided by 2 as 6c7h is the same as 7h6c = 1,326 unique hands

1326 * 1326 = 1,758,276:1

Well, in this case it's only 1326:1, since he didn't care what the first hand was, only that it was repeated on the second hand. If he had wanted two black 9's on two hands, then the math is accurate.

Also, this is presuming a completely randomized deck on the second deal, which (with theoretically perfect shuffles each hand) only happens after 7 shuffles of the deck. I've found that most people lose patience after 2-3 shuffles.

bjjensen

[bjjensen]

Originally Posted by Dazzler

Yes bjjensen's right. If you aren't fussed about which hand is replicated then the odds are 1326:1

If you stipulate the hand before the first is dealt then the odds are 1,758,276:1

Of course, you said more clearly in a sentence what it took me half a page to describe. Maybe that's why everyone tells me to shut up at the poker table .

bjjensen

[abby99]

This has happened to me just once, at a $3-6 LHE table at the Mirage. Kh2h, back to back, from two different decks. But here's an interesting twist: I won both hands with a K-high flush (hearts, of course). Somebody want to tackle the odds on that? (This really did happen a couple of years ago.)

[OutOfCrown]

Originally Posted by bjjensen

Well, in this case it's only 1326:1, since he didn't care what the first hand was, only that it was repeated on the second hand. If he had wanted two black 9's on two hands, then the math is accurate.

Correct, for one player. Since there are multiple players at the table, and presumably we were going to be wondering about this event if any of them got the exact same hand twice in a row, the chance of this happening somewhere at the table is far better than 1326:1 (someone else can do that math)