Back-to-back Quads?

[Joe Taylor]

Had an interesting situation occur in a recent tournament I was in, wondering what the odds were of it happening.

Got dealt pocket 7's, flopped a set, turned quads.
Very next hand: Get dealt pocket 7's, flop quads.

From a pocket pair, what are the chances of having quads by the river? And then of course, what are the chances of hitting back-to-back identical quads two hands in a row?

Weird one, I couldn't work out the math, figured somebody here could help me.

[jason_t]

If you have a pocket pair the probability that you have quads by the river is

(2 choose 2) * (48 choose 3) / (50 choose 5) = .0082.

[DavidC]

Here's the funky part...

The chances of it happening twice are exactly the same as the chances of it happening once, if you count directly after the first event occurs. [img]/images/graemlins/smile.gif[/img]

If, before the first event happens, you predict that the event will occur twice, THEN you get a huge number.

--Dave.

(I asked a similar question long long ago, but it was about getting like 8 [img]/images/graemlins/heart.gif[/img]q [img]/images/graemlins/spade.gif[/img] twice in a row or something. Same question, though.)

[Joe Taylor]

Very true, very true. The first hand completing quads is the catalyst event, so the odds at that point are all revolving around the chance of hitting that *specific* quad hand. Still funky to think about it that way, though.

Thank you both for your help. My calculations were showing it at 1 in 122.5, but I didn't trust my rusty statistical skills. Turns out I had the right idea. [img]/images/graemlins/smile.gif[/img]

So, the probability of two quads in a row given P(quads | pocket pair) = .0082 is .0082^2 = .00006724 = approx. 14871 : 1 against, assuming we take it before either set of quads come.

As discussed above, the Lack of Memory Property states that if we've already gotten one set of quads, the probability of another on the next hand is still .0082.

[Joe Taylor]

The interesting "twist" on this one was that the quads were identical (7777) to the previous hand. But unless I'm missing it completely, that would put the probability at 193323:1 against it happening. (Since it isn't just quads, but a *specific* quad hand that needs to occur.)