Set over Set

[mblax10]

I've seen flopped set over set many times recently. I realize this is just an abnormailty, but am curious if this is as rare as some people make it out to be.

My probability skill are lacking, could somebody kindly tell me the odds of two players flopping a set in Hold Em?

Also if it's not too hard could you put it in relative terms. (i.e. Set vs. Set is 1000 times less likley than being dealt pocket Aces or 100 times less likley than hitting a runner runner flush.)

Please and thank you to all responders

[Cobra]

This is a web site that has many probability questions answered. Look for set over set.

http://www.math.sfu.ca/~alspach/mag86/

Cobra

[Reef]

1 / 7.5 * 1 / 7.5 assuming you both hold pocket pairs preflop

[Falker11]

I think you need to rethink your math. The probability is not simply the mutiplication of the probability that each had will hit a set. If you look on the website above you will see where your caclcualtion deviates from his.
Falker

[The4Aces]

the calculation would have to include the probability that you get a pocket pair. Then the probability that you each will get one card in the next 5 cards toward your set.

[Reef]

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the calculation would have to include the probability that you get a pocket pair. Then the probability that you each will get one card in the next 5 cards toward your set.

[/ QUOTE ]

my answer was only odds of being over setted on the flop given you both hold pocket pairs. If it's still wrong, I apologize

[knsmith85]

[ QUOTE ]
[ QUOTE ]
the calculation would have to include the probability that you get a pocket pair. Then the probability that you each will get one card in the next 5 cards toward your set.

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my answer was only odds of being over setted on the flop given you both hold pocket pairs. If it's still wrong, I apologize

[/ QUOTE ]

Even if this were the calculation the original poster were looking for (I don't think it is, but that's not really a big deal), the significent problem with this is that the events here are not independent. Now, often times with hold 'em, we can assume independence even when it's not the case, because it has such a minute effect on probabilities. However, in this case, one person hitting their set significantly changes the odds of the other person hitting their's.

Let's just assume that the first card on the flop gives Hero a set. Now that 7.5 number you used is totally irrelavent for player two. It is ~(1 - (1 - 45/47)*(1 - 44/46)). When one player DOESN'T get their set on the first card, the odds for the other player become ~(1 - (1 - 48/50)*(1 - 47/49)*(1 - 46/48)). Those two probabilities are significantly different from each other.

Also, this provides a decent example of the independence discussed above. When we're calculating the odds of a set, we aren't counting the chances of a 4 of a kind coming up. However, since 4 of a kind will occur so rarely, even when holding a pocket pair, it is not really significant in this "pizza-box" calculation.