Odds of hitting trips on the flop from a pair

[Sandsifter]

Just curious what my odds were of getting trips on the flop from a pocket pair?

[Juff]

(2/50)+(2/49)+(2/48)

12.25%, or 1 in 8.

[BruceZ]

[ QUOTE ]
(2/50)+(2/49)+(2/48)

12.25%, or 1 in 8.

[/ QUOTE ]

You can't just add these probabilities since hitting on any of the 3 cards are not mutually exclusive (you can hit on more than one). Even if they were mutually exclusive, it would be 3*2/50 because each card has the same chance before any of them are dealt. Since these are not mutually exclusive, this counts the quads twice, so for trips or better we must subtract from this the small probability of quads to get 3*2/50 - 48/C(50,3) =~ 11.8% or 7.5-to-1. This also includes full houses. We can also get this by taking 1 minus the probability that we don't hit, or 1 - (48/50 * 47/49 * 46/48) =~ 11.8% or 7.5-to-1.

For trips only, this would be 2/50 * 48/49 * 44/48 * 3 = 10.8% or 8.3-to-1. We multiply by 3 since we can hit on any of the 3 flop cards.

[OrianasDaad]

I guess combinatorics sin't the best way to do that one. What would be the equation for that?

I get (2C1 * 48C2)/50C2, which is giving me a different answer.

[ QUOTE ]
Just curious what my odds were of getting trips on the flop from a pocket pair?

[/ QUOTE ]

I think the probability is 1 minus the probability of not getting either of two other cards you need on any of the three flop cards. That would be 1 - (48/50)*(47/49)*(46/48) = 0.118, or aout 7.5-to-1.

[BruceZ]

[ QUOTE ]
I guess combinatorics sin't the best way to do that one. What would be the equation for that?

I get (2C1 * 48C2)/50C2, which is giving me a different answer.

I guess combinatorics sin't the best way to do that one. What would be the equation for that?

I get (2C1 * 48C2)/50C2, which is giving me a different answer.

[/ QUOTE ]

If you divide that by C(50,3) instead of C(50,2) you get trips or full house, but not quads. You can include full houses and quads as 1 - C(48,3)/C(50,3). Excluding full houses and quads is best done as I did it rather than with combinatorics.

[spaminator101]

2/50 help you so
48/50*47/49*46/48=88/100or
12%of the time
more like 11.5% actually

[BeerMoney]

Bruce, it may be a good idea to make a sticky for this forum, and include this.

[ QUOTE ]
We can also get this by taking 1 minus the probability that we don't hit, or 1 - (48/50 * 47/49 * 46/48) =~ 11.8% or 7.5-to-1.


[/ QUOTE ]

I hope I don't get flamed for the following dumb question, but it's something I've been thinking about and I need someone smarter than me to help me out [img]/images/graemlins/smile.gif[/img]

Do the odds of flopping a set with a pocket pair change with the number of opponents at a table? IOW, if we had a 22 person hold 'em game (theoretically possible with a big enough table), would I have longer odds to flop a set since more cards are used and the probability of another 2 being dealt/discarded are higher?

I hear the argument that you should count only unseen cards, but I don't understand how card distribution doesn't affect your odds. Can someone help me out please? TIA,

Sysvr4

[uuDevil]

[ QUOTE ]
I hear the argument that you should count only unseen cards, but I don't understand how card distribution doesn't affect your odds.

[/ QUOTE ]

What do you know about the distribution of cards with 21 opponents that you don't know with 1? Nothing. So it doesn't matter.

Also see this thread.