Odds of Flopping 3 of a kind

[Dovberman]

If I have a pair in the hole, is this a correct way to calculate the odds of flopping three of a kind ?

(2/50 * 46/49 * 45/48)

Plus

(46/50 * 2/49 * 45/48)

Plus (46/50 * 45/49 * 2/48)

Which really means that I need to consider all the ways
that the flop could occur.

Thanks

[BugsBunny]

The easiest way to figure this is to calcualte what the chances are of you not getting the card you need and then subtracting this from 1.

So in this case there are only 2 cards that will help you. The chances of you not getting one of those 2 cards are:

48/50 * 47/49 * 46/48 = 0.8824489796

Subtracting this from 1 we get
1 - 0.8824489796 = 0.1175510204
or about 11.76% is the chance you flop a set or better when you hold a pp.

When figuring it the way you tried to figure you have to also adjust for the fact that the event you're calculating for has already occured.

So you'll connect on the first card 2/50 times or .04
That means you won't connect .96

Now you'll connect on the 2nd card 2/49 times, when you haven't connected on the first. So this part becomes: 2/49 * .96 = 0.0391836735

So adding the first 2 gives the time you'll connect on the first or second cards: 0.0391836735 + .04 = 0.0791836735
That means you won't connect 0.9208163265 of the time on the 1st 2 cards.

You'll connect on the 3rd card: 2/48 * 0.9208163265 = 0.0383673469

0.0383673469 + 0.0791836735 = 0.1175510204

Note that this matches the number we got using the much simpler method.

[BruceZ]

[ QUOTE ]
If I have a pair in the hole, is this a correct way to calculate the odds of flopping three of a kind ?

(2/50 * 46/49 * 45/48)

Plus

(46/50 * 2/49 * 45/48)

Plus (46/50 * 45/49 * 2/48)

Which really means that I need to consider all the ways
that the flop could occur.

Thanks

[/ QUOTE ]

2/50 * 48/49 * 44/48 * 3 = 10.8% or 8.3-to-1.

We multiply by 3 since the matching card can come in any position with equal probability. Note that this does not include quads or full houses.

[Dovberman]

The method you described is simpler. The result is slightly different than the one described by BruceZ. For practical purposes, it probably does not matter.

Thanks,

Dovberman

[BruceZ]

[ QUOTE ]
The method you described is simpler. The result is slightly different than the one described by BruceZ. For practical purposes, it probably does not matter.

[/ QUOTE ]

His method includes full houses and quads, while mine only counts sets, hence the difference.

[Dovberman]

I was using incorrect numerators to represent the cards that did not match the hole card. (2/50 * 46/49 * 45/48).
You are right. There are 48 cards remaining that do not match the hole card. Why are there only 44 remaining unmatched cards on the third drop ?

Thanks,

Dovberman

[BruceZ]

[ QUOTE ]
Why are there only 44 remaining unmatched cards on the third drop ?

[/ QUOTE ]

Because 3 of them pair the board to give you a full house, and 1 gives you quads.