Question About Probabilities

[Q.E.D.]

I have a problem that keeps coming up that seems basic but I can't figure out. There are some circumstances where you account for different combinations of possible outcomes and some where you don't and I can't figure out what separates the two situations.

For example, say you have AA in holdem and you want to know the probability of flopping exactly one ace and the probability of flopping no aces.

For exactly one ace you have Axx, xAx, xxA flops where x are non ace cards, or (( 2 x 48 x 47 )/ ( 50 x 49 x 48 )) x 3 combinations.

For no aces you have (48 x 47 x 46) / (50 x 49 x 48) and don't have to consider possible combinations.

This seems pretty basic but I'm constantly coming across situations where I'm not sure if I need to account for combinations. What's the determining factor?

Thanks

[ncray]

Combinations are just another way of doing the problem. Sometimes they make life easier.

For one ace flopping your calculation equals 141/1225.

Using combinations, you get C(2,1)*C(48,2)/C(50,3) = 141/1225

No aces you get 1081/1225

Using combinations it's C(2,0)*C(48,3)/C(50,3) = 1081/1225

[T50_Omaha8]

And using combinations it's quite easy to make sure there's no mistake. For example, the probability you flop 2 aces is C(2,2)*C(48,1)/C(50,3) = 48/19,600 = 3/1225.

Now you can sum the three probabilities (since the probability of flopping three aces is zero):

3+141+1081/1225 = 1225/1225 = 1

as it should.


Combinations also make more complicated flop calculations--like flopping 4 to a straight or flush--MUCH easier.

[BruceZ]

[ QUOTE ]
I have a problem that keeps coming up that seems basic but I can't figure out. There are some circumstances where you account for different combinations of possible outcomes and some where you don't and I can't figure out what separates the two situations.

For example, say you have AA in holdem and you want to know the probability of flopping exactly one ace and the probability of flopping no aces.

For exactly one ace you have Axx, xAx, xxA flops where x are non ace cards, or (( 2 x 48 x 47 )/ ( 50 x 49 x 48 )) x 3 combinations.

For no aces you have (48 x 47 x 46) / (50 x 49 x 48) and don't have to consider possible combinations.

This seems pretty basic but I'm constantly coming across situations where I'm not sure if I need to account for combinations. What's the determining factor?

Thanks

[/ QUOTE ]

In the first case you multiply by 3 because the A can come in any of 3 positions, and 2/50 * 48/49 * 47/48 by itself assumes that the A is on the first card only. In the second case, all of the cards are non-aces, so all possible orderings are already taken into account.

You can also do the first case as 2*C(48,2)/C(50,3), and the second as C(48,3)/C(50,3). Since the denominator in each case counts flops without regard to order, the numerator does the same.

Combinations often make things easier, but some problems like this one can be done both ways, and some problems are much easier with fractions, or equivalently permutations. Often it is a good idea to do it both ways when possible to make sure you get the same answer.

[AaronBrown]

Another hint is to always compute all the probabilities, to make sure they add up to 1. In that case, it's often easy to see a mistake.

In this case you can pick zero Aces 1 way, one Ace 2 ways and two Aces 1 way.

You can pick three cards out of 48, 48*47*46/6 ways, two cards out of 48, 48*47/2 ways and one card out of 48, 48 ways. So we can get zero Aces 1*48*47*46/6 ways, one Ace 2*48*47/2 ways and two Aces 1*48 ways.

1*48*47*46/6 + 2*48*47/2 + 1*48
=48*(47*46 + 47*6 + 1*6)/6
=48*(2,162 + 282 + 6)/6
=48*2,450/6
=48*49*50/6

Which is the number of ways you can pick three cards from 50.