I NEED A PROBABILTY LESSON

[LSU POKER PIMP]

I am fairly new to poker and an having some trouble calculating odds, I understand that if i have 3 outs to catch on the river i have a 3/46 chance of catching it, but how do you calculate if you have both the turn and the river to catch it. would this be 6/46.5?

[GrekeHaus]

That's not quite right, but for game time estimates, it's usually close enough.

The easiest way to calculate the probability is to take 1 minus the probability of not hitting either.

P(hitting 3-outer) = 1 - (44/47)*(43/46) = 0.124
while 6/46.5 = 0.129

[gamble4pro]

When dealing with odds, you have to clearly define the events first. Be careful with the words "and" and "or".
When you are saying "both the turn and the river to catch it" did you mean "by river" (this means "or") or "each of turn and river card to be one of your outs" (this means "and")? Here is the calculation for each case:
Let n be the number of outs. Let's denote by O an expected card.
-For the event "each of turn and river card to be one of your outs", the favorable combinations of the board (turn and river) are OO, in number of C(n,2)= n*(n-1)/2, from C(47,2)=1081 possible. Therefore, the probability is n*(n-1)/2162 (For your example with n=3, this is 0.277%)
- for the event "at least one out hit by river" (i.e. hit on turn or hit on river), the favorable combinations of the board are OO and Ox (x different from O), in total number of C(n,2) + n(47-n), from 1081 possible. The probability is (n*(n-1)+2n*(47-n))/2162 (For your example with n=3, this is 12.488%).

[rfsch]

I think I read in a Phil Gordon book that the "rule of four" will give you a quick approximation. Thus multiply your outs by four to get your answer. In your case you had three outs x 4 =12 (that's obviously not 12.88 via the long hand method but perhaps a useful shortcut)