PDA

View Full Version : What is probablility of hitting runner, runner on turn and river


TheNoocH
03-10-2005, 07:37 PM
Here's the situation...Player 1 holds AK...Player 2 holds A9...
Flop comes K rag rag...
Turn comes 9
River comes 9...
player 2 wins with a set...

what are the odds/probability???

Thanks...just happened to my brother....

kingstalker
03-10-2005, 09:26 PM
1 in 330.

elitegimp
03-10-2005, 10:12 PM
[ QUOTE ]
1 in 330.

[/ QUOTE ]

this is the right answer, but to include a little work here are two ways of doing this problem:

1) You know of 7 cards already dealt -- Two As, 2 Ks, 1 9, and 2 rags. Therefore, there are 45 cards left in the deck, and 3 of them are 9s. So the probability that the turn is a 9 is 3/45 = 1/15.

If the turn is a 9, there are 2 9s left in the deck and 44 cards to deal from, so there is a 2/44 = 1/22 chance that, if the turn is a 9, the river is also a 9.

So to get the total probability, multiply these two together: 1/15 * 1/22 = 1/330.

2) There are 45 cards left unseen, so there are 45 choose 2 = 45*44/2 UNIQUE ways to deal the turn and river (so 8s 8c is the same as 8c 8s). 45*44/2 = 45*22 = 990. There are 3 9s left in the deck, so there are 3 choose 2 = 3 unique ways to deal the two 9s. 3/990 = 1/330.

Siegmund
03-10-2005, 10:24 PM
Yes, 1/330 to get three nines.

Depending exactly what rags come, there may also be chances of a runner-runner straight with the nine to win, or a wheel to tie. (If one rag is a 4/3/2 and the other rag is a 6 or higher, no chance at all for A9 except three nines.)

Rushmore
03-11-2005, 05:26 AM
The probability of hitting runner runner on the turn and the river is 100%, because there's noplace else to hit runner runner.

cwes
03-11-2005, 06:07 AM
[ QUOTE ]
The probability of hitting runner runner on the turn and the river is 100%, because there's noplace else to hit runner runner.

[/ QUOTE ]

You might want to do a little bit of thinking about conditional probabilities.

You are giving the probability 'Pr(hitting rr on the turn and the river | hitting rr)', they are talking about 'Pr(hitting rr on the turn and the river)'.

Rushmore
03-11-2005, 01:21 PM
[ QUOTE ]
You might want to do a little bit of thinking about conditional probabilities.

You are giving the probability 'Pr(hitting rr on the turn and the river | hitting rr)', they are talking about 'Pr(hitting rr on the turn and the river)'.

[/ QUOTE ]

I know.

TheNoocH
03-11-2005, 02:41 PM
thanks all...and thanks elitegimp for showing the details...very helpful...
so if put another way 1 in 330 we're talking about 1/3 of 1 percent (0.3%)??? talk about a bad beat...thanks again