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avisco01
12-23-2005, 12:16 PM
I tried doing this myself, but I'm not that big of a math guy per se. Supposing the following situation:

AK vs. JT

Flop: Q X X

What are the odds of an ace hitting the turn, and king hitting the river, giving the JT the win. Since we see both opponent's hole cards, how would this calcuation go? Obviously, if the king hits the turn, the odds against the JT making a straight on the river increase as it will become an OESD. However, what are the odds, given this specific situation, that the player holding JT will go perfect, perfect hitting an inside straight draw?

soko
12-23-2005, 04:20 PM
assuming that X is never an A or a K

3 outs on the turn, 3 in 45 or 1 in 15
3 outs on the river,3 in 44 or 1 in 14.6

1/15*1/14.6=1/220

1 in 220

12-23-2005, 04:37 PM
[ QUOTE ]
assuming that X is never an A or a K

3 outs on the turn, 3 in 45 or 1 in 15
3 outs on the river,3 in 44 or 1 in 14.6

1/15*1/14.6=1/220

1 in 220

[/ QUOTE ]


Shouldn't there be 6 outs on the turn?

Turn: 6/45 or 1/7.5
River: 3/44 still 1/14.6

It works out close to 1 in 110 times if I did my math right.

ohnonotthat
12-25-2005, 07:38 AM
Stop guessing. /images/graemlins/confused.gif

If you know your opponent's hand there are 990 possible turn-river combinations. (45*44/2!)

Of these, 9 will give you a straight - any mix of an A and a K will do the trick.

You got the correct answer using the wrong method; in other words, you got a bit lucky.

2 + 2 equals 4

2 x 2 also equals 4

Nonetheless, the "+" and the "x" are not interchangable; they just happen to be here.

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