Hi all,

I was just wondering about the scenario below



It says he is 44.44% to win but he has 11 outs (any remaining club and 3 Aces)
so with 45 cards left in the deck 11/45 = 0.2444444444444

so I'm assuming the other 20% must come from runner runner 25 56 or 77

so My questions are how do I work out runner runner odds? how do I get 44.4% like above? how do I get that extra 20% from runner runner

Sorry if this is a silly question im quite new to poker just trying to learn to do odds

Thanks All

There are two cards left to come, so (34/45*33/44) = .567 flush or ace do not come in, or .433 they do come in. You also need to account for running sevens, backdoor straight and times flush beaten by full house to get exact odds.

The 11/45 is just the probability of hitting one of the 11 outs on the turn. We have 2 chances to hit it, and the easiest way to compute the probability of hitting an out on either card is to take 1 minus the probability of missing it on BOTH cards or

1 - (34/45)*(33/44) =~ 43.33%.

As for the runner runner outs, there are 3*3=9 ways to make each straight (25, 56) without a club (since we are already counting the club outs) times 2 straights gives 18 favorable pairs of cards. Adding the 3 ways to hit running 7s gives 21. But now we have to subtract for the 3*2 = 6 ways to make AQ which are losers, minus 4 more for KK, KK, KQ, KQ which make us a flush only to lose to a full house. This leaves 21-6-4 = 11 extra favorable pairs of cards. This is out of a total of 45*44/2 = 990 total possible combinations for the turn and the river, so these extra 11 back door possibilities are only worth an extra 11/990 = 1.11%. Adding this to the 43.33% gives the 44.44% that the program computes.

You don't need to be this precise, but trying to match the program is a good exercise in learning to recognize the back door possibilities, as well as tainted outs.

I was about to post my combinatoric solution when you beat me to it using the easy way. But the advantage to counting the combinations is that we can reproduce and understand exactly what Poker Stove counted, when it shows us 440 wins for the underdog and 550 wins for the favorite.

So we start with the C(45,2) or 990 possible ways to deal the turn and river. Poker Stove deals all those out and just counts the winners (and ties but there are none here).

Of those, the ways get a flush are:
990 - C(37,2) = 324, then less the ways we lose with a flush. So as you said those are
KK, KK, KQ, KQ leaving us with 320 winning flushes.

Then we have the wins when an Ace comes but no Queen and no flush (already counted flushes). An Ace without the 8 flush cards will come out C(37,2) - C(34,2) = 105 times, but we also lose the 6 AQ hands, leaving us 99.

Then we also win with the 3 runner-runner 7s and the 18 runner-runner straights (simple to count directly). So adding up:


Of 990 possible turn and river.
We win:

320 flushes
99 ways AA
3 ways trip 7s
18 straights (nine ways for 5,6 and nine ways for 2,5)
=========
440

440/990 = 44.444%

And that's how Poker Stove gets 440 wins, by directly counting all the winning outcomes.