PDA

View Full Version : Probability 101

Chemboss
11-10-2003, 04:23 PM
I was recently arguing with a fellow player on the odds of any pair vs. any non pair. He insisted that it was 5/8 (5 draws * 6 winning cards/48 remaining (Ignoring straights and flushes)) in favor of the non-pair. I was unable to convince him that this was wrong. I think that actual odds are about 53% in favor of the pair pre-flop. I could not explain his fault because I did not know how this is calculated exactly. How can these odds be calculated in multiple draw situations?

Also, are there any good books/webpages that concentrate on probabilities encountered in poker?

Thanks.

mosch
11-10-2003, 06:22 PM
Well, the exact odds depend on which pair and which non-pair you use, and what their relationship to each other is.

PokerCalculator (http://koti.mbnet.fi/jraevaar/pokercalculator/) is useful for calculating the odds on these showdowns.

I just ran a sim of heads up play, one player plays all pairs, other player plays all non-pairs, 100,000 random hands. The pair player won 69154 times, it was a draw 987 times, and the non-pair player won 29859 times.

With 22 versus AK, 22 won 51695 times, drew 647 times, and lost 47658 times.

Bozeman
11-10-2003, 06:28 PM
Basically, his method multiply counts all those times he hits more than once, and also in the real case the unpaired hand loses when the pair also hits.

The simplest way to calculate chance of hitting once or more is to take one minus the chance of always missing.

For a six outer, this is (assuming 2 irrelevant cards ar in his opps hand):
1 - 42/48 * 41/47 * 40/46 * 39/45 * 38/44 = (1-C(42,5)/C(48,5) ) = 50.3%

For two outs (for the pair) out of the 4 cards that may not match the unpaired hand, you get

1-C(45,4)/C(47,4)=16.5%

So you will hit and the pair not hit about 42% of the time. The rest comes from str8s and flushes (and two higher pair on board).

Craig