Deb Below is an old letter on calculating poker probabilities: Note: a "blaze" is a hand of 5 face cards without a 3 of a kind or better, e.g., QQJJK. Well, you're right in that 3008 to 1 is the wrong odds for a blaze (says me), but it's the other way, the odds are actually 6016 to 1. The reason they're wrong has to do with symmetry in taking combinations. They probably had some graduate student calculating these. I had to point out a similar error my statistics professor made a couple quarters ago (previous e-mail). I expect the reason a blaze doesn't beat 3 of a kind etc. is the same as why a pair of 7's with an ace doesn't, while the odds of getting a pair of 7's with an ace is small (113 to 1 vs. 46 to 1 for 3 of a kind), it's just a special pair and people might get miffed if it beat their 3 deuce's and an ace. Now to the math: Recall your combinations formula (N) N! (J) = ------- or for example J!(N-J!) there are 6 ways to make combinations of 2 things from a set of 4: {a,b,c,d} ab, ac, ad, bc, bd, cd (Note: order doesn't count that would be permutations where cb != bc. Poker hands: there are (52) ( 5) = 2,598,960 possible ones (no jokers). So in general you just count the number of ways you can make a hand and divide into the big number above. (flip the fraction to get a probability instead of odds). Example: Any 4 of a kind: Choose your value, then take combinations of all 4 suits for that card, then choose any other card you want from the remaining 48: (13)(4)(48) ( 1)(4)( 1) = 13*1*48 = 624: 2,598,960/624 --> 4164 to 1 or .00024 probability Exactly 3 of a kind (don't count 4 of a kind or full house): Choose the value, choose 3 out of 4 of the suits, choose 2 suits from the 12 remaining, 1 card from the each of the 4 values for each of these 2 suits: (13)(4)(12)(4)(4) ( 1)(3)( 2)(1)(1) = 13*4*((12*11)/2)*4*4 = 13*4*66*16 = 54,912 2,598,960/54,912 --> ~46 to 1 or .02112845 probability BLAZE: (Don't count 3 or 4 of a kind or full house) the temptation is to pick one of the 3 values (J,Q,K), then take 2 of the 4 suits from it then choose from the 2 remaining 2 of the 4 suits then one suit from what's left. Thus: & & (3)(4)(2)(4)(1)(4) (1)(2)(1)(2)(1)(1) = 18*12*4 = 864 for 3008 to 1 WRONG !!! ...as a quick sanity check will show: there are only (12) ( 5) = 792 TOTAL combinations from among the 12 J,Q and K cards so how did we get 864 without counting the 3 or 4 of a kinds or full houses (192, 24, 144 = 360 of these respectively). We forgot about combinational symmetry. There are only (3) (2) = 3 ways to choose which values will be pairs (J,Q; J,K; Q,K), {we counted (J,Q AND Q,J)} not 6 as multiplying the numbers under the &'s above gave us. We should have chosen 2 values to pair, selected 2 of 4 suits from each then one suit from the remaining value: (3)(4)(4)(4) (2)(2)(2)(1) = 3*6*6*4 = 432 for 6016 to 1 odds. We can verify this by taking all combinations of (J,Q,K) and subtracting off the ones that are better than a Blaze (3,4, FH): (12) ( 5) = 792 - (192 + 24 + 144) = 432. Be careful if you try a regular 2 pair also. Flushes and straights are different. Awake ???? - C