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