PF Math for Dummies

Benman · #1 08-05-2006, 08:58 AM

What's the "average" rank of the highest card in a two card hold'em hand? Is there a difference between the mean, median, and mode?

numeri · #2 08-05-2006, 10:34 AM

There are equal numbers of each, and they are all equally likely, so the mean is 8. The median would also be 8, since the cards are 2, 3, 4, 5, 6, 7, 8, 9, T, J, Q, K, A, and 8 is directly in the middle. There is no mode.

Benman · #3 08-05-2006, 10:38 AM

No, I was asking about the average rank of the HIGHEST card in a two card hand. It's got to be something like Jack or Queen.

numeri · #4 08-05-2006, 10:49 AM

Ahhh...I need to read more carefully. [img]/images/graemlins/laugh.gif[/img]

bigpooch · #5 08-05-2006, 11:12 AM

There are 1326 hold'em hands, and for each pair there are
6 combinations and each specific nonpair XY, there are 16
combinations. Thus, there are 6 hands with deuce being the
highest, 6+16=22 hands with a three being highest, ...

A quick computation shows that a jack is the median and
clearly the ace is the mode (more hands with an ace as the
highest of two cards). If you could reasonably define the
"mean", that would turn out to be a jack.

AaronBrown · #6 08-05-2006, 08:41 PM

Bigpooch is correct. If you define the mean with respect to the number of the card (J = 11, Q = 12, K = 13 and A = 14) then the mean will be 1/1,326 times the sum, from i = 2 to 14, of i*[6 + 16*(i - 2)]. This comes from Bigpooch's formula for the number of combinations with highest card equal to i.

The sum can be rearranged into 16 times the sum of the integers squared from 2 to 14 minus 26 times the sum of the integers from 2 to 14. The formula for the sum of the integers from 1 to N is N*(N+1)/2, and for the sum of the integers squared is N*(N+1)*(2N+1)/6. To get the sum from 2 to N we have to subtract 1 from these numbers. So we get:

16*(14*15*29/6 - 1) - 26*(14*15/2 - 1) = 13,520.

13,520 / 1,326 = 520 / 51 = 10 + 10/51 or 10.2 for the average higher card.

bigpooch · #7 08-06-2006, 03:08 AM

Of course your calculation is correct. I obtained the
result simply by letting the rank EIGHT be zero, NINE as +1,
SEVEN as -1, etc. Then, it's clear that the pairs have
"sum" zero. Also, there are 16 more hands with NINE as
highest compared to EIGHTs and 16 less hands with SEVEN as
highest compared to EIGHTs, so the computation is then
(1/1326) (16*(+1)*(+1)+16*(-1)*(-1)+16*(+2)(+2)+...
16*(+6)*(+6)+16*(-6)*(-6)) = (1/1326)(2*16)(1^2+2^2+...6^2)
= 2*16*(7*13)/(26*51) (using 1^2+2^2+6^2 = 6*7*13/6)
= 16*7/51, etc.