Average Dealt High Card - Gambling and Probability

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Average Dealt High Card

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07-17-2018 , 02:27 PM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 4,987

Dabbling in a little heads-up research, I wondered what’s the average dealt high card one could have (I know, I know, the median is the more applicable stat).

Being lazy-and maybe not too adept at an analytic solution here- I simply set up a 13 x 13 matrix in Excel, used its MAX function to get the high card for each of the 169 hand types and computed the average (Ace=14).

Could one of the math whizzes show an analytic or logical solution and we’ll see how we compare. I’ll also show the median.

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07-17-2018 , 03:08 PM

md46135

grinder

Join Date: Mar 2015 Posts: 420

If I understand it correctly, this is what you mean:
A is a high card for A2, A3, A4, A5, A6, A7, A8, A9, AT, AJ, AQ, AK and AA. That is 12 unpaired hands and 1 paired hand.

K is a high card for K2-KQ and KK. 11 unpaired hands and 1 paired hand.

So, the way to calculate this is 1/1326* ( 14*(12*16 +6) + 13*(11*16+6) + ... )
In general I think it would be like this http://www.wolframalpha.com/input/?i...)+),+i%3D0..12

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07-17-2018 , 06:08 PM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 4,987

Formula seems good on first look, but we get slightly different answers.

You (Wolfram) got 520/51≈10.1961

and I got 10.1538.

Because of symmetry, median will be same if saying the Ten or Jack is the median makes sense.

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07-17-2018 , 06:45 PM

whosnext

Carpal \'Tunnel

Join Date: Mar 2009 Posts: 6,732

Are you simply taking the straight average over the 169 cells?

That would not be quite right since unpaired hands and paired hands occur with different frequencies.

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07-17-2018 , 06:57 PM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 4,987

Ahhh, my error. Gave the pairs the same weight as non-pairs. When I weighted got same answer

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