When to rasie UTG vs three at hold'em?

[TheCat]

You are Heads up at the end of a Hold’em tournament. Then, putting aside any questions of playing the player but just concentrating on the cards, it must be correct to raise whenever you hold one of the top 50% of hands and correct to reraise with one of the top 25%. This just makes common sense, at least it does to me.

However when there are 3 or more players how would you determine the mathematically correct percentage of hands to raise with when you are UTG? This appears to be a really difficult problem.

A similar if not exactly analogous problem that is solvable is:
You throw two six sided dice. What is the average value of the highest dice? Answer 4.47222…
Here’s how , list every possible combination of two dice. Place the highest of the two in the right hand column and the lowest in the left. Add up the columns and divide by the number of combinations in the table, which is 36, see the working below.

This suggest you could use a similar method for three handed hold’em. In a three column table list every possible way 52 cards can be dealt to two players. Which is C(52,2) * C(50.2) * C(48,2) =
1,326 x 1,225 x 1,128 = 1,832,266,800
Ok, stop now this is too big a task even for a computer.

Hopefully I’m missing something obvious or you know some maths I don’t that makes this nut much easier to crack.

Any observations, ideas or full solutions welcome.


Low die, High die
6 6
5 6
4 6
3 6
2 6
1 6

5 6
5 5
4 5
3 5
2 5
1 5

4 6
4 5
4 4
3 4
2 4
1 4

3 6
3 5
3 4
3 3
2 3
1 3

2 6
2 5
2 4
2 3
2 2
1 2

1 6
1 5
1 4
1 3
1 2
1 1
====
Total lowest 91
Total Highest 161

Average high dice = 161/36 = 4.472222222
Average low dice = 91/36 = 2.527777778

Average for two dice = (91 + 161)/36 = 7