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

[AaronBrown]

The two-handed problem is a little more complicated than that. First of all, you don't want to get too predictable, but let's put that aside (maybe that comes under the heading of playing the player rather than the cards). Second, you have to consider what's in the pot already, and what might get put in later. But leave that aside for now as well and assume there is nothing in the pot and no further betting.

Say you raise every time you have better than a 50% chance of winning, from your perspective. Your opponent looks at his hand as asks, do I have a 50% chance of winning given that the other guy thinks he has better a 50% chance? If not, he'll fold and you won't win anything. If he does, then he'll bet, and you'll lose more than 50% of the time.

Despite these complaints, there is some truth in what you say. One of you is going to win the pot. If you have better than a 50% chance, you're ahead of the game, and you want to make the pot bigger. So a bet says your hand is in the top half of hands. A raise says your hand is in the top 75%, that is the top 50% of hands that beat the bettor's worst hand.

Using the same logic with three players, you want the pot bigger if your hand has better than one chance in three of beating both opponents' hands. If one opponent bets and the next one folds, you want a hand with a 50% chance of beating one opponent, given that he thinks his hand has better than one chance in three of beating two opponents.

That's the basic mathematical principle. You need to apply it strategically. Just because you want the pot bigger doesn't mean you bet or raise, sometimes you can get a bigger pot by checking or calling. Also, you might prefer to wait until a later round to make the pot bigger. That's generally true if you expect to get more information from the flop than your opponent will, suited connectors versus a small pair, for example. Then, of course, there's pot odds, implied pot odds and deception.

Then you have to translate the probabilities to hands. For two players it's easy, just take the half of the hold'em hands ranked by probability of winning heads up. When you add the third player, you drop the lowest pairs (I would play 3's against one opponent, but not two), medium suited connectors (87s, 76s are good against one but not two), Ace/low card (I would play A4 to A7 against one but not two), face card/9 (I'd play offsuit Q, 10, or J with a 9 against one but not two) and borderline suited high card/low card (K6s, Q5/6/7s, J7s, 10 7s; all play against one but not two).

[TheCat]

The answer is 66.67% I calculated this in a horribly complex way don't ask me exacly how but it's related to throwing two dice and finding the average highest die. This converges to 66.67 as the sides on the die increase.

This method is only barely practical for the two opponent case. It just goes mental for three or more.

Anyway I found an empirical way of doing it. You get your computers random number generator to produce one number between one and a hundred for each opponent. You hold onto the largest of these values and average then over 1,000,000 runs.

So... To have at least a 50% chance of having the best hand against P oppents, you need a hand in the top N%

Where N= (1- (1/(P+1) ) ) x 100. Why it should be this I have no idea.

If anyone has any insights here I don't, I'd be really keen to know.

[AaronBrown]

You have solved for an answer to a slightly question than the one you are asking.

If you have a hand in the top 1/(P+1) of all hands, then the chance of it being the best hand is [1-1/(2*P+2)]^P. For P = 1 this is 75%. For P = 2 this is 25/36 = 69%. For P = 3 this is 343/512 = 67%. As P gets large, this approaches exp(-0.5) = 61%. This is true regardless of the distribution of odds of winning.

Having the best hand is necessary but not sufficient to have better than a 50% chance of winning. With one opponent it is the same thing, but with more it's different. You might have the best hand with a 34% chance of winning, against two opponents who each have 33%. You have the best hand, but not a 50% chance of winning.

To compute how many of the times your best hand also has better than a 50% chance of winning, you need to know the distribution of odds. You assumed it was uniform from zero to one. We could have a game where the best hand always won, in which case having the best hand means 100% chance of winning, so the best hand is always better than 50%. Another game might have very small advantages, so all the players are near 1/(P+1) chance of winning. If P>1, that means you never have better than a 50% chance of winning, even with the best hand.

As we get deeper in to the hold'em deal, having the best hand is more likely to mean better than a 50% chance of winning. With two cards dealt to each player, there is still a lot of uncertainty. Only in a small game with a top hand will anyone have better than a 50% chance of winning. Just before the river card is dealt, one player often has 100% chance of winning, and it's almost always true that someone has better than a 50% chance.