Article on Calculating Equity --- Question?

[PJM1206]

Trying to understand ------

Below is the second example given in the atricle on calculating equity. I was confused on how he handled the tight solid opponent that he thought had AA,KK, AK. The formula he used only took into account the other opponets range of hands and not this one. Shouldnt this range of hands be factored in as well?

You raise under the gun with

T T
A tight, very good, very aggressive player next to act 3-bets. Everyone folds around to the BB who calls. The BB is very loose and fairly aggressive. He's a decent thinker, but he loves to "gamble it up." He will push strong draws and marginal hands, but he can recognize when it becomes clear that he's behind. You call, and three people see the flop of

K T 4
The BB bets out, you raise, and the solid player 3-bets. The BB calls without hesitating. You 4-bet, and the solid player caps it at five bets. BB calls again without hesitating, and you call.

The turn is the A. BB checks, you check, and the solid player bets. The BB now check-raises. What's your equity and what should your play be?

Well, first of all, let's start with our assumptions. (I'm not going to go through all the explanation of how I came to these, since it's not the purpose of this article; it's getting the math right that I'm most concerned with.) The very solid player to our left almost certainly has AA, KK or AK. In fact, so much so it's safe to say 100% of the time. (Based on how the action has gone so far, AA and KK are more likely than AK, but he definitely has one of these hands). The BB is about 70% likely to have QJ, 20% to have 44, and 10% likely to have AT or KT or AQs, or some other cheese. So again, let's bust out the formula:

E(x)p(x) + E(y)p(y) + E(z)p(z)= E(o)

You'll notice this time the formula has one additional component. That's because we have three groups of hands to deal with here, instead of the two groups in the last example. There can be as many components as you feel is necessary, as long as the sum of all the p(...)'s equals 1.

Poker Calculator tells us that our equity against group x is 4.61%. Our equity against group y is 30.95%, and 30.42% against group z. Plugging these numbers into our formula, we get:

(0.7)(0.0461) + (0.2)(0.3095) + (0.1)(0.3042) = 0.1246

So your equity in this pot is 12.46%. Let's see where that puts you in the hand. Before the turn action, there were 12.25 BBs in the pot. The best-case scenario costs you 3 BBs to see a showdown, and even that is optimistic, especially if the board pairs on the river. But, assuming only three bets to showdown, we're getting 18.25-to-3 to call down, or roughly 6-to-1. With 12.46% equity in the pot, your opponents have 87.54% equity, which makes you just over a 7-to-1 underdog. You should fold.

[unfrgvn]

test

[Gazzbut]

Ive been wondering the very same thing.

The article doesnt actually explain how to use these calculations with mulitple opponents.

[Rian Mullins]

>> Shouldnt this range of hands be factored in as well?

It looks like they already are. He gets 4.61% chance of you winning the pot if player "C" has QJ, which is about the average chance of winning for the 3 cases when player "B" has AA, KK, AK, which I get as 2.4%(1/42), 2.4%(1/42), and 9.5%(4/42) respectively, which average to 4.76%(2/42), which is within rounding errors the same as 4.61%, I assume.

Likewise for his second case where player B has 44, the odds of winning for you when Player C has AA, KK, AK are 1/42, 1/42, and 37/42 which average to 13/42 = 30.95% , which is the number he gave.

Why he gave the 3 possible hands of player B equal footing is curious. It's the difference between 37 "outs" (not really "outs" if you have the best hand) and 1 out, which is a rather large difference. And Given just the cards on the board, the odda of having AK is three times that for AA or KK i.e. 9 ways to have AK (3 remaining aces X 3 remaining kings ), 3 ways to have KK ( 3 kings left, choosing 2 of them to include is the same as choosing the remaining one to exclude, and there are three ways to do that ), and 3 ways to have AA. And I think AA and AK for player B would be played in EXACTLY the same way here, as they are basically the same hand here.

[JacksonTens]

This would be a massive laydown if it occured.

JT