morton's theorem... brilliant stuff

[NickTheKid 0]

More over the top equations no one is ever going to use on the table.

[reedmcneal 0]

Just to be clear, Sklansky says his theorom applies to 2-way and most multiway pots, not all. He even gives a counter example, where in multiway pots it doesn't apply.

[Patricnz 0]

good stuff.

[Makata 0]

Now each of those fish only have a 10% chance of winning against your powerhouse, but collectively they have a 90% chance of beating you. In real terms, one of them will flop two pair, a straight, a flush or trips and BAM! (It's a minefield out there - If it wasn't for luck I'd win every hand)

You're doing your math wrong. This would only be correct if it were never the case that if one fish had a better hand than you, no one else did. This of course is impossible for 9 players, let alone 2. In actuality, if you want to say you're a 9 to 1 favorite to each player individually, then you simply have to beat all 9 individually to win .. which is 0.9^9 or about 0.38. In other words, even if all 9 players has a 10% chance of beating you, they collectively only have ~62% chance of beating you. This is why raising a hand such as AA, even from early position still remains profitable in NL, because even if everyone chases you, 32% of the time you win 9 fold your investment.Easy way to show your math is off is to assume instead each player had a 75% chance of beating you (if you were holding 23o, and I'm sure 75% is generous). By the time the second person called you're already be guaranteed to win -50% of the time.

There was one very important piece of information that you guys missed that someone posted.Does any of this information change what you are going to do?

Yes. I think this could help reaffirm how powerful nut draws are in multiway pots, especially ones where most if not all of the outs cannot be counterfieted by any other hands (excluding of course when they hold those cards), such as if you had an open ended, but someone else had a 4-flush.Say for instance you had QK on a flop of J23, 5 way, you being second seat. It's checked around and the turn is a T, completing the rainbow. Say the pot size is 5 BB, and first position leads out with a bet, giving you 3.5 to 1 on your call. Your read on all 4 of the other players, including BB is that they will usually call or lead out if they have a good hand but manage to become almost clarivoyant on the river and will only call your bets if you are beat, giving you 0 implied odds. You're getting 3.5 to 1, but your chance to make it is only 4.75 to 1, thus it seems like you should fold.However, once you call, the other 3 players will be getting 4.5 to 1 (or better) and could end up calling a variety of draws, both correctly and incorrectly. You need only 2 to end up calling to boost your pot odds to 5.5 to 1 to make your original call correct.Fairly contrived example sure .. but if nothing else I think this article drives home the point that in lower limit games, especially ones with alot of loose players, that nut draws can often take you alot further than textbook premium hands. The reason being that while AA is great against 1 opponent, it becomes progressively weaker against many. Nut flush draw however, is just as strong against 1 opponent as 20. In both cases your pot investment decreases as players go up, but only in the second does EV increase dramatically.

[PMJackson21 0]

Not only are Andy's posts incredibly insightful, but the responses/discussion of it are just as educational. If you go through the RGP archives, you can find Caro's detailed response to this, as well as Sklansky's and many others. Ha, I love how Caro refers to himself in the 3rd person sometimes...Reminds me of 'Jimmy' on Seinfeld.Thanks for the post AK; I had read it before, but not in a long while, nor had I looked into the follow-up discussions between Sklansky, Caro, Badger, Abdul, etc etc.Oh, and Sklansky says he knew this (Morton's Theorem) when he wrote 'Theory of Poker', but he dumbed TOP down for the 'average joe'. Ha.Patrick

[bobbytheo3 0]

There was one very important piece of information that you guys missed that someone posted.Does any of this information change what you are going to do?You can sit here and theorize all day long about wether player A or B laying down when the pot is 15 bb or 17bb makes it better for you or not. It still changes nothing about the hand and how you are going to play it. It might be nice to go through and figure out the math about something like this but to me it's a waste of time. I won't change a loose player into a tight smart one and I won't change a rock into a maniac at will depending on if I want them to call or not. This is the reason players are classified as such. A loose player will play that middle pair to the river hoping to catch, and a rock will lay down a missed AK if the price isn't right to draw. What you do in the hand is constant regardless if you think someone is drawing or not.

you can't change what your opponents do, but you can make strategy adjustments.we've seen that the most insensitive and best draws benefit the most from loose games and loose calls, so:1. being suited makes your hand's value go way up (SSHE is right)2. having nut draws is important, so that can affect your post-flop play. this can also affect your strategy in order to maximize your expectation (whether to add one more caller or to raise it up and have double stakes with one less caller, etc.).3. protecting your hand with vulnerable sensitive made hands is important. think about betting into an aggressive player, or checkraising to make the field face two, things like that.i think there are some applications, we just have to think about what those are.aseem

without doing the math, (maybe you could do that for me in a hypothetical similar to the ones youve posted), it seems that this also may mean that player C (an insensitive draw guy) might be inclined to raise Player A's bet (TPTK guy) in this situation if he thinks player B (who already called A's bet) will again make a correct call of the raise, and the extra bets still keeps the pot in the paradoxical region. in other words, it might make re-raising with insensitive draws a more appealing option since correct calls by your opponents add to your expectation, and so does the possiblity of incorrect folds by your opponents. it seems like it would be impossible for both your opponents folding or calling could add to your expectation, but could the existence of the paradoxical region make it so?

[pokerkid 0]

First of all, i really liked this post. Very interesting and it just gave me something to think about.My questions/comments refer more to strategy.First of all, it was mentioned that player B's call would hurt you. I don't think it hurts you, it just helps player A. In a tournament that would hurt you, but in a cash game, if one of player A's outs hit, you lose the pot anyways and it doesn't matter to you if a few more of player B's bets get sent to player A's br.Please respond with questions or comments to what i just said. I really don't mind if someone proves me wrong as long as i understand the concept more thoroughlyHaving said this, isn't the correct play on the turn to still bet?You might as well make the other player's payfor a card that most likely will not help them. In this one instance, you might as well get as much money into the pot as possible because it is most likely that your hand will hold up.Besides the preflop strategies already mentioned, any other tips on how to play low limit hold em more effectively?Any strategies besides preflop strategies?

[akishore 0]

First of all, it was mentioned that player B's call would hurt you. I don't think it hurts you, it just helps player A. In a tournament that would hurt you, but in a cash game, if one of player A's outs hit, you lose the pot anyways and it doesn't matter to you if a few more of player B's bets get sent to player A's br.

well, it DOES hurt you. you can see the formula for your expectation depending on whether B folds or calls. when the pot is in the paradoxical region, you actually do lose expectation when B calls incorrectly.

Having said this, isn't the correct play on the turn to still bet?You might as well make the other player's payfor a card that most likely will not help them. In this one instance, you might as well get as much money into the pot as possible because it is most likely that your hand will hold up.

yes, you should still bet since you have incomplete information and can't change what player B will do.

Besides the preflop strategies already mentioned, any other tips on how to play low limit hold em more effectively?Any strategies besides preflop strategies?

i'd be interested in this, too. also, i'd be interested in any changes to post-flop strategy.aseem

[pokerkid 0]

When the original author calculated your expeceted value he wrote

To figure out which action on player B's part _you_ would prefer, calculate your expectation the same way: E(you|B folds) = 37/46 * (P+2) E(you|B calls) = 33/46 * (P+3) Your expectation depends in each case on the size of the pot (ie, the pot odds B is getting when considering his call Setting these two equal lets us calculate the potsize P where you are indifferent whether B calls or folds: E(you|B calls) = E(you|B folds) => P'_you = 6.25 Big bets.

should the equation beE(you B Folds) = 37/46 * (P+2) - 9/46 * (1)E(you B Calls) = 33/46 * (P+3) - 13/46* (1)because the expectation equations for player b calling were:E(player B|calling) = 4/46 * (P+2) - 42/46 * (1) when assem said to go back and look at the expectation formula, this caught my eye...i'll probably type up another response about my point earlier once i've had time to think about it.

[shedrackle 0]

First off I'd like to start by saying I admire Akishore's posts not only of the current subject but his insight on former topics as well.I do think that there is some merit into the thought process and mathematics explained in Morton's theory. However, in my 10 years of experience playing poker in various locations I have never seen anyone bring paper or a calculator to the table to make such algebraic calculations outlined in the theorem. I'm not saying this at all as a flame. I'm mentioning that the theorem has merit and for those that chose to nitpick apart the thought process he used is like saying the entire theorem is wrong. No one is going to calculate to a finite detail the exact odds or percentages for each and every person in the hand. So, IMHO what we need to do is understand the fundamentals of the theorem. Which is what I believe Akishore was pointing out in the first place.

[akishore 0]

When the original author calculated your expeceted value he wrote

To figure out which action on player B's part _you_ would prefer, calculate your expectation the same way: E(you|B folds) = 37/46 * (P+2) E(you|B calls) = 33/46 * (P+3) Your expectation depends in each case on the size of the pot (ie, the pot odds B is getting when considering his call Setting these two equal lets us calculate the potsize P where you are indifferent whether B calls or folds: E(you|B calls) = E(you|B folds) => P'_you = 6.25 Big bets.

should the equation beE(you B Folds) = 37/46 * (P+2) - 9/46 * (1)E(you B Calls) = 33/46 * (P+3) - 13/46* (1)because the expectation equations for player b calling were:E(player B|calling) = 4/46 * (P+2) - 42/46 * (1) when assem said to go back and look at the expectation formula, this caught my eye...i'll probably type up another response about my point earlier once i've had time to think about it.

no, the original author was correct.he's calculating your expectation AFTER you've bet, not OF your bet. since "P" was the pot size at the beginning of fourth street, "P+2" is the pot size including your bet and player A's call, and "P+3" is the pot size including your bet, player A's call, and player B's call.so, 37/46 of the time (if B folds) or 33/46 of the time (if B calls), a safe card will land and you'll win that whole pot. the rest of the time, you win nothing (and lose nothing, since you're not putting in another bet).aseem

[pokerkid 0]

I can't seem to understand why the expectation equation for player B is:E(player B|calling) = 4/46 * (P+2) - 42/46 * (1) but the expectiation equation for you is either E(you|B folds) = 37/46 * (P+2) orE(you|B calls) = 33/46 * (P+3) In the equation, why do you substract the percentage of cards (the thing i put in bold) that player B will lose on, but you don't subtract the percentage of cards that you will lose on?

[akishore 0]

I can't seem to understand why the expectation equation for player B is:E(player B|calling) = 4/46 * (P+2) - 42/46 * (1) but the expectiation equation for you is either E(you|B folds) = 37/46 * (P+2) orE(you|B calls) = 33/46 * (P+3) In the equation, why do you substract the percentage of cards (the thing i put in bold) that player B will lose on, but you don't subtract the percentage of cards that you will lose on?

because your bet is already in the pot, and you're simply calculating your expectation AFTER you've bet, not the expectation OF your bet (think about that).if you've already bet, the pot grows by 1 bet to become P+1. when player A calls, it becomes P+2.so if B folds, the river will be a safe card 37 out of 46 times, so you win that pot (P+2). if it's a flush card, you're not losing another bet (remember that for simplicity we said there would be no action on the river) since you've already put in that bet.if B calls, the pot becomes P+3. if it's a safe card (33/46 cards), you'll win the whole pot (remember, your bet is already in the pot, so it's no longer your money), and if it's not a safe card, you don't gain or lose anything more (again, remember that your bet is already in the pot, so it's not your money anymore).make sense?aseem

[pokerkid 0]

Alright, it makes more sense now. Thanks for clarifying.