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Multiway odds calculation
imaybthebest
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April 25, 2016 - 4:03 am
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The flop comes two tone. If the villain is playing any two suited cards, the odds of him having a flush draw is around %10 (I use Flopzilla). What if i am playing 4 way? How do i make the calculation? Do i just add %10 + %10 + %10 and say “Odds of someone having a flushdraw is %30”. I know it doesn’t work that way. I know i should have studied math better at high school but i didnt 🙂 Can somebody help?

chaos
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April 25, 2016 - 4:37 pm
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First, establishing “if the opponent has any two cards” is completely useless the vast majority of the times. You should always be given your opponent a range and then used THAT range as the calculation. Of course if your villain has a 80% VPIP then he is close to playing any two but that’s not the norm.

I’d say multiway pot calculations are extremely complex. They are difficult to figure out when you DO know the cards your opponents have, they are almost impossible when you don’t. Theoretically you could take their ranges, put them on pokerstove, check your equity against them given a particular flop and decide.

In practice I think there are way too many variables in this scenario to try to calculate odds of someone in particular having a draw and acting based on that, specially given there’re many other scenarios that are also likely and that you should consider (someone with a set does also have a “draw” to a full house, someone may have two pair and thus another “draw” to a full house). The term “draw” is usually applied (mostly) when it refers to a hand that goes from having almost no value to being a very strong hand if it hits.

On heads-up play this is quite important, on multiway pots this is less important as the equity of your hand will almost ALWAYS be less than the combined equity of your rivals. As that is always (or almost always) true, calculating the odds of someone having a draw is less important than evaluating if you hand is currently ahead of all the rest and whether your hand has the possibility of staying (or improving to) being the best hand.

There’s possibly other people here that may have better advice but in my case, on multiway pots (specially with more than 3 people including me), my concern is to figure out where I am, if I have a value hand and trying to narrow down their ranges. There’s only two ways gaining information in poker (other than cheating), betting or checking and observing everyone else reaction so based on my hand,  my odds, my positions and what others do I’ll decide my next action. I think on multiway pots on the flop it is a lot more about your hand, the board texture and the action than it is about trying to range other people. After all, with another 3 players each having a 20% calling range you have a mixture of really strange hands, specially since the range will get wider with each call (meaning UTG may have a 15% but button may be 30% and the BB can be playing any two) 

imaybthebest
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April 25, 2016 - 5:38 pm
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chaos, I was trying to figure out the method to calculate odds in these situations. I wasnt trying to ask strategy question. Your answer was very helpful anyway 🙂 Thanks for that. 
I will talk about another situation. Odds of someone having QQ+ AK (if i dont have any blockers) are %2.56, It is folded to me on the CO, What are the odds of BU, SB, or BB having QQ+ AK? Is it %2.56 times 3?

joelshitshow
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April 27, 2016 - 12:08 am
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A lot of it is just math, sure. But this can get you in the right direction. The odds the remaining players have QQ+/AK kind of works like this:

The probability that exactly one has it is the probability that one has it and 2 don’t. The probability that exactly two has it works similarly. And then there is the probability that all 3 have it. Find each of those three numbers and add them together. That’s the odds that at least one of them has QQ+/AK.

joelshitshow
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April 27, 2016 - 12:09 am
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You could also find the probability that each of them DOESN’T have that range. That might be easier, actually because you can take that number and cube it.

imaybthebest
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April 27, 2016 - 9:47 am
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Thank you very much joelshitshow. So;

For the odds of someone having QQ+ AK when it is folded to me on CO:
Probability of one of them not having QQ+ AK = %97,44 = 0,9744
Probability of all 3 of them not having QQ+ AK = (0,9744)*(0,9744)*(0,9744) = 0,9251
Probabilty of at least one of them having QQ+ AK = 1 – 0,9251 = 0,0749 = %7,49

For the odds of someone having a flusdraw on a two tone flop 4 way (including me) (i am making the calculation against wide ranges. i assume the villains are fishy and plays any two suited card)
Probability of one of them not having a flushdraw = %90 = 0,9
Probability of all 3 of them not having a flusdraw = 0,9 * 0,9 * 0,9 = 0,729
Probabilty of at least one of them having a flushdraw = 1 – 0,729 = 0,271 = %27,1

Are these calculations right guys?

Foucault

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April 27, 2016 - 12:24 pm
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The first part looks right. The second part, about the flush draws, you need to be more specific about what exactly you are trying to calculate. With Flopzilla, you should be able to give each player a starting range and then determine how often that player will have a hand of a given type on the flop. It’s not enough to just say people will play any two suited cards (which I don’t think is a reasonable assumption anyway, even for fishy players), you need to look at their entire starting range. FWIW, I don’t think this is really an important number to calculate with such precision, though. All you really need to know is that, before there’s any action on the flop, the odds that any given player has a flush draw is not particularly high. Once a player puts money into the pot on the flop, the odds go up considerably, because he would probably fold many of his non-flush draw hands.

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