Sklansky: What's the odds on you helping me with the odds?

Nice post; this is getting good.

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If none of the first nine players has an Ace, the probability that the last player does goes up from 1 - (48/52)*(47/51) = 15% to 1 - (30/34)*(29/33) = 22%. With fewer non-Aces available, the chance of getting an Ace goes up.

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You're right, of course, though I'll make a slight modification: since the button is dealt last, but receives the 10th and 20th card - not the 19th and 20th - the probability becomes 1 - (39/43)*(29/33) = 20%. My 15% suffered from a simliar weakness: bearing in mind the order of dealing, I now calculate that the SB has about a 16.5% of landing an Ace, with consistent increases around the ring to the button's 20.3%. That's 3.8% more often, which is often enough over the long haul to be a factor, for sure.

One mitigating factor is the kicker: if a MP player is dealt an Ace as his/her first card, the remaining players are now drawing to only the three remaining. That brings the button's odds of having an Ace down to 15.4%. If MP is dealt A-3, and folds, then the button is less likely to have an Ace than before, but we have no way of knowing that.

So the button has a 20.3% chance of having an Ace if none have been folded, and a 15.4% chance if one has. I've calculated that there are 198 Ace hands (4 combinations of 12 Ax suiteds, 12 combinations of 12 Ax unsuiteds, and 6 combinations of AA). Let's say a a loose player calls with all except A6-A2 unsuited. Those are 60 mucks out of 198 total, or 30%. So really, the expected probability that the button has an Ace is more like (70%*20.3%) + (30%*15.4%) = 18.8%. The average probability that each player has an ace is around 18.3%, so our margin has decreased significantly.

There's much more that you've done there which is quite interesting, but that's all I have time to really delve into now. Food for thought though, so thanks for that.

[Dave G.]

Here is something I posted in the ML forum a while ago regarding the issue of unseen cards and probabilities. Someone had stated that they might not have odds to go for their draw because "some of the cards might be in their opponents hands", which is what you're saying here.

This is an incorrect way to view your outs. You don't know what your opponents have. In fact, it can be proven that this doesn't even matter; your odds don't change at all.

It's 4am and I'm in a mathematical mood, so I don't know if this post is going to be useful for anyone, but I feel like doing it anyway. So here goes.

I'll use the more simple example of a flush draw. The flop comes with 2[img]/images/graemlins/diamond.gif[/img]'s. You hold 2[img]/images/graemlins/diamond.gif[/img]'s, giving you a flush draw. There are 2 people in the pot against you (for simplicity I'll limit it to two; the calculations get too long for more than that).

There are 47 unseen cards, and 9 unaccounted for [img]/images/graemlins/diamond.gif[/img]'s. We say you have 9 outs. 9/47 = 0.191489 which is your probability of hitting a diamond on the turn.

With 2 other people, there's a decent chance that someone has one of your outs. Lets say we take this route, and assume that one of our opponents does indeed have a [img]/images/graemlins/diamond.gif[/img], and we discount our outs to 8. Once you do this, you are no longer dealing with a deck that has 47, or 46 unseen cards. That is, we don't just calculate 8/46 and that's our chance of hitting our diamond on the turn. We are considering the odds of all our opponents cards when we say that someone must have a diamond. So the new number of unseen cards, that is, the cards we are free to draw at, is 43. (2 opponents, 2 cards each).

Here's the correct way to do the new calculation. The probability that an opponent holds 0 diamonds is 38/47*37/46 = 0.6503. We'll call this p0. p2 (someone holds 2 diamonds) = 9/47*8/46 = .0333 and p1 = 1 - p0 - p2 = 0.3163.

Call m the probability of there being a certain number of total diamonds between our two opponents. If there are 0 diamonds between our opponents, there are still 9 diamonds left in our new 43 card deck, and the probability that neither opponent has one is equal to p0*p0. So, m0 = 9/43*p0*p0. (Neither opponent has a diamond).

Similarly,

m1 = 8/43*p0*p1 + 8/43*p1*p0. (Either player has a diamond, but the other does not).
m2 = 7/43*p0*p2 + 7/43*p1*p1 + 7/43*p2*p0 (Either one player can have two diamonds, or both players have one).
m3 = 6/43*p1*p2 + 6/43*p2*p1 (one player has one and the other has 2)
m4 = 5/43*p2*p2 (both players have 2 diamonds)

Now we have a list of probabilities. If we calculate m0+m1+m2+m3+m4, which is the total probability of all of these events, we get a probability of 0.191489. Which, as it turns out, is exactly the same as the initial probability with 47 unseen cards - 9/47.

It doesn't matter if you think your opponent has X card in his hand. Your odds of drawing do not change, because he will only have X card in his hand with some probability. And by considering that someone has that card in their hand, you have to take your opponents cards out of the deck when calculating odds.

[spaminator101]

if you want sklanksky personaly to help you you better go to the science math and philosophy forum

[switters]

Dave - your math is flawless, as near as I can tell - but you are missing a key point...

You are right that "in a vacuum", my opponent will only have 2 [img]/images/graemlins/diamond.gif[/img]s with a predictable probability, and that the math will always wash out to be the same as if I assumed there were 9 [img]/images/graemlins/diamond.gif[/img]s left in the deck. But what if I have extra information about my opponent that indicates the cards in his hand better than strict probability would allow?

for example, let's say I open raise from middle position with 7 [img]/images/graemlins/diamond.gif[/img] 8 [img]/images/graemlins/diamond.gif[/img], I'm called by Total Unknown Guy and three-bet by Rocky Tight Guy. both me and TUG call the three-bet.

flop: 2 [img]/images/graemlins/diamond.gif[/img] J [img]/images/graemlins/diamond.gif[/img] Q [img]/images/graemlins/club.gif[/img]

I decide to bet out my flush draw, and Total Unknown Guy raises me. Now, Rocky Tight Guy *calls* the two bets cold. Let's say I know rocky tight guy well enough to know the following:

* a three-bet pre-flop means AA, KK, AKs, or QQ, and nothing else.
* on this flop, he would raise with a set of queens to protect his hand. He would three-bet with AA or KK one time to see if he's still ahead, and then shut down (fold or check-call, depending on the action) unless he improved on the turn.
* BUT, he does understand pot odds, and since we can't ever make it more than 2 bets to him cold... if he held a big draw to the nuts (12, as far as he knows), he would call any number of bets, two at a time, all the way to try to hit.

now, this is a *little* contrived, although I have certainly played against players this tight and this readable, and they're pretty common at the 10/20 - 20/40 level. The point is, I think in this situation I could say with about 90% certainty that the Rocky Tight Guy is sitting on A [img]/images/graemlins/diamond.gif[/img] K [img]/images/graemlins/diamond.gif[/img], and since Total Unknown Guy is likely to hold a Q or a J on that flop, I'm probably dead to running 8's and 7's - if I'm lucky.


The math you did is still really useful, though -- you showed how you can arrive at the "all the cards are unseen" numbers by figuring out the probability that your opponents are on a flush draw, using no additional knowledge. If I can use extra table knowledge, however, to say that player X is 50% likely, given his play, to have 2 of my [img]/images/graemlins/diamond.gif[/img] outs, rather than the 3% that pure probablity allows for, I can plug that number in instead, and recalculate my odds accordingly.

[Dave G.]

What you say is true, and the conclusion you draw basically hits on the main point that we should consider with every draw that doesn't give us the stone cold nuts, although there's some things to clarify.

The correct way to handle this situation then is just as you would any other draw where your draw doesn't give you the nuts and will lose some of the time - you devalue your flush outs. As you suggest, this could be done by using a modified probability in the calculation, but there's a couple of things wrong with this approach:

<ul type="square">[*]I can't imagine anyone doing this at the table.[*]The probabilities between uber rocks hand and the other guys hand are not independent.[*]Your read might be incorrect.[/list]
If you think it's 50% likely, just call your flush as 5 outs intead of 9, and proceed accordingly. (I'd need a real solid read on the guy to devalue a flush so much. Flush over flush is rare enough for me not to worry about it in most cases.) You wouldn't say that he has 2 [img]/images/graemlins/diamond.gif[/img]s and so you only have 7 outs. If he has those two diamonds, your flush isn't worth anything except for the double gap straight flush, and you should fold your hand.

The other problem with this is that it ignores the chance that the other guy has one or two diamonds himself. If he has something like A[img]/images/graemlins/diamond.gif[/img]Q[img]/images/graemlins/spade.gif[/img] he's going to raise this flop. Infact he could raise a lot of similar hands, perhaps trying to get the rock behind him out. He cold also raise KT with no diamonds hoping to fold backdoor flush draws. He might not know what he's doing and raise with a flush draw himself.

What this all comes down to is that once you start fiddling with the probabilities that one player has a certain card or cards in his hand, that affects the probabilities of every other players hand. You can't take into account one players cards and ignore everyone elses; the probabilities are not independent.

The math I posted shows that in the end, it doesn't really matter one way or another; the probabilities average out to the same total probability in the end. For your method of doing the calculation to work out correctly, you'd need to come up with a pretty good approximation for both players possible hands and use them both in the calculation. Since a [img]/images/graemlins/diamond.gif[/img] is no more likely to play before the flop than any other suit, this is pretty difficult.

Thanks

Wow...

That was a pretty productive joust.

(Though part of me can't help but think that a fella such a D.S. did all of that math and then some and gave us a tidy 'condensed' version in his books)

Thanks... I'm gonna take a couple of weeks and digest all of this now.