Pots odds theory

I have always calculated pot odds, and I, like every other player who posts here, use pot odds to make most of my drawing decisions. However, lately I've questioned the fundamental assumptions of pot odds. The calculation of pot odds relies on the assumption that faced with no evidence to the contrary, all of your drawing cards are live. Unless the cards that you need have been exposed, you assume that the cards you need are live, not in another player's mucked hand. But is this a resonable assumption? Why don't we calculate the average number of cards that we can expect to be dead on a random deal? We can use computers to deal out 10,000 hands and track the number of spades that are dealt out on average. We can then incorporate this average number into our pot odds calculation. For instance, if the computer calculates that 6 spades are dealt out on a random deal to 9 players, we can incorporate that into the odds for drawing to the flush. If you hold A4 to a 9 K 3 flop, the traditional pot odds to make your flush are about 5:1. But if we knew that there are likely to be 4 additional spades dead (2 in your hand, and 4 in other people’s hands on average, making the average of 6 that the computer generated), we can push the odds back to more like 9:1 or so. The average would provide an additional tool for calculating pot odds that might be beneficial. The usefulness of the average would rely, I guess, on the steepness of the curve. I'd love to hear the thoughts of the experts on this.

[Solami17]

I am pretty sure someone just posted this same idea not too long ago. Try the search feature

You assume all unseen cards as equivalent. Yes, some would be in the muck or burned. But that doesn't change the basic odds that are outs/number of unseen cards. Not to mention the deviation would be so high its absurd to use as an average.

[Kurn, son of Mogh]

If you hold A4 to a 9 K 3 flop, the traditional pot odds to make your flush are about 5:1.

1) the odds against making your flush on the turn are 38:9 or 4.22:1

2) In the example above, you have 9 outs to a flush, 1.5 outs to your A (SSHE says discount overcard outs by 50%), and a backdoor str8 draw (worth maybe 0.5 outs). Thus by typical calculations, you have the equivalent of an 11 out hand.

3) The only time discounting flush outs even remotely makes sense is when you have 3 of the suit on the board and you hold a draw to the one card nut flush in a multiway pot. While you cannot assume all of your opponents hold at least one of the suit, you can assume almost all of them do (you should be able to infer from the action if someone already has a made flush). Usually in these situations the pot is big enough to warrant calling anyway.

I think that you gain very little advantage from your approach. besides, if you're going to subtract the good dead cards from the calculation, why not subtract the bad dead cards as well. So if you assume 3 dead flush cards in 8 dead hands, shouldn't you also assume 13 dead non-flush cards in those same hands, making you river odds 25:6 or 4.2:1?

In other words. Don't overthink the problem. An unseen card is an unseen card is an unseen card.

[CORed]

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I think that you gain very little advantage from your approach. besides, if you're going to subtract the good dead cards from the calculation, why not subtract the bad dead cards as well. So if you assume 3 dead flush cards in 8 dead hands, shouldn't you also assume 13 dead non-flush cards in those same hands, making you river odds 25:6 or 4.2:1?

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You've hit the nail on the head. Your suggested approach is correct, and will give you the same results as treating all unseen cards as equivalent. The OP has a fundamental misunderstanding of probability.

[soko]

Everything thats been needed to say has already has been said, I'm just going to repeat it

If the cards you need are distributed equally throughout the deck and opponents hands, so are cards that will not help you, keeping the odds the same.

Oh, I see. I knew that I was figuring the problem wrong somewhere, I just didn't know where. Thanks for the responses, you guys cleared it up for me.

[winky51]

Well there is also something else.

If you know what wat of your opponents hold then your chance to hit in some cases is actually greater.

Say for example there was a raise by a rock and the action on the flop with all low cards makes it appear this rock had AA-QQ. Lets say you have a low end OE straight draw to the nuts and its rainbow. Don't worry about preflop calls just follow the logic plz. Lets say you have 76s and the flop is 54T rainbow. If you are sure the rock has a big pair like AA-QQ then those are 2 known cards and your % chance to hit goes up. Its not much maybe 2-3% but it does go up.

[BasketballNYC]

There is one area where the initial poster is correct. Not with flush draws but with straight and overcard draws. SSHE makes the point that overcards should often not be treated as full outs, but not because they are in other peoples hands or been folded but because you may hit and still be behind to two pair or a set (not a problem with the nut flush draw).

However, if I have AQ and raise preflop and between limpers and cold callers there are 6-7 of us to the flop which comes all rags, do I have a full six outs? (assuming I need them) I doubt it. There may not be an Ace left in the deck to be dealt. I understand the theory of unseen cards, but logic would dictate that if that many people are seeing the flop, some of my aces are dead. This would not make me say that I have no outs, but my A outs should certainly be discounted. This also applies to Broadway draws. If I need an A or 9 to make my hand and we have as many players as I stated before, I cannot assume that all 4 of my aces are live again. I would value this draw at perhaps 6 outs instead of 8.

[Xhad]

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I am pretty sure someone just posted this same idea not too long ago. Try the search feature

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Yeah if poker theory ever gets a FAQ this should be one of the questions; it's posted every other week at least.

To the OP: Average spades dealt to 9 random hands and average spades in your opponents' hands those times you flop a spade draw are not the same thing. The second number is lower. Also, the traditional calculation does not assume your cards are live, the usual calucalation on the turn is 9/47 where 47 is the number of cards you haven't seen, including those in everyone else's hands.