[DaveHNK]

New Year's Resolution was to stop playing like a moron, so I spent time over the holidays reading Theory of Poker, Small Stakes Hold 'Em and HOH. But one thing I can't stop wondering is why in the calculation of odds don't we discount the cards that are already dead?

Consistently, when you have four outs on the turn, you're odds are listed as roughly 10.5-to-1. The calculation is 4/46, right? Four cards to save you and forty-six unseen cards.

But because the deck isn't replenished (once a card is dealt, it is no longer possible to come back out), the 46 figure is incorrect. Rather the number needs to be discounted by the possibility that once of those 4 outs has already been dealt -- either to an opponent, or burned before the flop, turn, or river.

I don't have a good probability text handy , but I seem to recall that given a random distribution, the probability that any one of your outs is dead should be something like [out!/(dead-out)!]*out! and this figure should be divided against the "probability" of a card coming on the river which would be the number of outs divided by live cards, multiplied by the probability that any of the outs is dead to the power of the number of outs.

Aside from the exact math -- which I desperately need to check -- there are two implications: the odds commonly tossed around are probably a bit on the high side, meaning that marginally profitable calls are probably actually losing calls (you should be sure the odds are reasonably in your favor if you're behind), and the odds are probably worse the more people there are at your table. Basically, at a nine-handed table, there are eight more dead cards than at a five-handed table (the two cards dealt to your additional opponents).

In light of the posts on schooling, etc., I'm just wondering why this hasn't been factored in to the calculation of pot odds, since on of the aspects of schooling -- not only that your opponents have a greater chance of hitting on their live junk, but also they may have your needed outs -- is valid whether or not they fold or call at any point in the game.

[mrticsay]

I think there's lots of detailed, wordy writing about this but the explanation is pretty simple- you only know the cards that are exposed to you. The cards you are looking for could be anywhere- another hand, muck, burn or in the deck. Since you don't know where it is, its all the same from an odds perspective- the card(s) you want are in the unexposed cards.

Now, if some cards had been exposed...let's so someone's muck was exposed...then you can consider those cards and modify the odds. Was one of your outs shown, then reduce the numerator? If not, reduce the denominator.

Now, if you felt you had a "read" on someone's hand, then you could choose to adjust the odds as well, but then its a guesstimate.

-Michael

Originally Posted by DaveHNK

New Year's Resolution was to stop playing like a moron, so I spent time over the holidays reading Theory of Poker, Small Stakes Hold 'Em and HOH. But one thing I can't stop wondering is why in the calculation of odds don't we discount the cards that are already dead?

Consistently, when you have four outs on the turn, you're odds are listed as roughly 10.5-to-1. The calculation is 4/46, right? Four cards to save you and forty-six unseen cards.

But because the deck isn't replenished (once a card is dealt, it is no longer possible to come back out), the 46 figure is incorrect. Rather the number needs to be discounted by the possibility that once of those 4 outs has already been dealt -- either to an opponent, or burned before the flop, turn, or river.

I don't have a good probability text handy , but I seem to recall that given a random distribution, the probability that any one of your outs is dead should be something like [out!/(dead-out)!]*out! and this figure should be divided against the "probability" of a card coming on the river which would be the number of outs divided by live cards, multiplied by the probability that any of the outs is dead to the power of the number of outs.

Aside from the exact math -- which I desperately need to check -- there are two implications: the odds commonly tossed around are probably a bit on the high side, meaning that marginally profitable calls are probably actually losing calls (you should be sure the odds are reasonably in your favor if you're behind), and the odds are probably worse the more people there are at your table. Basically, at a nine-handed table, there are eight more dead cards than at a five-handed table (the two cards dealt to your additional opponents).

In light of the posts on schooling, etc., I'm just wondering why this hasn't been factored in to the calculation of pot odds, since on of the aspects of schooling -- not only that your opponents have a greater chance of hitting on their live junk, but also they may have your needed outs -- is valid whether or not they fold or call at any point in the game.

[Ramstone]

Originally Posted by mrticsay

I think there's lots of detailed, wordy writing about this but the explanation is pretty simple- you only know the cards that are exposed to you. The cards you are looking for could be anywhere- another hand, muck, burn or in the deck. Since you don't know where it is, its all the same from an odds perspective- the card(s) you want are in the unexposed cards.

Now, if you felt you had a "read" on someone's hand, then you could choose to adjust the odds as well, but then its a guesstimate.

Yep. That last bit is something of a leak of mine, though its pretty small admittedly. Note to self: I dopnt have nine outs if im chasing a one card flush and he's already made his hand.....

[DaveHNK]

Actually I figured it out just after posting it. What actually happens is that the discount of the probability that your card is dead is off-set by the increased probaility of hitting if your card is still alive.

In the easiest example of one out in a nine-handed game (19 dead cards), the probaility that your out is dead is 19/27. To simplify let's call that 50%, the factor by which we have to discount the probaility that it will hit if live.

But of course that probaility is not longer 1/46, but rather now 1/27, or 26-to-1. Discount that by 50% and you're back with the original 45-to-one (off because of rounding).

Will try and remeber to keep the thought process on basic and not start mulling over four-out problems to begin with next time!

Appreciate your quick feedback!

[DDD424]

Originally Posted by Ramstone

Yep. That last bit is something of a leak of mine, though its pretty small admittedly. Note to self: I dopnt have nine outs if im chasing a one card flush and he's already made his hand.....

Uh oh, I'm confused . If you would have the nut flush when you're chasing the flush, wouldn't you have nine outs?

[Poboy]

I think he means he if he is chasing a 4 card flush and thinks his opponent has a made flush, then two of his outs are in his opponent's hand. Thus he only has 7 not 9.

[jldecarlo]

Originally Posted by DaveHNK

But because the deck isn't replenished (once a card is dealt, it is no longer possible to come back out), the 46 figure is incorrect. Rather the number needs to be discounted by the possibility that once of those 4 outs has already been dealt -- either to an opponent, or burned before the flop, turn, or river.

Let me give a few examples.

Take a freshly shuffled deck. What are the chances that the top card is a K? Clearly 4 in 52. Cut the deck (in effect, throw away half the cards without looking at them). Now, what are the chances that the top card is a K? Still four in fifty-two. Ask someone to fan the cards face-down. Pick one at random. What are the chances it's a K? 4in52 again.

What are the chances the bottom card is a K, remembering that you haven't looked at any of the other cards? 4 in 52.

Now take your suffled deck again. Starting at the top, turn over the cards one by one. Suppose you turn over the first 48 and NONE are K's. Now what are the chances the bottom card is a K? 100%

Suppose that as you are turning the cards over you turn over all for K's before you get to the last 4 cards. What are the chances one of the last 4 is a K? 0%

To beat a dead horse, suppose you take the cards off your deck one by one, but this time you DON'T turn them over and don't know what they are. Do you have any idea what the bottom four cards are? No.

Where the cards are doesn't matter. The only thing that matters is how many cards you know the value of.


Say you're drawing to an inside straight and you have 4 outs. You need a K.

You have two cards - you know what these are.
There has been a flop - you know these cards also.
10 cards were dealt to other players - you don't know what these are.
1 card has been burned already - you don't know what it is.
So 16 cards are in place but you only know the values of 5 of them. Where the other cards are is irrelevent. Only the fact that each of them is equally likely to have a K.


Hope this helps.

L

[jojobinks]

well put, JL.

Originally Posted by jldecarlo

Let me give a few examples.

Take a freshly shuffled deck. What are the chances that the top card is a K? Clearly 4 in 52. Cut the deck (in effect, throw away half the cards without looking at them). Now, what are the chances that the top card is a K? Still four in fifty-two. Ask someone to fan the cards face-down. Pick one at random. What are the chances it's a K? 4in52 again.

What are the chances the bottom card is a K, remembering that you haven't looked at any of the other cards? 4 in 52.

Now take your suffled deck again. Starting at the top, turn over the cards one by one. Suppose you turn over the first 48 and NONE are K's. Now what are the chances the bottom card is a K? 100%

Suppose that as you are turning the cards over you turn over all for K's before you get to the last 4 cards. What are the chances one of the last 4 is a K? 0%

To beat a dead horse, suppose you take the cards off your deck one by one, but this time you DON'T turn them over and don't know what they are. Do you have any idea what the bottom four cards are? No.

Where the cards are doesn't matter. The only thing that matters is how many cards you know the value of.


Say you're drawing to an inside straight and you have 4 outs. You need a K.

You have two cards - you know what these are.
There has been a flop - you know these cards also.
10 cards were dealt to other players - you don't know what these are.
1 card has been burned already - you don't know what it is.
So 16 cards are in place but you only know the values of 5 of them. Where the other cards are is irrelevent. Only the fact that each of them is equally likely to have a K.


Hope this helps.

L

[SpeakEasy]

As I was reading the initial question, I was thinking of saying something like JL said -- examples of why the LOCATION of the unknown cards is irrelevant. Very good examples.

DaveHNK, its important to remembr that the only difference between the cards from mucked hands and the cards still in the deck is their LOCATION. Some are in the muck, some are in the deck, but they are ALL unknown to you. The next card off the deck could be any of the unknown cards.\

Once you get over this theoretical aspect of poker (which is good to question and resolve in your head), you are ready to move on to solid pot odds calculations and improving your game.

[DaveHNK]

Originally Posted by SpeakEasy

As I was reading the initial question, I was thinking of saying something like JL said -- examples of why the LOCATION of the unknown cards is irrelevant. Very good examples.

DaveHNK, its important to remembr that the only difference between the cards from mucked hands and the cards still in the deck is their LOCATION. Some are in the muck, some are in the deck, but they are ALL unknown to you. The next card off the deck could be any of the unknown cards.\

Once you get over this theoretical aspect of poker (which is good to question and resolve in your head), you are ready to move on to solid pot odds calculations and improving your game.

Thanks SpeakEasy and JL, appreciate your additional insights. What I think was throwing me for a loop was that in this case the probability that the card was dead or alive is the same, so the discount is cancelled out by the increased probability of its hitting if it is still alive. In most of the probaility models I've had to deal with, the probability of the various outcomes was different, so you did need to calculate the possibility that your optimal outcome was already ruled out.

But you were definitely right about one thing -- until I got this straightened out, I was having difficulty really focusing on solid pot odds and its own nuances (just some examples of which we had above).