[d_p]

OK, this has been bugging me for some time. (I'm sure this is not anything new and it has been discussed to death a million times.) Is there a flaw in the way we are told to calculate outs?

Let's say you are playing THE against 8 other players. You have:
in the hole
and the board is:
.

How many outs do you have?

Conventionally we are told that we have 9 's left in the deck and there are 46 unseen cards, so you have a 19.6% chance of drawing a spade. So we need pot odds of about 4 to 1 to make a correct call.

Welllll..... it can be argued yes and no. Correctly speaking there are 46 unseen cards and there are 9 spades left among those cards.

BUT 16 of those 46 cards are either in-play or have been mucked. (Another three have also been burnt, but I won't go into that just yet.) That means there are only 30 cards that have not been dealt.

Now, again correctly speaking, of those 16 cards out-of-play, the expected number of 's is 1 in 4, or the expected number of 's left is actually 5, not 9 which gives you only a 5 in 30 chance of drawing a or 16.7% which is 3 percentage points less of a chance to draw a flush.

OK, now let's get back to the burn. Your expected number of spades in the three burn cards is .75. Now you only have 4.25 in 27 cards or a 15.7% chance to draw a flush on the river. That's 4 percentage points less than the "conventional" way we are told to calculate outs. In other words, your chance of drawing a has been reduced by 20% following this logic and you actually need 5 to 1 pot odds to make a correct call.


What think? How has this line of reasoning been de-bunked?

[scottwire]

The fact that cards are out means nothing unless you know for sure that someone has those spades.

There are 46 unknown cards. What difference does it make whether the cards are in people's hands or in the deck.

Either way they are randomly distributed and 9 of those 46 are a spade.

[d_p]

OK actually only two cards have been burnt, altering the percentages at the end are 4.5 of 27 which means the burn cards really don't affect the whole thing that much.

[d_p]

Originally Posted by scottwire

The fact that cards are out means nothing unless you know for sure that someone has those spades.

There are 46 unknown cards. What difference does it make whether the cards are in people's hands or in the deck.

Either way they are randomly distributed and 9 of those 46 are a spade.

"Randomly distributed" is precisely why it makes a difference. 16 of those cards have zero chance of being dealt to you. If the cards are in fact randomly distributed as you say, there are 4 's among them meaning you have zero chance of getting those 4 's.

[scottwire]

Sometimes all 9 spades will be in your opponent's hands.
You should never draw to a flush again.

If you break out the percentages you will get the same distribution over between your cards out and the cards left in the deck. The reason you see this as being a different outcome is that you are using whole cards rather than fractions of cards in your distributions.

[Matthew]

Originally Posted by scottwire

The fact that cards are out means nothing unless you know for sure that someone has those spades.

There are 46 unknown cards. What difference does it make whether the cards are in people's hands or in the deck.

Either way they are randomly distributed and 9 of those 46 are a spade.

Bingo.

Another way to look at probability. Shuffle a 52 card deck - what's the probability the top card is the A ? 1 out of 52 possibilities. Now shuffle the deck, burn the top card - what's the probability the next card is the A ? 1 out of 52.

In order for the 2nd card to be the A the 1st card can't be it, so there are 51 /52 it isn't the A . There are 51 cards left now so the probability the 2nd card is the A is 1 / 51 - multiply the 2 events together and the probability the A is the 2nd card is 1 /52.

Odds are always calculated using the number of cards known and unknown regardess of whether they are in play or not.

[SpaceMonkey]

[quote="d_p"]

Originally Posted by scottwire

"Randomly distributed" is precisely why it makes a difference. 16 of those cards have zero chance of being dealt to you. If the cards are in fact randomly distributed as you say, there are 4 's among them meaning you have zero chance of getting those 4 's.

Sorry, you're way off base here. The burn cards and your opponent's cards are unknown. You only know what 6 out of 52 cards are in play. That leaves you with the 9 remaining spades, etc.

But by all means feel free to fold your draws when getting "conventional" odds.

[99%evil]

An easier way to look at it, it either will hit or it won't.

[jojobinks]

Originally Posted by 99%evil

An easier way to look at it, it either will hit or it won't.

um...okay.

but back to the discussion at hand. either you've thought about this in a way that no poker writer ever has, or you're just wrong. i vote for the latter.

sometimes all 9 are live, sometimes they're not. you can't know that or account for it. you can't pretend you can account for it. so what we know is how many unseen cards there are, and how many outs you have in those unseen cards.

what we didn't discuss is what your spades are. are you drawing to the nuts here? if not, then you have to discount your 9 outs after all.

[jojobinks]

Originally Posted by Matthew

Bingo.

Another way to look at probability. Shuffle a 52 card deck - what's the probability the top card is the A ? 1 out of 52 possibilities. Now shuffle the deck, burn the top card - what's the probability the next card is the A ? 1 out of 52.

In order for the 2nd card to be the A the 1st card can't be it, so there are 51 /52 it isn't the A . There are 51 cards left now so the probability the 2nd card is the A is 1 / 51 - multiply the 2 events together and the probability the A is the 2nd card is 1 /52.

Odds are always calculated using the number of cards known and unknown regardess of whether they are in play or not.

break it down, my brother.