Flush Outs (the Cautious way)

[Rallenkov]

In every book and in every Outs Chart I see people count 9 Outs on a flush draw.
My Question is, would it not be better to discount 1 Out for every 2 Players.

Theoreticaly every 4th card should be of the same suit as our hole cards (a little less since we already see 4 cards of that suite but since this is the cautious way we go with every 4 for this example).

[pzhon]

[ QUOTE ]
In every book and in every Outs Chart I see people count 9 Outs on a flush draw.
My Question is, would it not be better to discount 1 Out for every 2 Players.

[/ QUOTE ]
This is a common question. The answer is no.

If you have no information about what they are holding, the possibility that those players are holding cards of your suit is precisely balanced by the possibility that they are holding cards of other suits.

Suppose you are trying to draw a spade from a full, freshly shuffled deck. You have 13 "outs" in 52 cards. But wait! What if the bottom card is a spade? There is a 1/4 chance of that, so you might decrease the 13 by 1/4. You can only do this if you decrease the 52 by 1, and (13-1/4)/51 = 13/52. The 1/4 chance that you are drawing 12 out of 51 is precisely balanced by the 3/4 chance that you are drawing 13 out of 51.

[Pooter]

What about discounting the out that will pair the boared?

[qpw]

One way to convince yourself that you should not alter your expectation on the basis of unknown cards dealt to others is to take the situation to the extreme where it is transparantly obvious what the situation is.

Suppose you hope to get dealt the ace of spades from a properly shuffled fair deck.

You know that if you pick a card at random, you will have one chance in 52 of getting the one you hope for.

Now imagine that there are 51 other people in the room plus a dealer who deals one card to each person.

The other 51 get their cards. What is the probability you will get As?

Obviously exactly the same as it was before: 1 in 52.

From that you should be able to see beyond any doubt that cards dealt to others (and not known) are irrelevant.

[PantsOnFire]

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What about discounting the out that will pair the boared?

[/ QUOTE ]
This is valid. However, you probably shouldn't discount it fully unless there is 100% chance that card will make him a full house. Since 100% is unlikely, you might want to go with 50% and even that is conservative. So your flush draw is now 8.5 outs which really doesn't change much except for the closest of decisions. And in NL poker, if the river is a flush card that doesn't pair the board and villain has a set or two pair, you will likely get some payoff as implied odds so I don't see any reason to discount this out at all.

Discounting two straight outs that would three flush the board is probably more common. Even in that case though you would still probably do a 50% so you would go from 8 to 7 outs which may affect your decision.

I think discounting an out or two is only really applicable in limit holdem where math rules. In NL holdem, you can easily make up the difference with implied odds. The only time I use discounted outs in NL is when I'm facing an all-in and trying to decide if I have the right odds to call.

[iveyleague24]

<font color="blue">Warning: Math </font>

On the flop, there are two more cards to come. By that point, the deck has been reduced to 52 - 3 - 2x cards, where x is the number of players participating including yourself. The average amount of flush outs that were dealt to the opposing players can be represented by (9/47)2(x-1)
This leaves 9 - (9/47)2(x-1) flush outs remaining in a deck of 52 - 3 - 2x cards, or a

[9 - (9/47)2(x-1)]
------------------ * 100%
[49 - 2x]

chance of hitting your flush on the turn.

You can verify that this expression is equivalent to 9/47 (the traditional odds of hitting your flush on the turn) for all x. (Even 23). The same principles apply on the turn...etc etc.