Am I overcounting my outs?

[Aaron_]

This is a question that's been lingering in the back of my head for some time now. I finally sat down and quickly worked out the math on paper. (If this has been asked already, please point me in the right direction -- thanks in advance.)

Why don't we discount outs for those that have likely been dealt out to players already?

For example, I have A[img]/images/graemlins/spade.gif[/img]K[img]/images/graemlins/spade.gif[/img] and flopped a 4 flush. Normally, I would count 9 outs for the remaining spades in the deck. But it is not likely that all nine of the spades are available to me. Playing a 10-handed ring game, it is probable that 4.5 of my spades are either being held by someone, or have already been mucked (18 cards dealt out, 1/4 of them are spades = 4.5 spades).

Originally, counting 9/47 cards are spades makes me a 4.2:1 dog to the turn. But assuming that 4.5 spades are no longer available (and adjusting accordingly), I'm a 5.4:1 dog (13-4-4.5/52-5-18 = 4.5/29).

Do we ignore it because the difference is negligible, or did I miss something?

Thanks

No, we ignore it because while some of your spades may have been dealt, so have other cards.

If the cards that have been dealt are proportionally distributed in suits, then the proportion of the spades left in the deck should be the same.

So if 10 hands are dealt, 20 cards, there would be, on average, 5 spades there. But if you're going to remove those five spades from the calculation, then you should not count the probability of drawing one of the remaining 6 spades (two on the table, five dealt) out of 52 cards, but out of 52-20-3 cards, which is roughly the same (~20%).

Hope that helps.

Cheers,
FP

Edit: But there are times when you must start to figure that some of your outs are "gone": If you have read opponents as holding cards that you are looking for. Say you have AhKh, and the board is Kd-Ks-8s-3s and two people are raising, and you read them to have a flush or a flush draw. In this situation, you might want to figure one of your outs to have been dealt (the As), and you're therefore drawing to a (maximum) three-outer (Ad, Ac, Kc), not four.

Edit 2: Before someone points out that there are more outs than three (because the board can pair up again) - that was not the point I was trying to make. Just that a read can reveal that some of the cards you're hoping to catch are likely held by another player.

outs don't depend on how many cards are already dealt, it's just your cards and the board, the rest of the deck is 47 (count the burn cards?)after the flop, right? so, if you subtracted 2 cards from each person, after the flop with a burn card and 9 total players, there would only be 30 cards left in that case, but that's not how outs are determined. i guess you have to assume there are still 47 unseen cards and calculate from there. but hey, what do i know? i'm not very experienced with odds or probabilities, maybe this would be better in the probability forum.
i have wondered the same thing, like, if you assume you opponent has a pocket pair but you need one of those cards to complete a straight...stuff like that..interesting.
i hope someone clears us both up.

[BugsBunny]

[ QUOTE ]
Do we ignore it because the difference is negligible, or did I miss something?

[/ QUOTE ]

You're missing something. The math works out the same either way. You are not looking at 4.5 spades as being likely in the remaining 18 cards that have been dealt. This would be true if you were to deal 18 cards off the top of the deck, but in this situation the spades in the 47 remaining cards are diluted (there's less than 25% spades remaining)

The correct assumption would be:
9/47 * 18 = 3.4468085106 spades in the 18 cards

That changes the rest of your numbers as well.

13 - 4 - 3.4468085106 = 5.5531914894 spades remaining

5.5531914894/29 = 4.2:1 dog, same as originally figured without adjusting for dealt cards.

If we had used your original logic the correct way to do it (which is obviously wrong) would have been:
We've dealt out 23 cards. of those 23 cards 23 * .25 = 5.75 should be spades. We can account for 4 spades so that means there are 1.75 spades left in the remaining 18 dealt cards. 9 - 1.75 = 7.25 spades remaining in the 29 undealt cards left. You can see where that would get us - no place good.

[Aaron_]

Thanks bugs.

[bernie]

You're also forgetting about counting your A or K as outs.

b

[ohnonotthat]

a hand I watched a friend play some years back - NLHE with 5-10 blinds.

*

Note: However much I may know now or THINK I know now I was as green as Kermit back then.

*

Everyone at the table had over 500; most had well over 1,000.

Four players limped and my friend limped on the button with 5-4s.

The BB then made it 40 and the four limpers plus my friend called.

The flop came 9 - 3 - 2 rainbow.

The BB made a bet of ~ the pot and all folded to my friend who called.

The Ace came on the turn and my friend won a huge pot - beating AA.

- I probably wouldn't be telling the story otherwise. :-)

Later over dinner I told him I was suprised at his call preflop but was shocked at his call on the flop.

"Didn't you have a pretty good idea that he had AA", I asked him, "and if he did you only had 6 outs".

He replied, "if I didn't think he had AA I wouldn't have called".

I stared at him with what I'm sure was a very confused look on my face until the lightbulb finally flickered.

*

Sometimes having some of your outs tainted is exactly what you're looking for.

More importantly . . .


I really love that story !