Tell me I\'m wrong
 

I was talking to a guy who didn't seem to know a lot about poker but like to think he does. (You know the type) Well, we started talking about odds and he said you have to consider the other peoples cards(unseen to you) when figuring out your odds. Well this goes against everything I've ever read. I kind of wrote him off but later I was thinking about it and something wasn't setting right with me. So I tried to work out the math and this is the example I worked out.

Say you have a flush draw on the flop with two cards to the flush in your hand. Normally, your looking at 9 outs and 47 unseen cards which works out to about a 19% chance of hitting your flush on the turn. But here is what I tried.

You have 2 cards out of the 13 suited cards in your hand. (13-2=11)
And 2 cards out of the deck. (52-2=50)
If on average the rest of the suited card are distributed evenly through out the deck and there are 9 other players at the table then on average 4 of your suit will have been dealt out. (11/50*9*2 = 3.96 ~ 4)
11 of your suit, 50 cards in the deck, 9 players, 2 cards each.
So now 3.96 gets subtracted from the 11 and now there is only 7.04 of your suit, but there is also 18 more cards off the deck. So to update: 7.04 of your suit and 32 (50-18 = 32) left in the deck. Now the flop comes and there's your flush draw. 2 more cards get subtracted from your suit and 3 from the deck. Update: 5.04 of your suit and 29 left in the deck. Now your chance of making the flush on the turn is 5.04/29 = 17%

Now, I know what your thinking. All that work for a 2% difference. And I agree, you still should be looking for about 4 to 1 on your money for ok pot odds, but there still seems to be a discrepancy in the numbers. What do you think?

Re: Tell me I\'m wrong
 

All of the work you've done does not change the fact that on the flop there are 47 UNSEEN cards and 9 of them are your suit. That is that.

Vince

Re: Tell me I\'m wrong
 

You're wrong. (but I only said it 'cuz your title told me to)

Reverse implied odds - dude. You have to factor in the % of times that you will hit your draw - and still lose.

a couple examples:

a) Board: 4 h 5 h 6s
You have: Ac 7c
You have 8 outs to win the hand w/your straight...but only 6 outs if your opponent has a hand like Ah 6h.
b) Board: 4 s 5s Jc
You have: 8 s 9 s
Opponent: As Jd
You have 8 (not 9) outs for your flush = BUT you need to hit exactly ONE spade to win.

c) Board: 2c 8c 2d
You have: K c Q c
Opponent: A s 2 s
You can win with any club - but you need to avoid the last 2, any 8, ace or the pairing of the turn card on the river.

d) You have A s Q d
Board reads: K d Qs 5 h
w/all the betting goin on you think you figure you're behind and need to catch up...but how do you know how many outs you have?
If your opponent has A K you have TWO outs.
KJ gives you 5 outs
55 means you need runner runner for full house
KK means drawing dead.


Coolio?

Re: Tell me I\'m wrong
 

not sure if this is helpful. but not just reverse implied odds. but often you have to discount outs as you aren't really sure if they'll win for you.

you may have had nut flush or nut straight example where it's straight-forward.

but most times you have to give some consideration to your opponents cards... and in the nut flush example, what if your opponent has 2 of the suit too, so there are less in the deck for you to hit

Re: Tell me I\'m wrong
 

Yes, you are wrong. I'm too lazy to comb over your math too thoroughly but I did notice one elementary mistake: you can't assume your opponents have a normal distribution of "your" suit in their hands if two of your suit hit the flop. The fact that you caught your flush draw at all makes it slightly less likely others were dealt that suit, though I don't know quite how much it matters.

Intuitive explanation: It does not matter where the cards are that you have not seen, only that they exist. Look at it this way; would the calculation be any different if the dealer burned 19 cards and then dealt the 20th as opposed to just dealing the card off the top of the deck? Obviously not. So why does it matter if the first 18 cards went to other people instead?

Mathematical explanation: I started to do this but it got hairy and I don't like posting things if there might be errors if I'm tired. I'll post it tomorrow if you want.

Re: Tell me I\'m wrong
 

Xhad, thank you for your thoughtful intuitive explanation. Whether the dealer burns x cards, or puts x cards into the opponents’ hands, the chances of making the flush on the turn are the same.

Unless I’m in error, the OP demonstrated a somewhat surprising conclusion. If the flush cards are normally distributed, the odds against making the flush on the turn are affected. As the OP said, "Tell me I’m wrong."

Re: Tell me I\'m wrong
 

You're wrong because if your outs are evenly distributed throughout the deck and your opponents hands, so are junk cards that will not improve your hand. therefore, if you do the math you will come up with the same odds of hitting your draw if all your opponents cards are in the muck or shuffled back in to the deck, just as long as you don't know what the cards are.

Re: Tell me I\'m wrong
 

[ QUOTE ]
You're wrong because if your outs are evenly distributed throughout the deck and your opponents hands, so are junk cards that will not improve your hand. therefore, if you do the math you will come up with the same odds of hitting your draw if all your opponents cards are in the muck or shuffled back in to the deck, just as long as you don't know what the cards are.

[/ QUOTE ]
Soko, that’s what I thought: It’s intuitive. However, can you find an error in the OP’s math? I couldn’t.

Re: Tell me I\'m wrong
 

[ QUOTE ]
[ QUOTE ]
You're wrong because if your outs are evenly distributed throughout the deck and your opponents hands, so are junk cards that will not improve your hand. therefore, if you do the math you will come up with the same odds of hitting your draw if all your opponents cards are in the muck or shuffled back in to the deck, just as long as you don't know what the cards are.

[/ QUOTE ]
Soko, that’s what I thought: It’s intuitive. However, can you find an error in the OP’s math? I couldn’t.

[/ QUOTE ]

Here is the error: OP assumes that your opponents are dealt an average number of cards in "your" suit. However, you will be more likely to flop a flush draw those times your opponents are dealt fewer cards in your suit than average.

If OP wants to fix his math he's going to either need to apply Bayes' Theorem, or deal the flop first then determine the average amount of cards that appear in his opponents' hands.

Re: Tell me I\'m wrong
 

[ QUOTE ]
Unless I’m in error, the OP demonstrated a somewhat surprising conclusion. If the flush cards are normally distributed, the odds against making the flush on the turn are affected. As the OP said, "Tell me I’m wrong."

[/ QUOTE ]

Part in bold is the faulty assumption. You are more likely to flop a flush draw those times that your opponents are dealt fewer of your suit than normal.

Re: Tell me I\'m wrong
 

An unseen card is an unseen card is an unseen card. Calculating the probability of which unseen cards are in your opponents' hands is about as useful as calculating where they are in the deck.

Re: Tell me I\'m wrong
 

[ QUOTE ]
You have A s Q d
Board reads: K d Qs 5 h
If your opponent has KK means drawing dead.


[/ QUOTE ]

Coulnt you still catch runner-runner Aces to make a bigger full?

Re: Tell me I\'m wrong
 

My bad.

You have to hit a two outer in order to be able to hit your one outer. (What are those odds? 1 in 100 ? Less? - too lazy to run the odds right now.)

Re: Tell me I\'m wrong
 

A good example of discounting outs:

You have A c K c
Board Reads Jc 7c 2 s
You may think you have two overs and a flush draw for 15 outs putting as the stat favorite - but if your opponent has A d J s....

Re: Tell me I\'m wrong
 

the error in your example is in assuming what was dealt to your opponants; but set that aside.

there are specific times you can nail down with a comfortable degree of certainty that certain card are out of not out. The most obvious example, and maybe the most useful, is in counting the remaining aces. If you are at a loose table where people tend to play any ace, and the opening pot was not raised, then the cards mucked before the flop are unlikely to contain any aces. On the other hand . . when a player who was aggressive before the flop turns passive after a ragged flop like 9-3-2 . . . that is very likely AK or AQ, so if you played 45s and are thinking about hitting your straight, you would probably be wise to count one less ace.

Your post alluded to flush draws. I am not as strong mathmatically as many others, but here is what I worked out a long time ago, I think it is about right.

If: you have been dealt two clubs, and the flop contains two clubs, Then: it is about three-quarters likely that another player was dealt two clubs. Further, there is about a 1 1/2% chance that any given player was dealt two clubs. Obviously you need to observe their actions to decide if you think they really have the goods, but those numbers give you a starting point.

Re: Tell me I\'m wrong
 

[ QUOTE ]
would the calculation be any different if the dealer burned 19 cards and then dealt the 20th as opposed to just dealing the card off the top of the deck

[/ QUOTE ]

This is undoubtedly the correct way of looking at this situation but in your example you are not playing with a full deck. I have been accused of not "playing with a full deck" so often that it makes me wonder if the OP is wrong. The problem with me though is that I'm not usually playing cards when someone accuses me of this. I don't get it.

Vince

Re: Tell me I\'m wrong
 

Well, if you want to get beyond the basics, there's a somewhat arcane way in which your friend is correct. IF YOU HAVE SOME INDICATION WHAT CARDS ARE IN YOUR OPPONENTS' HANDS, then you can use that to change your estimate of probabilities of cards coming out of the stub.

For example, in Omaha 8/b, many more good hands have low cards (e.g. A2xx, preferably A23x or A24x) than high cards (usually all four above a ten). If you're playing O8 with knowledgeable players, you're in late position, and many have limped in, your high hand might go up in value. This is because you reason they wouldn't be in the pot without low cards; therefore, the stub must be rich in high cards, and it's more likely that you'll hit, or at least that they won't and you can steal the pot with a bet.

That's a very advanced application of this concept, and likely well beyond what your friend means. If you know there are 9 spades among 47 unseen cards, you have no indication whether people contesting the pot with you have zero, one, or two spades. After the flop, if there are two spades on board, and several have called a flop bet, then you can make an inference that someone else might have two spades, so you might need to shade your probabilities down slightly. So maybe it's more applicable than I thought. But if you can't put them on more or fewer spades, then the applicability to flush draws is nil. And for that matter, if you limit your inference to hands with many flop callers, then we're only talking about two suited cards out of several, hence probably not changing the probabilities in any appreciable way!

So you definitely don't want to go shading your flush draw decisions very far based on this concept. Leave this for the O8 writers.