NL10... required reading?

Originally posted by TinoLaan
When you have a flush draw, there's 9 cards in the deck that can make you your flush. Perhaps all 9 hearts were dealt to other players that folded. That is true. But on the other hand, perhaps nobody had a heart at all. There's no way for you to know.

read it loads of times mate....... and it still doesn't change the fact that if you deal 1/4, 1/3, 1/2 of the cards in the deck then over time that proportion of any outs gets dealt too? there will be times when you either have much better or much worse odds and you will never know which...... but the middle ground still make the assumption of 1:4 for a flush and around that for a straight draw will still be way out for a standard full ring game?

Yeah thanks tino for continuing the explanation.

@chubby yeah it becomes too difficult when we think about what the people that folded could have but i agree that it is a relevant consideration thinking about the cards in other players' hands. I agree that we have to work with the information that we have. At this stage I don't disagree with your reluctance to chase draws because of the outs that you are discounting in other players' hands.

It makes sense to me to think that the chances that the other players on the table will be taking your outs is the same chance that they have of taking the outs of the other player and blocking their precise hand so these values most likely will balance out. I still advocate the 4:1 for the flush etc but i sympathise with your reasoning.

Regards,
Luke

Just reposting this here in case someone else finds it useful as well.

Thread: ChubbyTrucker

Originally posted by TinoLaan

Let's use a thought experiment here, lets say we are playing a game with 4 people, and we have a 4 card deck. A K Q J, if we deal everyone a card, what are our chances of getting the A if we are dealt the card first. Well 1/4, that is simple. But now if we are dealt last do our odds of getting the A change? Much like the flush question we would have more cards dealt out
before we got to draw for the A. Well using simple probability what are the chances of someone else drawing the A? Player 1 has 1/4 + and player 2 has 1/4 + and player 3 has 1/4 = 3/4, so 3/4 times someone else gets the A, leaving us with 1/4 of getting the ace. Easy right? Notice how our odds are off the full deck, not just the remaining deck? This can be verified easily at home.

Just using this example because it's the easiest.

This is a bit of a follow up to what ChowChow and Tomaloc said, about how it averages out. Let me explain.

The odds of drawing to the A are the same for every player, despite who gets dealt the first card. Say you are first to get dealt a card. You have a 25% chance of getting the A. Now let's say you're second to get dealt a card. So are your chances of getting the A bigger now? Or smaller?

Let's apply a simple formula. We know that 25% of the time player 1 gets dealt the A, meaning 25% of the time we have no chance of getting the A. The other 75% of the time we have a 33% chance of getting the A.

Add that, and we get the following formula:

.25 * 0 + .75 * .33 = 0 + .25 = .25

In other words, the odds of drawing to the A are always 25%.

This same logic applies to the more complicated case of a draw in hold em.

Now to prove the point further, let's also do this for when we're third to act (in spoiler)

Spoiler

We're third to act, and we want to hit the ace. Now, there's three things that can happen before us.

1: Player 1 gets the A
2: Player 2 gets the A
3: Neither player 1 nor player 2 gets the A

Case 1: This is really simple. There's four cards in the deck, so there's a 25% chance of player 1 getting the A.

Case 2: For player 2 to get the A, means player 1 didn't get it. So player 2 has a 33% chance to get the A. But we also know that 25% of the time, player 2 can never get the A.

Case 3: There's a 75% chance that player 1 doesn't get the A. If we know player 1 doesn't get the A, there's a 67% chance player 2 won't get it. If this happens, there's a 50% chance we (player 3) get dealt the A.

Now to put this in a formula. It's basically the same as when we got dealt second.

Case 1: 25%
Case 2: .75 * .33 = .25 = 25%
Case 3: .75 * .67 = .5 = 50%

So now we conclude that 50% of the time player 1 or 2 gets the A, and 50% of the time there's a 50% chance we get the A. In mathematical terms:

.5 * 0 + .5 * .5 = 0 + .25 = .25 = 25%

So if we're third to act, there's still a 25% chance we get dealt the A.

Hopefully this little mathematical analysis helps you understand that regardless of the fact that other players have cards, your chances of drawing are always the same.