I need a little help with this. Are hand percentages figured on a full table(10players). I have looked around for an asnwer to this question and havent been able to find an asnwer.

Say for example you have a flush draw after the flop that gives you app. 37% chance of making your draw by the river.

No. Since you are unaware of any cards but your own, no other cards or number of players are taken into effect. Only your two hole cards and the cards to come on the board.

Thank you for your reply.

I dont understand how the percentages can be anywhere close to correct. Seems like it gives you too many variables to be accurate. I am not a math or theory whiz. I realize that the percentages are excepted and used by people with a much better understanding than I have , I'm just trying to get a better understanding of the game.

hi, welcome.
I am a math wiz
Well, at least as far as Probability.

The number of players on the table will not impact your flush draw chances of hitting. Now, of course, the ACTION during the hand might give you clues that other player(s) are drawing to the same flush. But in terms of simple probability, 10 handed, 3 handed, 16 handed, you still have the same chance to hit the flush.

Why?

Well, forget the math for now.
Let's just use common sense.
When you make a flush with 2 suited cards in your hand and 2 more on the flop, and one more on the turn or river, that occurred because prior to the deal the cards were in such an order as to end up that way. Say you are at an 8 handed table, and you are on the button.

You will receive the 8th and 16th cards from the deck.
The flop will be the 18th, 19th, and 20th cards
The turn will be the 22nd card
The river will be the 24th card.

So if at least 5 of those 7 cards are the same suit, you made a flush.

Now, stare at this for awhile, and forget about "well, but it 10 players are in the hand, more of my outs are gone" Just look at what I wrote above and let it sink in. It will hit you.


Ok you want math?

you simply sum across: ( the conditional probabilities that there are 0 to 9 of your outs in other players hands times the chance that given those cards in their hands, you catch the flush)
What you'll find is the the sum of this, is exactly equal to the assumption that all 47 cards are live after the flop. Because while sometimes the players have more of your outs than normal proportion would indicate, sometimes they have less.


I should get paid for this.

good answer Actuary, I wonder how long he will stare at it before it sinks in

The way I found it easy to explain it is, say, at a 10 handed table, we know that say 5 of our flush outs are distributed in the 9 mucked hands, for example, as they all folded face up, for whatever reason. So, that leaves only 4 flush cards left, and (52-23) 29 cards left in the deck that are unseen. This gives us a 13% chance to hit the flush on the next card, which is lower than the standard odds.

Now, say for example we know that no one had one of our flush cards, so now there's 9 left out of the 29 cards left in the deck. Now we have a 9/29, 31% chance to hit it on the next card, which is more than the standard odds we use.

All in all, the times that your flush cards are gone, and then time that they are all left in the deck will be normally distributed, meaning it all evens out. And, it all evens out to end up being 9 flush cards remaing, out of 47 unseen cards, as we haven't seen the other 9 hands now. 9/47 = 19%, which are the odds that we will hit the flush on the turn.

I suck at explaining in math terms, but this may make sense to those who are not as mathematically inclined as Actuary.

In my mind I understand what you are saying. Or the odds of the card you need coming out is equal to cards being in the correct order after the shuffle no matter the number players playing. Im thinking that I reworded that correctly.

Anyway, I didnt stare too long. Thanks for help. I think you explained it well.

Here's a simple way to look at it:

If, after the flop, you've got four spades to a flush, and you've seen 5 of the 52 cards in the deck, the remaining 9 spades are out there in the remaining deck to be dealt or the other players' hole cards. Don't worry about where they are. The 9 remaining spades are in the 47 cards that you haven't seen yet, and 2 of the 47 cards will be dealt on the turn and river.

To approximate your % to win the hand after the flop, multiply your # of outs by 4. If you hold the 2 3 of spades, and you put the other player on a high pair, you can multiply the 9 remaining spades by 4 to give you a roughly 36% chance of making a flush.

To approximate your % to win the hand after the turn, multiply your # of outs by 2. If you hold the 2 3 of spades, and you put the other player on a high pair, you can multiply the 9 remaining spades by 2 to give you a roughly 18% chance of making a flush.

The %'s are not exact, but they're close enough to get a good idea where you stand.

you got it.
logic should tell you that there will be 7 total cards in play for your hand: Your 2 plus the 5 on the board. There's no reason to think having more or less players at the table would cause those 7 cards to results in a pair/flush/boat/K high/two pair/ whatever more often.

*********************

Zach,

it's binomally distributed actually, probably a normal approximation would work though. And it's the weights of these probabilites that you use to multiply the corresponding flush odds, then sum all up. I know you know, just adding on here.


Sum of

Probability 0 Flush Cards Left in Deck given N opponents * Probability hitting flush with 0 flush cards left in deck

+

Probability 1 Flush Cards Left in Deck given N opponents * Probability hitting flush with 1 flush cards left in deck

+

Probability 2 Flush Cards Left in Deck given N opponents * Probability hitting flush with 2 flush cards left in deck

+
etc.. all the way to 9 cards possibily left, assuming we have 4 to the flush post flop

I have always used the system Dane described. It's the easiest by far.

Bleh, I told you guys I started sucking at math lately... I was going to just say randomly distributed, but thought I'd sound cool if I said normal...lol

Why multiply by .02 when it's so easy to divide by 46 and get a more accurate answer?

while the rule of 2 or 4 is certainly good to know/use, I don't think that was the point of the OP. He was interested in understanding the impact of more players in the hand vs less.