Calculating flush odds MY WAY.

Wow thanks for replies. I'm going to rereread this stuff.

[tpir90036]

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discuss...

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you can't just make stuff up until the numbers look right.

[OrianasDaad]

I asked basically the same question some time back in the probability forum. A search may bring it up. The math that I learned that week on this subject moved my game up a notch. Mabye two.

[binions]

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10 handed. 10/20 cent limit. u get a5 of spades. 5 calls = pot total 50 cent.

SB bets > 3 folds. You have 4 flush. the odds of ending up with flush at the showdown are according to D. Sklansky are 1.86 : 1.

My version: 20 cards were dealt out preflop. quarter of them or 5 cards will be in the long run spades. u picked up two. Lets assume the other 3 spades will be folded. and two other on the board.

Shoulnt you calculate the chance of flush at the showdown like this:

13 spades total. u have two and three were folded, 2 spades on board. 6 spades in the deck over. so 6/47 * 5/46 + 6/47 * 41/46 + 41/47 * 6/46.

so odds of ending up with flush are 3.09 : 1

discuss...

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Discuss? Are you serious?

Every other week, some lamebrain posts the exact same thing. What about the discards? There must be only 7 spades left in the deck. Yada Yada.

I guess it's a good sign.

I have a similiar question...

Is it ever common practice to take a away an out or so if you read another opponent for a flush draw in a 3 way pot?

A hand I played recently illustrated this. Essentially, I had the Ace of clubs and the flop was 3 low clubs. Early position bet and on the flop and turn both another player before me called and I called. If my read on the player before me is right and I believe he has the K or Q of clubs, should I ever take that into account when determining my outs? Or is it too likely that your read is simply wrong for it to have an effect on what you believe your odds of hitting the flush are?

Flame away if this isn't the right way to think about it.

[binions]

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Is it ever common practice to take a away an out or so if you read another opponent for a flush draw in a 3 way pot?

A hand I played recently illustrated this. Essentially, I had the Ace of clubs and the flop was 3 low clubs. Early position bet and on the flop and turn both another player before me called and I called. If my read on the player before me is right and I believe he has the K or Q of clubs, should I ever take that into account when determining my outs? Or is it too likely that your read is simply wrong for it to have an effect on what you believe your odds of hitting the flush are?


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Whenever you have enough info to put someone on a range of hands, you discount your outs according to that range to come up with your effective outs.

If you have a naked Ace on a 3 flush board heads up, you either have 7 outs (if you put the other guy on a made flush), or 8-11 outs if you put him on the K of the suit. You might have as many as 11 outs if your ace is live and he doesn't already have a flush. (This assumes your Ace high is behind. You might be ahead.)

And so, instead of 47 and 46 unseen cards, you would use 45 and 44 unseen cards if you have discounted your outs by your foe's range of hands.

For a discussion about using 45 and 44 unseen cards, see the Appendix in Barry Greenstein's Ace on the River.

[damaniac]

Nah this is an important concept. Someone posted a hand in SS a week ago where, for example, you had a gutshot and needed a 3 and a 3 only to win. Preflop, two people who you knew had just read and were applying SSHE limped in early position, and another TAG raised after them. The limpers were gone by the turn. The TAG bet, a fish called, and you were next to act getting 8:1. Normally this would be a tough call, needing to average 3 BB's on the river to make this even slightly profitable. But given that you essentially know that neither the raiser nor the limpers have a 3, your odds are 9:1, and it is an easy call.

Granted, you really don't want to take this too far. Just because someone is calling all along, doesn't mean they have a flush draw, for example. But when you know the players well, the action can help limit their holdings and allow you to more accurately calculate your odds.

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10 handed. 10/20 cent limit. u get a5 of spades. 5 calls = pot total 50 cent.

SB bets > 3 folds. You have 4 flush. the odds of ending up with flush at the showdown are according to D. Sklansky are 1.86 : 1.

My version: 20 cards were dealt out preflop. quarter of them or 5 cards will be in the long run spades. u picked up two. Lets assume the other 3 spades will be folded. and two other on the board.

Shoulnt you calculate the chance of flush at the showdown like this:

13 spades total. u have two and three were folded, 2 spades on board. 6 spades in the deck over. so 6/47 * 5/46 + 6/47 * 41/46 + 41/47 * 6/46.

so odds of ending up with flush are 3.09 : 1

discuss...

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Discuss? Are you serious?

Every other week, some lamebrain posts the exact same thing. What about the discards? There must be only 7 spades left in the deck. Yada Yada.

I guess it's a good sign.

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"Lamebrain"? I understand the veterans here have seen many of these posts before and can get tired of them. What I don't understand is why you don't either: A.) Answer it for the new guy who is trying to take a thinking approach to the game and came up with a question. or B.) Just skip replying. If you don't have the ability to explain a simple point, then just go back to your reading and leave this question to be answered by others. Everyone knew what this post was going to be by the subject, so if you don't have anything useful, just move on.
Sorry for the rant.

[AaronBrown]

As Thinkquick and several others pointed out, you have to account for both the spades and non-spades among the other pocket cards. If you do it correctly, you get exactly the same answer.

The usual calculation is to say there are 9 spades among the 47 unseen cards. The chance of getting a non-spade on the turn is 38/47, and if that happens, the chance of getting a non-spade on the river is 37/46. Multiply them together and get 1,406/2,162. This is your chance of not getting a flush. Subtract from 1 to get the chance of a flush, it's 756/2,162 = 34.97%.

To do it your way, you have to consider how many spades are in the other players' hands. The table below shows the possible numbers of spades from 0 to 9, the probability of that many spades being in the other players' hands and your chance of making the flush given that many spades in the other players' hands.

0 0.735% 53%
1 5.670% 48%
2 17.525% 43%
3 28.446% 38%
4 26.668% 32%
5 14.934% 26%
6 4.978% 20%
7 0.948% 14%
8 0.093% 7%
9 0.004% 0%

If you multiply the last two columns together and add them up, you get exactly the same 756/2,162 = 34.97% as the standard calculation. Doing it your way is handy for some purposes, but unless one of those applies, it's a longer way to get to the same result.

Thanks