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  #1  
Old 11-04-2003, 08:19 PM
magithighs magithighs is offline
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Default Math behind flush odds

I'm trying to figure out the odds for the flush draw (2 flush cards on the board and two in hand) to remember the principles of the calculation.

It's been a long time since I studied probability, but I remember that when calculating an "and" situation you need to multiply the odds of the events, and when calculating an "or" situation on two mutually exclusive events you add the probabilities.

Are these mutually exclusive events -- the flush card on the turn "OR" the river. So do you calculate the odds on the turn, and then add the odds on the river?

I'd like to know the math behind this -- not necessarily the answer, so I can better calculate "true" odds of events happening.

Many thanks for your help.
Magithighs
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  #2  
Old 11-04-2003, 09:25 PM
BruceZ BruceZ is offline
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Default Re: Math behind flush odds

It's been a long time since I studied probability, but I remember that when calculating an "and" situation you need to multiply the odds of the events, and when calculating an "or" situation on two mutually exclusive events you add the probabilities.

Are these mutually exclusive events -- the flush card on the turn "OR" the river. So do you calculate the odds on the turn, and then add the odds on the river?


Good question. Getting a flush card on the turn is not mutually exclusive of getting it on the river because you can get it on both. If you simply add these probabilities, you will double count the times you make it on both.

These events are also not independent because whether or not you make it on the turn changes the probability that you make it on the river. You cannot multiply these probabilities either. However, there are both mutually exclusive events and independent events that you can use to solve this problem.

The event of getting a flush card on the turn is mutually exclusive of the event of getting a flush card on the river given that you did not get a flush card on the turn. These probabilities can be added.

Also, the event of NOT getting a flush card on the turn is independent of NOT getting a flush card on the river given that you did not get one on the turn. These probabilities can be multiplied to get the probability of not getting it on either, and then subtracted from 1 to get the probability of making the flush.

This post gives 3 methods. The first uses independence. The second uses mutually exclusive events. The last adds probabilities of events that are not mutually exclusive, and the subtracts the double counted events to arrive at the right answer.
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  #3  
Old 11-04-2003, 09:37 PM
DPCondit DPCondit is offline
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Default Re: Math behind flush odds

P = 1 - ( 38/47 x 37/46 )
( 1 - P ) / P = odds to one against making it by the river
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  #4  
Old 11-05-2003, 05:02 PM
GuruCane GuruCane is offline
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Default Re: Math behind flush odds

I'm confused. How do you ultimately arrive at 1.85 to 1 for a flush draw when you have 9 outs? After the flop, isn't it 38/9 with 47 unseen cards? When calculating pot odds, isn't it then 4.222 to 1 on a flush draw after the flop as opposed to the 1.85 to 1 that you get by using the calculation that David Sklansky (and the prior poster) has given us? The 1.85 to 1 result would almost never make it prohibitive to call on the come. I would greatly appreciate your insight as this has bugged me for a while. [img]/images/graemlins/confused.gif[/img]
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  #5  
Old 11-05-2003, 05:19 PM
BruceZ BruceZ is offline
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Default Re: Math behind flush odds

How do you ultimately arrive at 1.85 to 1 for a flush draw when you have 9 outs? After the flop, isn't it 38/9 with 47 unseen cards? When calculating pot odds, isn't it then 4.222 to 1 on a flush draw after the flop as opposed to the 1.85 to 1 that you get by using the calculation that David Sklansky (and the prior poster) has given us?

4.2-1 is for making your flush on the turn, which is actually used more often. 1.86-1 is for making it by the river, as calculated by the 3 methods described in the posts linked above.
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  #6  
Old 11-05-2003, 05:55 PM
GuruCane GuruCane is offline
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Default Re: Math behind flush odds

Thanks for the reply. One further related question and another unrelated question. First: when calculating pot odds is it still correct to treat the turn and the river as independent calculations? I.e., 4.2 to 1 before the turn and 4.1 to 1 before the river? It just doesn't seem right to me that it's 1.86 to 1 as long as I see the final two cards (for calculating pot odds, that is). It would entice me to stay in when I know that I should muck the hand. I apologize for my ignorance on this as I'm not a prob/stat guy.

Second question: How do you calculate which hands are favorites against one another. E.g., I believe 77 is a slight favorite vs. AKo (54%?). How do you calculate this?

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  #7  
Old 11-05-2003, 06:03 PM
M.B.E. M.B.E. is offline
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Default Re: Math behind flush odds

[ QUOTE ]
The 1.85 to 1 result would almost never make it prohibitive to call on the come.

[/ QUOTE ]
In limit holdem, it usually is correct to call the flop with a flush draw.

A common example of when you apply the 1.85-to-1 result is when there's four-way action and you checkraise the flop with the nut flush draw. It doesn't matter how much went in the pot preflop; you're happy to put more money in now, as long as at least two opponents are matching everything you put in. If there's more than two opponents who will call your checkraise, you can do this with any flush draw, not only the nut draw.
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  #8  
Old 11-05-2003, 06:15 PM
GuruCane GuruCane is offline
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Default Re: Math behind flush odds

I believe that this absolutely correct. There is often so much action in low limit that you should almost always call with a flush (or even open straight) draw. However, my biggest issue is whether, when deciding whether to bet to see the river when I am on a flush draw, I need 4.1 to 1 to call or some other number. I'm sure Mr. Z or someone else will settle this. Thanks in advance.
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  #9  
Old 11-06-2003, 12:31 AM
M.B.E. M.B.E. is offline
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Default Re: Math behind flush odds

[ QUOTE ]
However, my biggest issue is whether, when deciding whether to bet to see the river when I am on a flush draw, I need 4.1 to 1 to call or some other number.

[/ QUOTE ]
I don't understand what you mean by "bet to see the river". Bruce already gave a thorough answer to your original question. When you flop a four-flush, the odds against completing your flush on the turn are 4.22-to-1. The odds against completing your flush on the turn or the river are 1.86-to-1.
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  #10  
Old 11-06-2003, 02:46 AM
DPCondit DPCondit is offline
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Default Re: Math behind flush odds

[ QUOTE ]

I believe that this absolutely correct. There is often so much action in low limit that you should almost always call with a flush (or even open straight) draw. However, my biggest issue is whether, when deciding whether to bet to see the river when I am on a flush draw, I need 4.1 to 1 to call or some other number. I'm sure Mr. Z or someone else will settle this. Thanks in advance.

[/ QUOTE ]

If you are all in, or your opponents are, then you can just consider the odds of making it by the river. However, that is usually not the case, so you must consider the likelihood of having to call another bet (or even raises) on the turn as well. You also need to figure how many bets can you get your opponents to put in on the river (or turn) if you hit. Then figure how many bets should be in the pot at the end if you make it, minus the bets you put in on the flop, turn, and river, and divide it by the likely bets you will put in on the flop turn and river. That is you implied odds. Then figure in whatever likelihood of makng your flush and still losing the hand.

No easy clean solution there, it depends a lot on the texture of the game and your specific opponents in this hand.

Don
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