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  #1  
Old 01-07-2005, 02:01 PM
Billy Baroo Billy Baroo is offline
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Posts: 114
Default Did I win this bet?

Hopefully the good folks at the 2+2 forums can help settle a bet between me and an acquaintance.

Here's the bet. An acquaintance bet me that I don't know how to calculate how flop outs correspond to odds. For instance, most of us know that a flush draw in hold 'em has 9 outs, and is about 1.9: 1 on the flop. His bet was that I couldn't do the math that proves it, and he doesn't believe my answer. We are using 2+2 to officially settle our bet.

Here was my answer (for the uninitiated, I have included how to do this with a 9 out flush draw as it is IMO a little easier to follow it that way):

((Outs/unseen cards on turn)+(Outs/unseen cards on river))
-((Outs/unseen cards on turn)* (Outs/unseen cards on river))
=% hand will be made

100-% hand will be made= x

x/% hand will be made= odds: 1

So did I win the bet? Below is an example of how the math works out for a 9 out flush draw.

((9/47)+(9/46))-((9/47)* (9/46))=35%

100-35= 65
65/35= 1.857
The odds of making a flush are 1.857:1 when you have 4 of the suit on the flop.
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  #2  
Old 01-07-2005, 02:39 PM
dtbog dtbog is offline
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Join Date: Jun 2004
Posts: 19
Default Re: Did I win this bet?

[ QUOTE ]
((Outs/unseen cards on turn)+(Outs/unseen cards on river))
-((Outs/unseen cards on turn)* (Outs/unseen cards on river))
=% hand will be made

100-% hand will be made= x

x/% hand will be made= odds: 1


[/ QUOTE ]

Your calculation is the following:

(chance to make on turn) + (chance to make on river, assuming that you missed the turn) - (chance that you made both).

This isn't right -- for one thing, your calculation of the chance that you made both doesn't make sense. If you did, in fact, hit your flush on the turn, then another spade coming would actually be 8/46 instead of 9/46. That's why I don't do these calculations that way.. you'd need some sort of conditional branching to account for the different possibilities for the river.

To truly calculate flop odds, you must do the following:

(38/47) = chance of missing on the turn
(37/46) = chance of missing on river, given that you missed on the turn

(38/47) * (37/46) = 65.0%

This is your chance of missing the flush. Nothing is double-counted; this accurately describes your chances of NOT having a flush at the end of the hand.

Obviously, 1.00 - 0.650 = 0.35, and this (35%) is your chance of making the flush by the river.

Yours actually gave a result of 34.96%, where the correct math gives 35.0% - a trivial difference in this case, but I suspect larger numbers will give a more pronounced discrepancy.

How much was the bet?

-DB
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  #3  
Old 01-07-2005, 02:50 PM
MortalWombatDotCom MortalWombatDotCom is offline
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Join Date: Dec 2004
Posts: 64
Default Re: Did I win this bet?

[ QUOTE ]
Hopefully the good folks at the 2+2 forums can help settle a bet between me and an acquaintance.

Here's the bet. An acquaintance bet me that I don't know how to calculate how flop outs correspond to odds. For instance, most of us know that a flush draw in hold 'em has 9 outs, and is about 1.9: 1 on the flop. His bet was that I couldn't do the math that proves it, and he doesn't believe my answer. We are using 2+2 to officially settle our bet.

Here was my answer (for the uninitiated, I have included how to do this with a 9 out flush draw as it is IMO a little easier to follow it that way):

((Outs/unseen cards on turn)+(Outs/unseen cards on river))
-((Outs/unseen cards on turn)* (Outs/unseen cards on river))
=% hand will be made

100-% hand will be made= x

x/% hand will be made= odds: 1

So did I win the bet? Below is an example of how the math works out for a 9 out flush draw.

((9/47)+(9/46))-((9/47)* (9/46))=35%

100-35= 65
65/35= 1.857
The odds of making a flush are 1.857:1 when you have 4 of the suit on the flop.

[/ QUOTE ]

your notation is a little sloppy (e.g. you freely convert between probability and percent chance, you use more significant digits than your earlier rounding justify, and the odds are roughly 1.9:1 against making the flush) but your basic method was sound.
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  #4  
Old 01-07-2005, 02:53 PM
MortalWombatDotCom MortalWombatDotCom is offline
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Join Date: Dec 2004
Posts: 64
Default Re: Did I win this bet?

[ QUOTE ]
[ QUOTE ]
((Outs/unseen cards on turn)+(Outs/unseen cards on river))
-((Outs/unseen cards on turn)* (Outs/unseen cards on river))
=% hand will be made

100-% hand will be made= x

x/% hand will be made= odds: 1


[/ QUOTE ]

Your calculation is the following:

(chance to make on turn) + (chance to make on river, assuming that you missed the turn) - (chance that you made both).

This isn't right -- for one thing, your calculation of the chance that you made both doesn't make sense. If you did, in fact, hit your flush on the turn, then another spade coming would actually be 8/46 instead of 9/46. That's why I don't do these calculations that way.. you'd need some sort of conditional branching to account for the different possibilities for the river.

To truly calculate flop odds, you must do the following:

(38/47) = chance of missing on the turn
(37/46) = chance of missing on river, given that you missed on the turn

(38/47) * (37/46) = 65.0%

This is your chance of missing the flush. Nothing is double-counted; this accurately describes your chances of NOT having a flush at the end of the hand.

Obviously, 1.00 - 0.650 = 0.35, and this (35%) is your chance of making the flush by the river.

Yours actually gave a result of 34.96%, where the correct math gives 35.0% - a trivial difference in this case, but I suspect larger numbers will give a more pronounced discrepancy.

How much was the bet?

-DB

[/ QUOTE ]

this is just plain wrong. simple algebra can demonstrate the two methods are the same. the difference was due to rounding error.
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  #5  
Old 01-07-2005, 02:56 PM
BruceZ BruceZ is offline
Senior Member
 
Join Date: Sep 2002
Posts: 1,636
Default Re: Did I win this bet?

[ QUOTE ]
Hopefully the good folks at the 2+2 forums can help settle a bet between me and an acquaintance.

Here's the bet. An acquaintance bet me that I don't know how to calculate how flop outs correspond to odds. For instance, most of us know that a flush draw in hold 'em has 9 outs, and is about 1.9: 1 on the flop. His bet was that I couldn't do the math that proves it, and he doesn't believe my answer. We are using 2+2 to officially settle our bet.

Here was my answer (for the uninitiated, I have included how to do this with a 9 out flush draw as it is IMO a little easier to follow it that way):

((Outs/unseen cards on turn)+(Outs/unseen cards on river))
-((Outs/unseen cards on turn)* (Outs/unseen cards on river))
=% hand will be made

100-% hand will be made= x

x/% hand will be made= odds: 1

So did I win the bet? Below is an example of how the math works out for a 9 out flush draw.

((9/47)+(9/46))-((9/47)* (9/46))=35%

100-35= 65
65/35= 1.857
The odds of making a flush are 1.857:1 when you have 4 of the suit on the flop.

[/ QUOTE ]

Your expression gives the right anwer (1.860-to-1) because of 2 errors which cancel. It should be

9/47 + 9/47 - (9/47 * 8/46) = 1.860:1.

The probability of the turn card being a flush card and the probability of the river card being a flush card are each 9/47 before either of these cards are dealt. If the flop card is a flush card, the probabilty is 8/46 that the turn card is also a flush card, so the probabilty of getting a flush card on both is (9/47)*(8/46). This happens to give the same answer as yours. Note these two other equivalent forms:

9/47 + (38/47)*(9/46) = 1.860:1

That is, 9/47 on the turn plus 9/46 on the river of the 38/47 of the time you miss on the turn.

1 - (38/47)*(37/46) = 1.860:1

That is, 1 minus the probabilty of missing on both the flop and the turn.
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  #6  
Old 01-07-2005, 03:06 PM
MortalWombatDotCom MortalWombatDotCom is offline
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Join Date: Dec 2004
Posts: 64
Default Re: Did I win this bet?

[ QUOTE ]
Your expression gives the right anwer (1.860-to-1) by a self-cancelling error.

[/ QUOTE ]

that's classic. also, the Odyssey was not written by Homer, but by another author of the same name.

hey! you editted your post after i quoted it. equally silly, but less quotable.
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  #7  
Old 01-07-2005, 03:13 PM
MortalWombatDotCom MortalWombatDotCom is offline
Member
 
Join Date: Dec 2004
Posts: 64
Default Re: Did I win this bet?

this thread is so much fun! [img]/images/graemlins/laugh.gif[/img]

sorry to hijack, but i've got this great way to figure out what 2*2 is. you multiply 2 by 3, substract 2, and you get 4!
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  #8  
Old 01-07-2005, 03:15 PM
MortalWombatDotCom MortalWombatDotCom is offline
Member
 
Join Date: Dec 2004
Posts: 64
Default Re: Did I win this bet?

[ QUOTE ]
this thread is so much fun! [img]/images/graemlins/laugh.gif[/img]

sorry to hijack, but i've got this great way to figure out what 2*2 is. you multiply 2 by 3, substract 2, and you get 4!

[/ QUOTE ]

sorry Mortal, you got the right answer, but you made 11 errors that cancelled themselves out.

the correct way to do it is 2*2 = (3 - 1)(3 - 1) = 9 - 3 - 3 + 1 = 4.
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  #9  
Old 01-07-2005, 03:30 PM
BruceZ BruceZ is offline
Senior Member
 
Join Date: Sep 2002
Posts: 1,636
Default Re: Did I win this bet?

[ QUOTE ]
[ QUOTE ]
Your expression gives the right anwer (1.860-to-1) by a self-cancelling error.

[/ QUOTE ]

that's classic. also, the Odyssey was not written by Homer, but by another author of the same name.

hey! you editted your post after i quoted it. equally silly, but less quotable.

[/ QUOTE ]

You're wrong and need to think about this a lot more. You in no way can justify the factor of (9/47)*(9/46) which appears in the poster's derivation. This is supposed to be the proability of getting a flush card on both the turn and the river which is (9/47)*(8/46). You also cannot justify the 9/47 + 8/46. This corresponds to nothing. You cannot add these probabilities because the 8/46 is on a different sample space than the 9/47, namely, it applies only after the flush card is dealt. You must add both probabilities before either card is dealt, which is 9/47 + 9/47, or else add 9/47 + (38/47)*(9/46) as I have shown above. Both of these are very common errors, BTW.

The fact that you can derive the poster's equation from my equation by alegebra in no way means that the original derivation makes any sense. He explained what his terms were supposed to represent, and that explanation was wrong. He made two errors which happen to cancel to give the correct numerical answer, just as I said. I also don't post things that are "silly", and you should have considered that before you made a huge ass of yourself. If you say "his basic method is sound" as you have done below, then you also do not know how to properly derive this formula.
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  #10  
Old 01-07-2005, 03:58 PM
MortalWombatDotCom MortalWombatDotCom is offline
Member
 
Join Date: Dec 2004
Posts: 64
Default Re: Did I win this bet?

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Your expression gives the right anwer (1.860-to-1) by a self-cancelling error.

[/ QUOTE ]

that's classic. also, the Odyssey was not written by Homer, but by another author of the same name.

hey! you editted your post after i quoted it. equally silly, but less quotable.

[/ QUOTE ]

You're wrong and need to think about this a lot more.

[/ QUOTE ]

quite possibly, and blatantly false, respectively.

[ QUOTE ]
You in no way can justify the factor of (9/47)*(9/46) which appears in the poster's derivation. This is supposed to be the proability of getting a flush card on both the turn and the river which is (9/47)*(8/46).

[/ QUOTE ]

the original poster didn't say that is what "9/47 * 9/46" was supposed to represent. you did. then you said that was wrong, which it was.

as for justifying it, i can, although i don't think it will be satisfactory to you. a reasonable way to solve this particular problem is to take (1 - probability of hitting on the turn) and multiply it by (1 - probability of hitting on the river given you missed on the turn) to get the probability of missing on both the turn and river, and subtract from 1 to get the probability of failing to miss on both the turn and the river. well, 1 - (1 - x)(1 - y) = 1 - [1 - x - y + xy] = x + y - xy, and it always will. the fact that the original poster used this formula without explaining how or why he derived it (and i suspect someone else derived it and he just saw the end product) doesn't make him wrong.

[ QUOTE ]
You also cannot justify the 9/47 + 8/46. This corresponds to nothing.

[/ QUOTE ]

not only can i not justify it, i can't even find it.

[ QUOTE ]
The fact that you can derive the poster's equation from my equation by alegebra in no way means that the original derivation makes any sense. He explained what his terms were supposed to represent, and that explanation was wrong.


[/ QUOTE ]

i repeat myself, but he didn't explain why the formula he used is or should be right.

[ QUOTE ]
I also don't post things that are "silly", and you should have considered that before you made a huge ass of yourself.

[/ QUOTE ]

well, i do post things that are silly. [img]/images/graemlins/tongue.gif[/img]

why should i consider such things before i make a huge ass of myself? will that make it easier?

[ QUOTE ]
If you say "his basic method is sound" as you have done below, then you also do not know how to properly derive this formula.

[/ QUOTE ]

ok, perhaps i should have considered my words more carefully. how about "whereas i would have used a different formula than you did, because i feel that mine gives a clearer insight into the techniques involved and permits a more straightforward and intuitive description of what each term means, your formula is algebraicly equivalent to mine and will produce the correct answer"?
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