PDA

View Full Version : Did I win this bet?


Billy Baroo
01-07-2005, 02:01 PM
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.

dtbog
01-07-2005, 02:39 PM
[ 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

MortalWombatDotCom
01-07-2005, 02:50 PM
[ 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.

MortalWombatDotCom
01-07-2005, 02:53 PM
[ 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.

BruceZ
01-07-2005, 02:56 PM
[ 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.

MortalWombatDotCom
01-07-2005, 03:06 PM
[ 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.

MortalWombatDotCom
01-07-2005, 03:13 PM
this thread is so much fun! /images/graemlins/laugh.gif

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!

MortalWombatDotCom
01-07-2005, 03:15 PM
[ QUOTE ]
this thread is so much fun! /images/graemlins/laugh.gif

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.

BruceZ
01-07-2005, 03:30 PM
[ 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.

MortalWombatDotCom
01-07-2005, 03:58 PM
[ 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. /images/graemlins/tongue.gif

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"?

Billy Baroo
01-07-2005, 04:00 PM
[ QUOTE ]
How much was the bet?

[/ QUOTE ]
$100

I'm still waiting to see if more people agree with mortalwombat, because as of now there isn't a concensus between posters. I'll admit that my background in mathematics is lacking, and I tried to figure out the problem on the fly (which is why the form was less than stellar)

[ QUOTE ]
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.

[/ QUOTE ]

To test this I worked out the math for a 15 out straight flush draw.

Using my method I get 54.1% or .85: 1
Using yours I get the same answer.

Billy Baroo
01-07-2005, 04:13 PM
From Bruce Z's post:
[ QUOTE ]
Your expression gives the right anwer (1.860-to-1) because of 2 errors which cancel

[/ QUOTE ]

Did the two errors cancel out just in this case, or will they always cancel out. In response to dtbog's post, I worked out the answers using both my method and yours for calculating a 15 out straight-flush draw. The answers were the same.

[ QUOTE ]
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 ]

As far as how I derived it, I am proud to say that I came up with that sloppy formula on my own. I can't really say why I thought it would work. It just intuitively made sense to me (whether it makes sense to others is another matter) and when I plugged in a 9 out flush draw it ended up with the answer I knew was correct beforehand. So I figured I had it right.

MortalWombatDotCom
01-07-2005, 04:35 PM
[ QUOTE ]
[ QUOTE ]
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 ]

As far as how I derived it, I am proud to say that I came up with that sloppy formula on my own. I can't really say why I thought it would work. It just intuitively made sense to me (whether it makes sense to others is another matter) and when I plugged in a 9 out flush draw it ended up with the answer I knew was correct beforehand. So I figured I had it right.

[/ QUOTE ]

well, if i were your probability instructor and you provided this as an answer on an exam, i would take off points, maybe all of them. personally, i think if you are going to memorize one of the aforementioned formulae, yours has the benefit of being the easiest to get close enough to correct in ones head at a poker table. it doesn't give any insight into how one would, for example, compute the odds of making a flush in a hold-em like game with an extra card on the end, whereas the other approaches mentioned in this thread extend more or less intuitively into variations of that nature. YMMV.

BruceZ
01-07-2005, 05:20 PM
[ QUOTE ]

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

[/ QUOTE ]

Then we agree on that.


[ QUOTE ]
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 ]

Well, it does if if the bet required him to show his work in deriving this formula. Rereading the original bet, it said "An acquaintance bet me that I don't know how to calculate how flop outs correspond to odds. " So if "calculating" simply means getting the correct answer by some formula which is not algebraically derived, then fine. If it means showing the correct derivation for this formula, which I thought was the point of this, then he did not show it, as you seem to agree. I never claimed he lost the bet. That is between him and his acquaintance.


[ 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 ]

[/ QUOTE ]


Typo on my part, meant 9/47 + 9/46.



[ 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"?

[/ QUOTE ]

Perfect. I already said that his formula produces the right answer and can be derived from mine by algebra. The only additional thing I said was that he arrived at his formula via 2 canceling errors, which you laughed at and called "silly". I make this claim because a) he didn't show any algebra or any valid derivation of his equation, and b) from vast experience in showing people how to do this problem countless times over the years, I know the standard errors when I see them, and I saw two of them here. From my experience, and from his subsequent statements, this is almost certainly the correct interpretation. I pointed out these errors so that the student would more clearly think about these kinds of problems in the future. That doesn't rate your assenine responses.

[ QUOTE ]

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

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

[/ QUOTE ]

Let me put it this way, you will treat the poster's on this forum with the respect they deserve, or you will not be part of this forum in the future. I see you are new. Learn which way is up before ridiculing those who know what they are doing.

Marm
01-07-2005, 05:42 PM
OK you guys are killin me!

Did he win the bet or not?

Simple answer!

MortalWombatDotCom
01-07-2005, 06:12 PM
[ QUOTE ]
Let me put it this way, you will treat the poster's on this forum with the respect they deserve, or you will not be part of this forum in the future. I see you are new. Learn which way is up before ridiculing those who know what they are doing.

[/ QUOTE ]

personally, i think your posts showed as much disrespect to the original poster as mine did towards you, and that is the reason i adopted that tone in the first place. i realize this is a matter of opinion and perhaps mine is the minority one.

in fact, acknowledging that two wrongs don't make a right, i apologize, BruceZ. i should have made my points more clearly and more politely.

if your contention is that i ridiculed you because i thought you didn't "know what you were doing", then i'm sorry, but that's just not true. i thought you knew what you were doing but that you failed to distinguish between "suboptimal" or "potentially misleading" or "possibly arrived at incorrectly" and "wrong."

if your contention is instead that the fact that you "know what you are doing" should be a shield against ridicule on any level, then i'm sorry, but i "know what i am doing" too, and so you should respect my ridicule shield if you want me to respect yours. but i don't think that is what you meant.

BruceZ
01-07-2005, 06:55 PM
[ QUOTE ]
personally, i think your posts showed as much disrespect to the original poster as mine did towards you, and that is the reason i adopted that tone in the first place.

[/ QUOTE ]

Really! How interesting that a person could have your opinion! Let's recap. First of all, I was not disrespectful to the original poster AT ALL. All I said was that his answer was CORRECT due to 2 errors which canceled. It was his derivation I was criticizing, not him personally, and there is a huge difference. You on the other hand, felt it was appropriate to respond to me with mocking sarcasm in THREE POSTS:

post 1:
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!

post 2 (responding to self):
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.

post 3:
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 [emphasis added], but less quotable.


Now calling someone's honest contribution silly and using sarcasm like this is clearly disrespectful, and if you can't see the difference, then I can't help you.

You were more disrespectful than I was in your post to dtbog when you dismissed what he wrote as "just plain wrong". He had carefully and correctly pointed out the same shortcomings in the original poster's terms that I did. You read the original post too fast if you thought the only problem was roundoff error.


[ QUOTE ]
in fact, acknowledging that two wrongs don't make a right, i apologize, BruceZ. i should have made my points more clearly and more politely.

[/ QUOTE ]

And in the event that the original poster actually did the algebra in his head, or is a genius and just intuited the correct answer clearly without any misconceptions, then I apologize TO HIM (but I really don't believe this is the case, and I don't believe that either he or you believes this is the case).


[ QUOTE ]
if your contention is instead that the fact that you "know what you are doing" should be a shield against ridicule on any level, then i'm sorry, but i "know what i am doing" too, and so you should respect my ridicule shield if you want me to respect yours. but i don't think that is what you meant.

[/ QUOTE ]

My contention is that nothing I post on this forum is likely to be correctly classified as "silly", especially in the few minutes that you spent considering it, and that is true no matter who you are or how much you know what you are doing.

MortalWombatDotCom
01-07-2005, 07:14 PM
[ QUOTE ]
[ QUOTE ]
personally, i think your posts showed as much disrespect to the original poster as mine did towards you, and that is the reason i adopted that tone in the first place.

[/ QUOTE ]

Really! How interesting that a person could have your opinion! Let's recap. First of all, I was not disrespectful to the original poster AT ALL. All I said was that his answer was CORRECT due to 2 errors which canceled. It was his derivation I was criticizing, not him personally, and there is a huge difference. You on the other hand, felt it was appropriate to respond to me with mocking sarcasm in THREE POSTS:

post 1:
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!

post 2 (responding to self):
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.

post 3:
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 [emphasis added], but less quotable.


Now calling someone's honest contribution silly and using sarcasm like this is clearly disrespectful, and if you can't see the difference, then I can't help you.

You were more disrespectful than I was in your post to dtbog when you dismissed what he wrote as "just plain wrong". He had carefully and correctly pointed out the same shortcomings in the original poster's terms that I did. You read the original post too fast if you thought the only problem was roundoff error.


[ QUOTE ]
in fact, acknowledging that two wrongs don't make a right, i apologize, BruceZ. i should have made my points more clearly and more politely.

[/ QUOTE ]

And in the event that the original poster actually did the algebra in his head, or is a genius and just intuited the correct answer clearly without any misconceptions, then I apologize TO HIM (but I really don't believe this is the case, and I don't believe that either he or you believes this is the case).


[ QUOTE ]
if your contention is instead that the fact that you "know what you are doing" should be a shield against ridicule on any level, then i'm sorry, but i "know what i am doing" too, and so you should respect my ridicule shield if you want me to respect yours. but i don't think that is what you meant.

[/ QUOTE ]

My contention is that nothing I post on this forum is likely to be correctly classified as "silly", especially in the few minutes that you spent considering it, and that is true no matter who you are or how much you know what you are doing.

[/ QUOTE ]

You are unconditionally right. I am unconditionally wrong. Please accept my second apology.

MortalWombatDotCom
01-07-2005, 07:21 PM
[ QUOTE ]
OK you guys are killin me!

Did he win the bet or not?

Simple answer!

[/ QUOTE ]

No. His friend won the bet:

[ QUOTE ]
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.


[/ QUOTE ]

if that really was his friend's bet, then, while he did provide a formula that provides the correct answer, he failed to provide proof of correctness. at least, he didn't provide proof in this thread. this is just my interpretation of the facts as they have been presented to me. please don't hurt me.

Billy Baroo
01-07-2005, 07:24 PM
From one of Bruce Z's posts:
[ QUOTE ]
Rereading the original bet, it said "An acquaintance bet me that I don't know how to calculate how flop outs correspond to odds. " So if "calculating" simply means getting the correct answer by some formula which is not algebraically derived, then fine. If it means showing the correct derivation for this formula, which I thought was the point of this, then he did not show it, as you seem to agree. I never claimed he lost the bet. That is between him and his acquaintance.


[/ QUOTE ]

The bet had nothing to do with algebra. It didn't matter if I calculated the odds by the position of the sun. The only thing that matters as far as the bet is concerned is that I have a method that works.

From my point of view, "the point of this" is whether I won the bet or the other party did, not whether I used correct algebra. As of now, the thread has not given a conclusive answer and neither of us is about to pay up. So we've agreed to post a poll and we will go with the majority decision.

Edit: The poll is posted in a seperate thread.

BruceZ
01-07-2005, 08:19 PM
[ QUOTE ]
Did the two errors cancel out just in this case, or will they always cancel out.

[/ QUOTE ]

They will always cancel out, if by "always" you mean for any number of outs when computing the probability of completing a hand by the river.

[ QUOTE ]
As far as how I derived it, I am proud to say that I came up with that sloppy formula on my own. I can't really say why I thought it would work.

[/ QUOTE ]

In other words, you could not prove it would work.

jayboo
01-08-2005, 01:01 AM
i use a little bit simpler method to calculate odds on the flop when in a hand. It works for different types of draws not readily memorized. You have AK with flop 10j3 which leaves u 2 overs and a gutshot. 6 outs on overs and 4 on the queen which equals 10 out multiply (out x cards to come X 2) 10 outs = roughly 40% chance of hitting. with 1 card to come its roughly 20% to hit. this is what i use on the fly when in hands and i find it is very fast and easy to use.
good luck /images/graemlins/wink.gif

Stork
01-08-2005, 09:16 PM
So Billy, if your thinking had been "I have 9 outs, times 2 more cards, times the 2 cards I hold in my hand= about 35, so I hae a 35% chance to hit the flush, or 1.9-1 odds to hit my flush."
Would you win the bet?

TomBrooks
01-08-2005, 11:15 PM
[ QUOTE ]
An acquaintance bet me that I don't know how to calculate how flop outs correspond to odds.

[/ QUOTE ]
Billy,

While your technique might not have been perfect, it provides an accurate solution every time.

Therefore:

A) If your bet was that you knew how to calculate, every time, the correct drawing odds with two cards to come given x number of outs, you win.

B) If your bet was that you could provide a mathematical proof of your method that would survive the technical scrutiny of mathematicians, you appear, based on the opinions of the two obviously mathematically knowledgable persons above, to have lost.

-TomBk

And I hope no one considers this either silly or assinine.

eurythmech
01-09-2005, 05:53 AM
OK, I don't know if there's need to post more...
But I read a few posts and people didnt look like they knew what the hell they were talking about =)

Your original post almost got it right, but for some reason you rounded off the percentage to exactly 35%.

You do it like this

1 - (38/47)(37/46)

Which means you take 1 (which is all possibilities, or 100% of the outcomes) and reduce all cases where you miss your flush out on both the turn and the river (ie, dont make the flush at all). Since it is necessary for you to miss on the turn AND the river we need to multiply those probabilities.

38/37 means 38 of the 47 possible cards for the turn are no good, and 37/46 is self explainatory.

You then get roughly 34.97% or in your strange American way of writing things roughly 1,8598:1

So the general formula for figuring out the probability for any number of outs to hit on either the turn or the river is

1 - ((47-x)/47)((46-x)/46) where x is the number of outs you have.

For instance, if you were to have 61 outs (like, if you include an Uno deck) you would have about 90.2% and also our formula inevtiably leads us to the following conclusion

((47-x)/47)((46-x)/46) = 1

(1-x/47)(1-x/46) = 1

1 -x/47 -x/46 + x/2162 = 1

1 - 93x/2162 + x/2162 = 1

x/2162 - 93x/2162 = 0

x - 93x = 0

x = 93x

x = 93

x = 47 + 46

If we have 93 outs we will make it 100% of the time!!!!!!!!