What are the odds? (nc)

[dfscott]

I've been taking some stabs at the 55s. I've played exactly 9 of them. Here is my finish distribution:

1. 1
2. 1
3. 1
4. 1
5. 1
6. 1
7. 1
8. 0
9. 1
10. 1

Needless to say, I'm not very excited about playing my 10th tonight. /images/graemlins/smile.gif

[octaveshift]

Assuming all players are equal, I *think* it would be:

.10*.10*.10*.10*.10*.10*.10*.10*.10 = 1 e-9

aka

Just about a 20 bajillion to 1.

[1C5]

Finish 8th on purpose the next one just so you can say you did it. /images/graemlins/cool.gif

[jcm4ccc]

Assuming you are an exactly average player, the chances of a distribution like that would be:

1 * 0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 = 0.003

[octaveshift]

[ QUOTE ]
Assuming you are an exactly average player, the chances of a distribution like that would be:

1 * 0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 = 0.003

[/ QUOTE ]

Are you sure? Because the odds that he would bust in each spot is would be exactly .10 each time, regardless of the finish before.

[jcm4ccc]

[ QUOTE ]
[ QUOTE ]
Assuming you are an exactly average player, the chances of a distribution like that would be:

1 * 0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 = 0.003

[/ QUOTE ]



Are you sure? Because the odds that he would bust in each spot is would be exactly .10 each time, regardless of the finish before.

[/ QUOTE ]

Yeah I'm sure.

First game: He can end up in any spot. The chances of doing that are 100%

Second game: He can end up in any of the 9 remaining spots. The chances of doing that are 90%.

Third game: He can end up in any 8 of the remaining spots. The chances of doing that are 80%.

etc etc etc

[octaveshift]

[ QUOTE ]

Yeah I'm sure.

[/ QUOTE ]

Thanks for the explanation. Makes perfect sense now.

[maddog2030]

[ QUOTE ]
Because the odds that he would bust in each spot is would be exactly .10 each time, regardless of the finish before.

[/ QUOTE ]

Correct. Any combination of finishes has the same probability of any other combination of finishes for the average player.

[Voltron87]

no, the chances of finishing in 1st, 4th, 10th, are not equal.

[maddog2030]

You missed a key line: "Assuming you are an exactly average player."

[jcm4ccc]

[ QUOTE ]
no, the chances of finishing in 1st, 4th, 10th, are not equal.

[/ QUOTE ] Really? Care to elaborate?

[Voltron87]

Checked my stats... 9th and 10th are well under 10%.

edit: do you actually finish in 10th the same number as 1st and 4th?

[dfscott]

[ QUOTE ]
Checked my stats... 9th and 10th are well under 10%.

[/ QUOTE ]

Of course, since as a 2+2'er, you should be above average, which would skew you off the perfect distribution of an average player.

[Voltron87]

you guys are so close with the "average player" idea... but they are not competing with a group of other average players.

[maddog2030]

No, but they are better than some and worse than others. In fact, perfectly so. Therefore "average". The average player at the $55s is better than the average player at the $6s.

[jcm4ccc]

[ QUOTE ]
you guys are so close with the "average player" idea... but they are not competing with a group of other average players.

[/ QUOTE ] Umm, over the long run, we all are.

[Voltron87]

are you defining average as breakeven?

[jcm4ccc]

[ QUOTE ]
are you defining average as breakeven?

[/ QUOTE ]

No, because that would be incorrect. The average player loses money.

[maddog2030]

Breakeven to the prizepool, meaning he's losing his rake (which doesn't get put into the game). It's a zerosum game. One man's loss is another's gain. All gains = All loses. Average = 0.

[dfscott]

[ QUOTE ]
Breakeven to the prizepool, meaning he's losing his rake (which doesn't get put into the game). It's a zerosum game. One man's loss is another's gain. All gains = All loses. Average = 0.

[/ QUOTE ]

Except for the roughly 10% vig (more at the bottom, less at the top). The "average" player has an ROI of -9% due to the vig.

[maddog2030]

[ QUOTE ]
Except for the roughly 10% vig (more at the bottom, less at the top). The "average" player has an ROI of -9% due to the vig.

[/ QUOTE ]

[ QUOTE ]
Breakeven to the prizepool, meaning he's losing his rake (which doesn't get put into the game). It's a zerosum game. One man's loss is another's gain. All gains = All loses. Average = 0.

[/ QUOTE ]

[Scuba Chuck]

Are you asking us to take odds on you finishing 8th tonight?

[dfscott]

[ QUOTE ]
[ QUOTE ]
Except for the roughly 10% vig (more at the bottom, less at the top). The "average" player has an ROI of -9% due to the vig.

[/ QUOTE ]

[ QUOTE ]
Breakeven to the prizepool, meaning he's losing his rake (which doesn't get put into the game). It's a zerosum game. One man's loss is another's gain. All gains = All loses. Average = 0.

[/ QUOTE ]

[/ QUOTE ]

That's why your post is somewhat contradictory. Everyone is losing the rake, so it can't be zero sum (unless you want to count the casino as a "player" in which case every game is zero sum).

[dfscott]

[ QUOTE ]
Are you asking us to take odds on you finishing 8th tonight?

[/ QUOTE ]

If someone spots me $55, I'll be happy to give it a shot. /images/graemlins/smile.gif

[freemoney]

poker is a zero sum game.

[dfscott]

[ QUOTE ]
poker is a zero sum game.

[/ QUOTE ]

In a zero-sum game, the payoff to one player is the negative of that going to the others.

This is not the case for online poker except for unraked games.

[freemoney]

the rake is not included within this idea.

[Freudian]

All in with 72o with 9 left. You can do it!

[dfscott]

[ QUOTE ]
the rake is not included within this idea.

[/ QUOTE ]

Edit: actually, after reading your other thread: you win.

[maddog2030]

[ QUOTE ]
That's why your post is somewhat contradictory. Everyone is losing the rake, so it can't be zero sum (unless you want to count the casino as a "player" in which case every game is zero sum).

[/ QUOTE ]

It is not contradictory. Read my post again. I said "meaning he's losing his rake (which doesn't get put into the game)." I defined my closed system (the game) as not including the rake, because the rake does not affect the game at all; it only affects your profits. Therefore I do not consider it to be a part of the game and simply a metagame factor.

If the only reason that I go to a friends house is to play a poker game that has no rake, I think you can agree it is zero sum. But would you say that it is no longer zero sum game if you count the gas money you spent to get over there? How is that different from a house fee? After all, both are fees directly related to you playing in that game.

Whether it's zero sum is completely dependent on how you define your system because you can almost always widen the system to include nonzero sum elements. Where to draw the line can get kind of hairy, but my initial reason for not including the rake because it does not affect the game structure at all. Whether you have a rake or not should not change your game strategy (except whether you should even play at all).

If you want to argue that my exclusion of the rake is faulty, then that's fine. It probably is, and in that case its a fixed negative sum game. Either way you still arrive at the conclusion the average player is losing to the rake.

[dfscott]

[ QUOTE ]
[ QUOTE ]
That's why your post is somewhat contradictory. Everyone is losing the rake, so it can't be zero sum (unless you want to count the casino as a "player" in which case every game is zero sum).

[/ QUOTE ]

It is not contradictory. Read my post again. I said "meaning he's losing his rake (which doesn't get put into the game)." I defined my closed system (the game) as not including the rake, because the rake does not affect the game at all; it only affects your profits. Therefore I do not consider it to be a part of the game and simply a metagame factor.

If the only reason that I go to a friends house is to play a poker game that has no rake, I think you can agree it is zero sum. But would you say that it is no longer zero sum game if you count the gas money you spent to get over there? How is that different from a house fee? After all, both are fees directly related to you playing in that game.

Whether it's zero sum is completely dependent on how you define your system because you can almost always widen the system to include nonzero sum elements. Where to draw the line can get kind of hairy, but my initial reason for not including the rake because it does not affect the game structure at all. Whether you have a rake or not should not change your game strategy (except whether you should even play at all).

If you want to argue that my exclusion of the rake is faulty, then that's fine. It probably is, and in that case its a fixed negative sum game. Either way you still arrive at the conclusion the average player is losing to the rake.

[/ QUOTE ]

Ok, you win, too.

I'm too tired to of being sucked out for real to worry about getting sucked out on a message board...

[maddog2030]

Do I get a cookie?

[dfscott]

[ QUOTE ]
[ QUOTE ]
Are you asking us to take odds on you finishing 8th tonight?

[/ QUOTE ]

If someone spots me $55, I'll be happy to give it a shot. /images/graemlins/smile.gif

[/ QUOTE ]

Oh, well -- best I could do was a 6 and 5. If I'd stop playing these things I'd probably start showing a profit.

[lorinda]

you guys are so close with the "average player" idea... but they are not competing with a group of other average players.


Although I can see where you are coming from, in my world the definition of an average player is one who finishes in all positions equally.

You are changing the definition to create your (interesting) argument.

If you want to talk about the finishing distribution of a player who is on average the 5.5th best in an event, that would be a different matter.

Lori

[astarck]

[ QUOTE ]
First game: He can end up in any spot. The chances of doing that are 100%

Second game: He can end up in any of the 9 remaining spots. The chances of doing that are 90%.

Third game: He can end up in any 8 of the remaining spots. The chances of doing that are 80%.

etc etc etc

[/ QUOTE ]

This seems WAY off.

While he needs to end up in 90% of the spots in the 2nd sng to accomplish part of this feat, he can still end up at any of the 10 spots. Just because he got Xth in the first sng, doesn't mean he can't get it again. And assuming he is an "average" player, meaning equal distribution in all spots than it would be .1x.1x.1x.1....

[jcm4ccc]

[ QUOTE ]
[ QUOTE ]
First game: He can end up in any spot. The chances of doing that are 100%

Second game: He can end up in any of the 9 remaining spots. The chances of doing that are 90%.

Third game: He can end up in any 8 of the remaining spots. The chances of doing that are 80%.

etc etc etc

[/ QUOTE ]

This seems WAY off.

While he needs to end up in 90% of the spots in the 2nd sng to accomplish part of this feat, he can still end up at any of the 10 spots. Just because he got Xth in the first sng, doesn't mean he can't get it again. And assuming he is an "average" player, meaning equal distribution in all spots than it would be .1x.1x.1x.1....

[/ QUOTE ]

So are you saying that the answer is .10 * .10 * .10 . . .

Because you're wrong if that's what you're saying.

[astarck]

[ QUOTE ]
So are you saying that the answer is .10 * .10 * .10 . . .

Because you're wrong if that's what you're saying.

[/ QUOTE ]

Assuming he does finish in each spot 10% of the time, thats what I'm saying.

[lorinda]

Assuming he does finish in each spot 10% of the time, thats what I'm saying.

Work out the chances of someone finishing in all the spots in a 3 player tournament.

Lori

[astarck]

Actually, I'm seeing the light.

What I was doing the calculations for would be in a certain combination.

jcm...you are correct.

[microbet]

[ QUOTE ]
Assuming you are an exactly average player, the chances of a distribution like that would be:

1 * 0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 = 0.003

[/ QUOTE ]

Hey, not a great thread, but what the heck, isn't it

1 * 0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 * 0.1 = 0.0003

He did specify a 10 player game right?

For the people that thought .1 * .1 ...., it is a good example of where you should check with common sense. I know these numbers are not exactly easy off the top of you head guesses but saying this happens one out of 10 billion times does seem a little off, doesn't it?

[jcm4ccc]

[ QUOTE ]
[ QUOTE ]
Assuming you are an exactly average player, the chances of a distribution like that would be:

1 * 0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 = 0.003

[/ QUOTE ]

Hey, not a great thread, but what the heck, isn't it

1 * 0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 * 0.1 = 0.0003

He did specify a 10 player game right?

For the people that thought .1 * .1 ...., it is a good example of where you should check with common sense. I know these numbers are not exactly easy off the top of you head guesses but saying this happens one out of 10 billion times does seem a little off, doesn't it?

[/ QUOTE ]

I was calculating for dfscott's particular situation. He did not complete that mythical tenth time. He only did it 9 times.

[microbet]

right right.

[dfscott]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Assuming you are an exactly average player, the chances of a distribution like that would be:

1 * 0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 = 0.003

[/ QUOTE ]

Hey, not a great thread, but what the heck, isn't it

1 * 0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 * 0.1 = 0.0003

He did specify a 10 player game right?

For the people that thought .1 * .1 ...., it is a good example of where you should check with common sense. I know these numbers are not exactly easy off the top of you head guesses but saying this happens one out of 10 billion times does seem a little off, doesn't it?

[/ QUOTE ]

I was calculating for dfscott's particular situation. He did not complete that mythical tenth time. He only did it 9 times.

[/ QUOTE ]

And unfortunately, I had to get two more 5s, a 6th, and a 2nd before I finally got my mythical 8th place finish (pushing 44 from the CO on level 4 finally did it).

Thank god these things have finally destroyed my BR enough that I can go back to the 33s.

As a sidebar, for a "no content" thread, this sure has a lot of traffic...

[1C5]

Soory to hear the 55s didnt work out yet. I actually took a different approach. I have 2 nice 2001FPs so right now I am 6 and soon 8 tabling the 22s and then will move to the 33s.

Not ready for the 55s just yet (maybe ready but not ready for the swings yet).
8 tabling is getting easier as I have 8 tabled 200 or so 11s now.