anyone know how to calculate standard deviation? - Page 2

splashpot · #11 11-15-2005, 01:56 AM

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Variance has nothing to do with % of people getting paid out. Since there are three more places I can finish in, that jacks up the variance. More possibilities, more variance.

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What? This is totally wrong. If there was a tournament that paid 10 places, but only had 11 entrants, the variance would be extremely low. More places you can finish.

Shillx · #12 11-15-2005, 01:59 AM

Heh I had it all right but I got rid of it.

For 6 max here would be your numbers (assume a 22 buyin)....

-22, -22, -22, -22, +18, +58

xbar = - $2
s = root ((4*20*20 + 20*20 + 60*60)/6) = $28.28 = 1.38 buy-ins

For 10 man SNGs

-22, -22, -22, -22, -22, -22, -22, +18, +38, +78

xbar = - $2
s = root ((7*400 + 400 + 1600 + 6400)/10)) = $33.46 = 1.52 buy-ins

Brad

Edit - The better the player you are, the more your varience goes up. So even if you are a losing player, you have quite a bit of varience in these games. It would not be uncommon for a good player to have a varience of 1.6 BI or higher for regualr 10 man SNGs. If you sucked ass (you finished OOTM everytime) your SD would be nil.

Re-edit: The SD assumes that you have played an infinate # of SNGs. Someone who has only played 6 SNGs with these results will have an SD square(6/5) larger then the number I listed.

HonchoOverload · #13 11-15-2005, 02:06 AM

[ QUOTE ]
[ QUOTE ]
Variance has nothing to do with % of people getting paid out. Since there are three more places I can finish in, that jacks up the variance. More possibilities, more variance.

[/ QUOTE ]
What? This is totally wrong. If there was a tournament that paid 10 places, but only had 11 entrants, the variance would be extremely low. More places you can finish.

[/ QUOTE ]

you're looking at this in the wrong way. If I have 100,000 tournaments of data of 6-table and 9-table tournaments for the same player, I guarantee you that the Standard Deviation and the variance itself (as the formula above proves) will be greater for 9-player tournaments.

Thanks shillx!

splashpot · #14 11-15-2005, 02:08 AM

Correct me if I'm wrong, but ITM% and payout% is what changes SD, not the number of people. For example, a 20 man SNG that paid 25% top 2, 15% 3-4, and 10% 5-6 would have the exact same SD as the normal 10 man party SNGs.

splashpot · #15 11-15-2005, 02:11 AM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Variance has nothing to do with % of people getting paid out. Since there are three more places I can finish in, that jacks up the variance. More possibilities, more variance.

[/ QUOTE ]
What? This is totally wrong. If there was a tournament that paid 10 places, but only had 11 entrants, the variance would be extremely low. More places you can finish.

[/ QUOTE ]

you're looking at this in the wrong way. If I have 100,000 tournaments of data of 6-table and 9-table tournaments for the same player, I guarantee you that the Standard Deviation and the variance itself (as the formula above proves) will be greater for 9-player tournaments.

Thanks shillx!

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The reason it's different is becuase 9 man SNGs have 50/30/20 payout and the 6 man ones have 65/35 payout. Not beacuse there are different amount of entrants.

Shillx · #16 11-15-2005, 02:14 AM

Let's say that there is a 100 man lotto. Entry is $1 and winner takes all. What is your SD in terms of buy-in?

xbar = 0

s = root ((100 + 100*100)/100) = $10.05 or 10.05 buy-ins

As you can see, a big top prize creates big varience. This is why MTTs have such huge varience (provided that you have a shot at winning). If they had a 10 man SNG where winner takes all (assume it is a $22), it would have much more varience then a typical Party 22. You will win 10% of the time.

ROI = -9.09%

s = root ((3600 + 32400)/10) = $60 = 2.73 buy-ins

splashpot · #17 11-15-2005, 02:16 AM

[ QUOTE ]
Let's say that there is a 100 man lotto. Entry is $1 and winner takes all. What is your SD in terms of buy-in?

xbar = 0

s = root ((100 + 100*100)/100) = $10.05 or 10.05 buy-ins

As you can see, a big top prize creates big varience. This is why MTTs have such huge varience (provided that you have a shot at winning). If they had a 10 man SNG where winner takes all (assume it is a $22), it would have much more varience then a typical Party 22. You will win 10% of the time.

ROI = -9.09%

s = root ((3600 + 32400)/10) = $60 = 2.73 buy-ins

[/ QUOTE ]
I don't understand where you're going with this. Yea, that all makes sense, but it has nothing to do with what I said. Do you agree or disagree with me?

Shillx · #18 11-15-2005, 02:19 AM

Yeah you are exactly right with what you said

HonchoOverload · #19 11-15-2005, 02:22 AM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Variance has nothing to do with % of people getting paid out. Since there are three more places I can finish in, that jacks up the variance. More possibilities, more variance.

[/ QUOTE ]
What? This is totally wrong. If there was a tournament that paid 10 places, but only had 11 entrants, the variance would be extremely low. More places you can finish.

[/ QUOTE ]

you're looking at this in the wrong way. If I have 100,000 tournaments of data of 6-table and 9-table tournaments for the same player, I guarantee you that the Standard Deviation and the variance itself (as the formula above proves) will be greater for 9-player tournaments.

Thanks shillx!

[/ QUOTE ]
The reason it's different is becuase 9 man SNGs have 50/30/20 payout and the 6 man ones have 65/35 payout. Not beacuse there are different amount of entrants.

[/ QUOTE ]

You may be right. My head sure hurts.