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# Sample Size....

• Bronze
Joined: 15.04.2010
.... so what sample size do I need to say that I am a winning player?

Currently playing \$4 Rush SNGs.

Sample 851 Tournaments
Profit \$1521.75
ROI +40%

Cheers

Jass
• 10 replies
• Bronze
Joined: 04.05.2009
seems enough for sng's coming from a guy who has no clue
• Bronze
Joined: 17.06.2010
The number of tournaments you need to conclude that you have strong statistical evidence that you are a winning player depends on your win rate and the standard deviation.

You play multitable SNGs. This makes a huge difference in the standard deviation. For 9 player SNGs, the standard deviation per tournament is about 1.55 buy-ins, so a rough 95% confidence interval after n tournament is your observed ROI +- 310%/sqrt(n). The standard deviation for MTTs depends on your playing style much more than it does in STTs. I would guess that the values for 135 player tournaments would range from 4 buy-ins to 8 buy-ins, so your confidence interval would look something like your observed ROI +- 1200%/sqrt(n).

That works out to about 40% +- 40%, so it looks like you are able to conclude that you have strong evidence that you are winning, but you don't know by how much.
• Black
Joined: 26.05.2007
Applying these formulas to big samples scare the **** out of me. Phzon, do you happen to know the ROI +- 310%/sqrt(n) equivalent for 6max (65/35)?
• Bronze
Joined: 17.06.2010
The standard deviation for 6-max tournaments is almost the same as for 9-max. For a 70-30 structure, it might be slightly higher. For a 65-35 structure, it might be slightly lower.

You can calculate your standard deviation from your finishing distribution and the prizes.

A = average amount won
B = average square of the amount won
Variance = B - (A^2)
Standard Deviation = Sqrt(Variance)

So, for example, if you play \$12+1 6-max tournaments which pay \$46.80 and \$25.20, and you place first 20% and second 18%, you could calculate the following:

A = 20% x \$46.80 + 18% x \$25.20 = \$13.896
B = 20% x (\$46.80^2) + 18% x (\$25.20^2) = 552.3552 square dollars
Variance = 552.3552 - 13.896^2 = 359.256 square dollars
Standard Deviation = Sqrt(359.256) = \$18.95 = 1.46 buy-ins.

That suggests that for similar distributions, a rough 95% confidence interval after n tournaments would be your observed ROI +- 292%/sqrt(n).
• Bronze
Joined: 15.04.2010
Originally posted by pzhon
The number of tournaments you need to conclude that you have strong statistical evidence that you are a winning player depends on your win rate and the standard deviation.

You play multitable SNGs. This makes a huge difference in the standard deviation. For 9 player SNGs, the standard deviation per tournament is about 1.55 buy-ins, so a rough 95% confidence interval after n tournament is your observed ROI +- 310%/sqrt(n). The standard deviation for MTTs depends on your playing style much more than it does in STTs. I would guess that the values for 135 player tournaments would range from 4 buy-ins to 8 buy-ins, so your confidence interval would look something like your observed ROI +- 1200%/sqrt(n).

That works out to about 40% +- 40%, so it looks like you are able to conclude that you have strong evidence that you are winning, but you don't know by how much.
Thanks pzhon
• Bronze
Joined: 14.06.2008
doing these calculations for 9player sng's would you prefer using the Adjusted EV line or just the green line?
• Bronze
Joined: 17.06.2010
If you use the all-in luck adjustment, the standard deviation is lower, and the confidence interval is tighter. Exactly how much lower depends on the speed of tournament you play and your playing style, but super turbo players see a lower variance by over 70%, hence a standard deviation which is about half of the size. The adjusted results after 1000 super turbos are about as reliable as the unadjusted results after 4000 tournaments.
• Bronze
Joined: 14.06.2008
I'm playing 9man turbos on full tilt..
I have 5.7% all in luck adjusted roi in 600 games..
I somewhere read a post I believe you made that said that with the all in luck adjusted roi you have close 2.2x the sample size...
So should I calculate it as if I have 1320 games played on a 5.7% roi???
Or should I rather just lower the standard deviation and confidence interval?
somehow my common sense tells me I'll get similar results no matter which one I do lol
• Bronze
Joined: 17.06.2010
Either way, you should get the same result.
• Bronze
Joined: 24.05.2009
Originally posted by pzhon
The standard deviation for 6-max tournaments is almost the same as for 9-max. For a 70-30 structure, it might be slightly higher. For a 65-35 structure, it might be slightly lower.

You can calculate your standard deviation from your finishing distribution and the prizes.

A = average amount won
B = average square of the amount won
Variance = B - (A^2)
Standard Deviation = Sqrt(Variance)

So, for example, if you play \$12+1 6-max tournaments which pay \$46.80 and \$25.20, and you place first 20% and second 18%, you could calculate the following:

A = 20% x \$46.80 + 18% x \$25.20 = \$13.896
B = 20% x (\$46.80^2) + 18% x (\$25.20^2) = 552.3552 square dollars
Variance = 552.3552 - 13.896^2 = 359.256 square dollars
Standard Deviation = Sqrt(359.256) = \$18.95 = 1.46 buy-ins.

That suggests that for similar distributions, a rough 95% confidence interval after n tournaments would be your observed ROI +- 292%/sqrt(n).
Mmm statistics can be beautiful