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MTTSNG Discussion and analysis of MTTSNGs.

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Old 06-02-2017, 09:05 AM   #1
csafischer
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roi and results in mmtsng

Hi everybody
I'd like to know how many sit mmt i have to play to know if i am a winner player.
This is my graph in last few month. I play very little because i have some problem, but now i think that i wanna restar to play it.

http://imgur.com/CB3e4b8

So, what's a good roi? An how many sit have i play to have a true indication about my ability?

Thank you very much.

Last edited by csafischer; 06-02-2017 at 09:32 AM. Reason: forgot to put pic
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Old 06-02-2017, 12:06 PM   #2
AceofSpades11965
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Re: roi and results in mmtsng

10k tournaments to have real picture, including upswings, downswings, some bad sessions, some really good session etc...
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Old 06-03-2017, 11:57 AM   #3
csafischer
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Re: roi and results in mmtsng

oh, I was hoping it took less sng :/
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Old 06-24-2017, 12:19 PM   #4
statmanhal
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Re: roi and results in mmtsng

I did a sample size analysis for a STT sit’n’go. 10k tournamemts for a MTT seems to me quite a stretch.

The table below displays the number of STT sit’n’go s you should play to determine with a 90% confidence that you are a winning player. It uses the central limit theorem to justify assuming a normal profit distribution with profit determined by the equation Profit = ROI*Total Buy-in (includes entry fee). The situation assumed is a 9-man SNG with a $9 prize-pool buy-in and a $1 entry fee. The prize structure is 50% for 1st, 30% for 2nd and 20% for 3rd. Given that data and the placing distribution when in-the-money (ITM), it is possible to calculate the standard deviation. Four placing distributions were analyzed

Uniform - each placing equally likely with relative weights of 1,1,1 when in the money.
Symmetrical – relative weights of 1,2,1, signifying that 2nd place is twice as likely as first and third places
First place most likely – relative weights of 3,2,1 signifying that first place is 1.5 times as likely as 2nd place and 3 times as likely as third place when in the money
Third place most likely – relative weights of 1,2,3

For each of these placing distributions, there is also a fourth result and that is out of the money meaning a loss of the total buy-in. This results in the First Place Most Likely placing distribution having the highest standard deviation and thus the largest sample size requirement. As one would expect, the largest sample size requirement to show you are a SNG winner is for the smallest ROI. The sample size range for the cases examined is 417 for a 10% ROI with first place money as the most frequently occurring prize while the smallest required sample size is just 17 when the ROI is a high 40% and third place is most likely.
[B]

  9-Man SNG/ Sample Size for 90% ConfidenceWR>0 vs. ROI
    Return on Investment, ROI 
ITM Placing Distrib. 10% 15% 20% 30% 40%
Uniform 1-1-1 350 162 92 42 24
Symmetrical 1-2-1 331 149 85 39 22
First Likely 3-2-1 417 189 109 50 29
Third Likely 1-2-3 275 124 70 31 17

Example: If you are equally likely to finish in first, second or third place if ITM, and if your ROI is 15%, to be 90% confident you are a winning player, you need to play 162 SNG tournaments.
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Old 06-25-2017, 04:08 AM   #5
LektorAJ
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Re: roi and results in mmtsng

Doesn't it depend on the Bayesian prior probability that you are a winning player?

So if someone has only been playing poker for one month and they post 15% ROI over 162 SNGs, there is still not a 90% chance they are a winning player.

Whereas I am a winning player in 25 euro 5-max SNGs, if I sat in the 50 cent 9-max games on pokerstars, even if I posted a negative ROI over the first 162 games it doesn't change the fact that I am highly likely to be a winning player at that level.

I haven't looked at your calculations in detail but it sounds like what you have calculated is something closer to "given you have an underlying ROI of 15%, how much do you need to play to have positive ROI 90% of the time."
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Old 06-25-2017, 09:52 AM   #6
statmanhal
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Re: roi and results in mmtsng

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Originally Posted by LektorAJ View Post
Doesn't it depend on the Bayesian prior probability that you are a winning player?
Yes, but only if you are a Bayesian. I simply applied classical confidence interval theory for a normally distributed variable assumed through central limit theorem.

Quote:
Originally Posted by LektorAJ View Post
I haven't looked at your calculations in detail but it sounds like what you have calculated is something closer to "given you have an underlying ROI of 15%, how much do you need to play to have positive ROI 90% of the time."
Technically, the more accurate statement would be, “given a sample ROI of 15% over n tournaments, then 90% of computed confidence intervals from similar samples of the same size would have a lower limit >= 0.”
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Old 06-25-2017, 01:10 PM   #7
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Re: roi and results in mmtsng

Quote:
Originally Posted by statmanhal View Post
Yes, but only if you are a Bayesian. I simply applied classical confidence interval theory for a normally distributed variable assumed through central limit theorem.


My second and third paragraph above are either objectively true or they're not, regardless of whether or not any particular person is a Bayesian. In fact they are true and therefore any method that gets the wrong answer isn't useful.

Quote:
Originally Posted by statmanhal View Post
Technically, the more accurate statement would be, “given a sample ROI of 15% over n tournaments, then 90% of computed confidence intervals from similar samples of the same size would have a lower limit >= 0.”
Well clearly that's nonsense then. There are formats like BTC where it's debatable whether they are beatable at all. Just because you run hot over some sample doesn't mean there's suddenly a 90% chance that its not only beatable but you are the man who can do it.

Or to take another example. If you have a 1000 players playing roulette, after 100 spins what percentage will have an observed winrate that leads you to conclude they are 90% likely to be winning?

Under Bayesian reasoning you would still say, no, none of them are winning players which of course is the correct answer.
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Old 06-25-2017, 07:12 PM   #8
statmanhal
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Re: roi and results in mmtsng

I think you may be misunderstanding my comments. When I said “you” have to be a Bayesian, the “you” is the analyst, not the player.

Two schools of statistics are classicists and Bayesians, where the latter accept the concept that a population parameter such as win-rate, can be considered to be a random variable while classical statistics does not. That’s why classicists use confidence intervals, which assign the random variable to the confidence interval and not the parameter. This has been ongoing for some 80 years and still is quite prevalent today.

What I did was to find the sample size, i.e., number of SNGs to play, so that there is a 90% probability that the lower limit of the computed interval is greater than 0, based on an observed sample win rate and standard deviation. If you don’t like that idea, that’s your prerogative.
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Old 06-28-2017, 04:19 PM   #9
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Re: roi and results in mmtsng

I thought I would try calculating the value of games needed for 180 MttSngs using the same rake percentage as statmanhal did for the 9s STTs above ie, $0.90 + $0.10 rake, with an exactly 10% roi. (I used a uniform 0.67902% for each itm place from 1st to 28th)

The variance per game for these $1 180's is 26.6762, so for a block of 4300 $1 games the variance is:
26.6762 * 4300 = 114707.5799
The stdev = sqrt(114707.5799) = 338.6851

If you play a block of 4,300 of these your expected mean gain is $430.00 but roughly 10% of the time you will get $0.00 or below. Phew, these big field tournies are pretty tough.

It seems with a true 10% roi in 9s you need about 350 games to be 90% confident of seeing a non negative block, but approx. 4300 games for the same level of confidence in 180's.

If a player with 10% roi plays blocks of 4300 180s games about 1 in 10 of these blocks will show a loss.

Last edited by BaseMetal2; 06-28-2017 at 04:28 PM.
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Old 06-29-2017, 04:40 AM   #10
LektorAJ
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Re: roi and results in mmtsng

Quote:
Originally Posted by statmanhal View Post
I think you may be misunderstanding my comments. When I said “you” have to be a Bayesian, the “you” is the analyst, not the player.
The right answer is just the right answer. It doesn't change depending on who the analyst is. If the analyst's beliefs don't get him to that right answer he isn't a good analyst.

Quote:
Originally Posted by statmanhal View Post
What I did was to find the sample size, i.e., number of SNGs to play, so that there is a 90% probability that the lower limit of the computed interval is greater than 0, based on an observed sample win rate and standard deviation. If you don’t like that idea, that’s your prerogative.
So if we make 100 bets alternating red and black at live European roulette and get 57 wins out of 100 for an ROI of 14%, the Vassar stats
http://vassarstats.net/binomialX.html page suggests that someone with that ROI will run positive 90% of the time over that run, so you say this means we can be 90% sure we are winning at roulette?

Based on the real ROI (-0.2702%) 5.8% of all players will run that hot over that sample and if those people meet a frequentist statistician the statistician will tell them there is a 90% chance they are beating the game?

If they meet a Bayesian statistician they will say there is a 0% prior probability the person can beat roulette by alternating red and black, and the results haven't changed it.

Only one of them can be correct.
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Old 07-01-2017, 07:21 AM   #11
maXXmilian
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Re: roi and results in mmtsng

Hello,

can someone please share what is the average duration of 2,5 and 3,5r 180s over some decent sample? Thanks.
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Old 07-16-2017, 01:34 PM   #12
Uhrenknecht
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Re: roi and results in mmtsng

Quote:
Originally Posted by maXXmilian View Post
Hello,

can someone please share what is the average duration of 2,5 and 3,5r 180s over some decent sample? Thanks.
it stands right in the lobby. 180man take max 115 and 3.5$R about 120-140 afaik.
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