Winrate Question

[Stork]

How many hands must be played in order to accurately know one's winrate within +/- .1BB/100, assuming respective standard variations of 10BB/100 and 20BB/100?

[Stork]

Too late to edit now, but if it matters, assume a true winrate of 2.5BB/100

[BruceZ]

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How many hands must be played in order to accurately know one's winrate within +/- .1BB/100, assuming respective standard variations of 10BB/100 and 20BB/100?

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It depends on what level of confidence you want. I will compute this for 95% confidence and 99% confidence, but you probably won’t achieve 0.1 bb/100 accuracy in your lifetime at these levels. The actual win rate doesn’t enter into the calculation. It would if you had specified the accuracy as a percentage of the win rate.

The standard error (SE) of the win rate is always SD/sqrt(N/100), where SD is the standard deviation for 100 hands, and N is the number of hands. For 95% confidence, we need 0.1 bb to be about 1.96 standard errors. For 99% confidence, we need 0.1 bb to be about 2.58 standard errors. These are from a table of the standard normal distribution, or the Excel function =NORMSINV((95%+1)/2) for 95% confidence, and =NORMSINV((99%+1)/2) for 99% confidence, etc. Call this number s, for the number of standard errors. Then we have:

0.1 = s*SD/sqrt(N/100)

N = 100*(s*SD/0.1)^2

We can use this equation to get the following results:

95% confidence:
For SD = 20 bb/100, N =~ 15.4 million hands.
For SD = 10 bb/100, N =~ 3.8 million hands.

99% confidence:
For SD = 20 bb/100, N =~ 26.5 million hands.
For SD = 10 bb/100, N =~ 6.6 million hands.

So as you can see, it takes a hell of a long time to achieve 0.1 bb/100 accuracy. You may want to settle for a wider margin of error, or a lower level of confidence.

[Stork]

Bruce, I'm pretty sure I've told you this before, but, you rule.

What about finding true winrate within +/- .5BB/100, with a 95% confidence interval?

Not to be rude, but he gave you the formula AND instructions for using them....

[BruceZ]

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Bruce, I'm pretty sure I've told you this before, but, you rule.

What about finding true winrate within +/- .5BB/100, with a 95% confidence interval?

[/ QUOTE ]

Just use the same equation, and change 0.1 to 0.5. This gives:

SD = 20 bb/100, N = 614,633 hands
SD = 10 bb/100, N = 153,659 hands

Note that decreasing the accuracy by a factor of 5 from 0.1 to 0.5 decreases the number of hands by a factor of 5^2 = 25. This will be true for any other factors. Also, changing the standard deviation by a factor of C will change the number of hands by a factor of C^2. The same is true for the number of standard errors s. You can use these relationships to compute any other result from the results that we already have.

[AaronBrown]

While this reply is excellent, as always, I would add the important caveat that makes several assumptions, the most important of which is that your win rate is constant. That means constant over time (you don't get better or worse, or have good and bad runs), at different games and limit structures and versus different other players.

As a practical matter, if you had a long time series of hands, the first thing you'd do is see if there are larger variations in the observed win rate than would be expected by random chance assuming a constant actual rate. Only if the series passed tests for randomness would you use BruceZ's calculation.

I've never seen a long series of hand statistics that did not exhibit some systematic variation of win rate. That adds considerably to the problem of estimating (or even defining) the long term average.

[BruceZ]

[ QUOTE ]
While this reply is excellent, as always, I would add the important caveat that makes several assumptions, the most important of which is that your win rate is constant. That means constant over time (you don't get better or worse, or have good and bad runs), at different games and limit structures and versus different other players.

As a practical matter, if you had a long time series of hands, the first thing you'd do is see if there are larger variations in the observed win rate than would be expected by random chance assuming a constant actual rate. Only if the series passed tests for randomness would you use BruceZ's calculation.

I've never seen a long series of hand statistics that did not exhibit some systematic variation of win rate. That adds considerably to the problem of estimating (or even defining) the long term average.

[/ QUOTE ]

Refer to the discussion I had with Copernicus about this issue starting with this post (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Number=303111&page=&view=&sb =5&o=&vc=1).

[Stork]

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Not to be rude, but he gave you the formula AND instructions for using them....

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You are absolutely correct, I merely skimmed the answers, I didn't look at the equation at all. Looks like Bruce was awesome enough to answer my second question anyways.
Aaron Brown definetely brings up a good point though in that winrate is not a fixed number which makes me wonder if figuring out ones winrate is a waste of time.

Considering how many hands it takes to get a close to accurate measure of one's true winrate (to minimize the fluxuations in winrate, assume for someone who plays the same game with the same cast of players), I wonder how likely it is that some B&M pros who play in the biggest side games are just on a good streak.