Number of hands and the "long run"

[Shaman]

In limit holdem, what is the minimum number of hands played that constitute the "long run"?

I read somewhere that the minimum long run is 2,000 hours at brick-and-mortar. At 35 hands per hour, this means 70,000 hands. Accurate?

[Reef]

I'm guessing 500k - 2M

[jason1990]

Thinking of this as a probability/statistics question, you must first define the "long run" before you can answer it. I like to define the "long run" (for a winning player) as the last time that player has a negative balance. Unfortunately, with this definition, you never know whether or not the "long run" has happened. But you can do probability estimates. For example, a player with a 2 BB/100 win rate and a 15 BB/100 standard deviation has about a 95% chance of reaching the "long run" in under 22,500 hands.

[arcticfire]

If it takes you 500K-2M hands to figure out how your doing or will do in "the long run" I'd say something is wrong.

[Marc B]

I like this... I wonder, are there any S&M books or essays that have a rough formula for calculating the likelihood you will see the long run in X hands? Clearly the formula would have to take into account std dev, and assumes you are playing the same 'game' over that # of hands.

[jason1990]

The probability of the long run (with this definition) occuring before N*100 hands is the same as the probability that

-W*sqrt{N}/S < X < W*sqrt{N}/S,

where X is a standard normal random variable, W is the winrate, and S is the standard deviation. You can use software or tables to get at these values. I wrote an article in which I derived this formula, but the derivation uses concepts such as Brownian motion, continuous time martingales, stopping times, optional sampling, and Girsanov transforms. If you really want to read it, I can email it to you.

[Munga30]

Look for Homer's Confidence Intervals post in the Micro forum FAQ. That will get you started.