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View Full Version : Luck and variance in large tournament fields

Burno
12-22-2004, 06:34 AM
Most of us understand that poker is a long term game and that the long run takes a long time to reach. However, over the course of several years it is very easy to reach the long run as far as ring games are concerned.

Now obviously this doesn't hold true for large field MTT's. From my understanding, it's impossible to reach the long run in an average career(live tournaments). So luck must play a part, but how big?

So here's my problem. If someone more mathematically inclined than I could assist me I would be quite appreciative.

Player A is 1.5 times as good as the average player

Player B is 2.0 times as good as the average player

Assuming the average field is 500, what is the likelihood that Player A will have a more successful career(wins and final tables) than Player B? Let's assume a career is 300 events. What about if it were over only 100 tournaments? 50?

I'm not sure if this can be calculated, but I'm interested to hear any thoughts on it nonetheless. Thanks.

TomCollins
12-22-2004, 10:37 AM
If you want a mathematical answer, you will need to define "more successful". If you just say final tables, or just wins, it should be fairly easy.

Burno
12-22-2004, 10:51 AM
TomCollins,

Final tables or wins would definitely be an adequate measure of success. I'd be grateful if you or anyone else could walk me through the calculation.

knsmith85
12-28-2004, 03:59 PM
[ QUOTE ]
TomCollins,

Final tables or wins would definitely be an adequate measure of success. I'd be grateful if you or anyone else could walk me through the calculation.

[/ QUOTE ]

I'm a novice, but the biggest problem I notice is the way you are quantifying (or rather, aren't quantifying) how much better player A is than player B. It is certainly feasible to calculate if you say:

"Player A has an average finish of (50th percentile) with a standard deviation of 20 (percentile).

Player B has an average finish of (40th percentile) with a standard deviation of 15 (percentile)."

If we stated the problem like this, it would ALMOST be easy to do using a simple hypothesis test. However, doing such a test would require us to assume a normal distribution of placing, which we obviously can't do in this situation.

The key is that there is no bias (I think this is the term we want to use) in the normal distribution, while there is in the non-normal one.

Now that the idea of using a 2-sample T-test is out, we need to figure out another way to do this calculation. Probably the best way would be a computer simulation, but that's not really going to help us understand how to approach something like this.

So, ruling out those two methods, I don't really know how you'd do it. Anyone care to enlighten me (and correct me on something I've said as I know I must've stated something incorrectly above).