ALERT: Poker is a LONG TERM game!

[KingOtter]

I want to say something that bears repeating. We've got a lot of people here that are newer than I am, and I'm not seeing the mantra about poker being a long-term game repeated as much as it was when I first came about.

If you play for 3K hands and win 20BB/100, you are not a winning player.

"Why not?!" You may ask, "Look at that winrate!"

Because 3K hands is short-term. 5k hand is short-term. 10k hands is short-term. 20k hands is short-term in terms of win-rate.

Even at 20K hands, a good run of 50BB over 500 hands could change your win-rate 20% (assuming 1BB/100, at 2BB/100 it goes up 10%).

Crushing a limit isn't about going to any table and walking away with 100BB. Being a winning player doesn't mean that you can win, no matter what. Poker is still gambling. In the long and short of it you still have to get better cards than the other guy, most of the time.

Being a winning player is all about:
1. Adjusting to your opponents play.
2. Making GOOD poker decisions, every time.
3. Putting your opponents in positions where THEY make mistakes.
4. Understanding that a day is short-term... a week is short-term... a month is short-term.
5. Patience.

If you're complaining that you can't beat a limit and you have less than 20k hands as evidence, you're wasting our time.

KO

[jrz1972]

Good post.

Here's something to think about. How many hands do you need to play before you can be 95% confident that your true winrate lies within one BB (one way or the other) of the winrate PokerTracker is reporting?

Assume a standard deviation of 18, and let X be the number of hands you've played (in hundreds). You need to find the X such that your standard error is 0.5.

18/root(X) = 0.5

=> X = 1296.

Remember, that's in hundreds, so you would need to play 129,600 hands just to nail down your winrate to one point either way.

How many people have played 129,600 hands at their current level?

[Entity]

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Assume a standard deviation of 18,

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That's a monstrous SD. Damn. That's higher than my 10/20 6m SD. :P

Rob

[jrz1972]

[ QUOTE ]
[ QUOTE ]

Assume a standard deviation of 18,

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That's a monstrous SD. Damn. That's higher than my 10/20 6m SD. :P

Rob

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Seriously, that's my std. dev. in .5/1 full ring (~60k hands).

My std. dev. in 6-max is smaller.

[KingOtter]

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Assume a standard deviation of 18,

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That's a monstrous SD. Damn. That's higher than my 10/20 6m SD. :P

Rob

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Seriously, that's my std. dev. in .5/1 full ring (~60k hands).

My std. dev. in 6-max is smaller.

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It's mine, too.

KO

[Isura]

[ QUOTE ]
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Assume a standard deviation of 18,

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That's a monstrous SD. Damn. That's higher than my 10/20 6m SD. :P

Rob

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My SD at 1/2 and 3/6 6max is around 19...

</font><blockquote><font class="small">In Antwort auf:</font><hr />
That's a monstrous SD. Damn. That's higher than my 10/20 6m SD. :P
Rob

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Even for SD 14 you still need 212K hands

[lautzutao]

Is this the "leet's strike back" post? Sounds like the same kind of whining you're referring to.

To be 95% confident, you need to multiply SD with 1,6449, giving you 350K hands for SD 18.

[numeri]

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18/root(X) = 0.5

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This is on the right track, but not quite right. You've actually found X (or n) needed to get your standard error within 0.5. What you want is for your true rate to be within 0.5.

A confidence interval for your rate (say x) is:

x +/- Z*SD/sqrt(n)

The error in this case is Z*SD/sqrt(n). For a 95% confidence interval, Z = 1.96, so it's actually

1.96*18/sqrt(n) = 0.5

So n = 4,978 or ~500k hands. Yikes!