Is there a certain number of hands or sessions one needs to play to get an idea?

10K hands for a very loose idea of if you can beat a level.
100k for a better idea of what your win rate is.

There are win rate and standard deviation calculators online you can play with. If you do you will see even a winning player, while in the minority, can have a 100k downswing though.

agreed, 100k hands seems like a decent cutoff, just remember that when it comes to probabilistic events nothing is definitive

There is no fixed number of hands that will tell you if you're a winning player. However, you can estimate the likelihood that you are a winning player as a function of your observed win-rate, standard deviation and hands played.

This site will do it for you without making you learn any formulas.

http://pokerdope.com/poker-variance-calculator/

As an example, if you play 10,000 hands with a true win-rate of 0 BB/100, crushed over the sample at 20 BB/100, and had a SD of 80 BB/100, the probability of having this result given the premise that we are a breakeven player is .6%, so it's 99.4% likely that we're a winning player.

You can find all these stats in PokerTracker.

If you suck, not that many

Thanks for this, I recently started transitioning to cash as my temperament is better suited for it I think. My MTT bb/100 was 21/100 over 60k hands at $5ABI and I recently put up 27/100 over 6kish hands at 10NL so I wanted to move up and I wanted a good idea of how long it would take for me to find out if I could hack it at 25NL and 50NL as I'm decently rolled for both (35-50 BIs on different sites), but I don't want to go off the deep end.

Set a floor.

"If will lose $XXXX, I will drop back to 5NL (or 2NL) level"

I like the 10k number, as it is what I have aimed to get in first and think later.

Depends on winrate to get an idea of if you are a winning player or not, higher the winrate over a given sample size the more likely is you are beating the limit and vice versa for losing. For a pretty accurate idea of winrate in terms of bb/100 100k+ - the more the better.

However, there are a few caveats in my opinion such as when you are playing online and starting at a new limit or a new site the lack of any notes and stats on any opponents definitely negatively impact your winrate - for example when I started at NL2 I was pretty much break even for the first 5-10k hands and then moved up to 9 bb/100 by the end of 100k hands and similar start to 5nl before consistently beating it over a decent sample size.

Use a confidence interval calculator.

Std dev should be ~100 for 6max NLH

you can input varius hand amounts, confidences, and winrates

https://www.mathsisfun.com/data/conf...alculator.html

Focus on getting your chips in and your opponents chips in when you have a edge in spots. If you have AA preflop you got a edge in that spot (unless opponent has AA too).

Don't focus on winning or losing money. Look for edges. If you can't find them look harder or go play mahjong.

Sample Size Needed to Assure You Are a Winning Player

I have posted several versions of the following, which uses normal distribution confidence interval theory to develop a required sample size. It has not been challenged as far as I know, yet I haven’t seen any great acceptance either. Use or ignore as you see fit.

Assume you have sample data that indicates you have a positive win rate of W/100 with a standard deviation of S/100. To determine what sample size you need to assure with a specified confidence that you are a winning player, you can use the following formula:

N =100*( Zc*S / W)^2

where Zc is the normal distribution factor corresponding to a one-sided C% confidence interval: Z80 = 0.84, Z90 = 1.28, Z95 = 1.645. Note that N increases with C and S, and decreases with W.

Example: You have an estimated win rate of 4 per 100 hands with a corresponding standard deviation of 75/100. To be 80% confident you are a winning player,

N = 100*(0.84*75 / 4)^2 = 24,800.

For C=90%, N = 57,600 (approx.) and N = 95,100 for 95% confidence.

The above assumes your win rate is distributed normally (valid by the Central Limit Theorem) and that your win rate and standard deviation are relatively constant over the sample period. If your win rate is not constant but is increasing and the standard deviation is non-increasing, then the sample size result can be considered to be conservative.- i.e., it is more than you need.