Hi guys, I hear a lot of people on this forum say stuff like "don't let the downswings get to you, as it will all even out over a long enough sample" If this was the case then shouldn't the answer to the famous probability question "if i flipped a coin and it came up heads 50 times in a row, what is most likely to come next?" be tails? Given that all probability events should equal out over a long enough sample. Thanks for any help.

no, what people are saying is that after the flip has come up heads 50 times, then the probability is still 50/50.

In other words, if you are a winning player, but due to variance have been losing over a period of time, you should expect to be winning if you keep on playing. Anyone thinking they are owed an heater are wrong, but if they are thinking that they expect to win going forward (as they are a winning player) is right. So during a downswing all you really can do is keep on playing and eventually you will come out on top.

It's still 50-50. Sure, in the long run, it will probably even out, but your logic ignores how long the long run really is. In 50,000,000 flips it may even out after those 50 straight heads, but that might not even be the long run.

"In the long run, we're all dead." - John Maynard Keynes

When variance is discussed here I take it as say we go all in pre with 80% chance of winning (say AA vs KK) but we loose twice in one session, the correct play is to always keep doing as it's a massively +EV play and thus will be hugely profitable in the long run.

If you run 20 buy ins below EV over x000 hands, and then play another y000 hands (where x and y are large) you should expect to still be 20 buy ins below EV at the end of those y000 hands (in other words you'd expect to win exactly your EV in that second bunch of hands).

BUT, the amount you're running under EV is getting smaller as a % of your expectation.

The evening out comes not in the coin magically forcing itself to land tails in order to make up for an excess of heads but rather in the difference between the number of heads and number of tails becoming smaller in relation to the number of tosses as the number of tosses gets larger.

You ask a valid question. Kurn is right - the NEXT time you flip the coin it's 50-50, period.

I think the point that people are making is that if you have a patch, especially a LONG patch, of runbad (i.e. you get QQ all in pre heads up vs. AKs and lose 10 times in a row, or get AA in vs. KK and lose 3 times in a row), it's very highly probable (but not GUARANTEED) that you will EVENTUALLY have a patch, or possibly even a LONG patch, of rungood (i.e. QQ holds up against AK 10 times in a row, or AA holds up against KK 20 times in a row).

Try it for yourself - flip a coin 100 times and see what kinds of stretches you encounter.

EDITED to say, "Don't let the downswings get to you" is still good advice. If you manage to get a flopped set in against a flopped flush draw you made the right play regardless of the result and should pat yourself on the back. If you see a villain open shove 55, 88, JJ, AK, then he openshoves in front of you while you have QQ and on that one hand he happens to have AA, you still made the right play and should pat yourself on the back. It's easier said than done but it's true.

OP, saying that the downswings and the upswings will eventually "even out" is a bit misleading and inaccurate IMO as you could infer propositions such as yours.

After a given x number of hands, you'll have had either more upswings than downswings or the opposite. Actually, the higher your sample, the less likely downswings and upswings will have exactly compensated each other.
Take a few graphs with several million hands and look at the difference between the winnings and the net EV line after 10, 100, 1000, 10 000, 100 000 and 1 000 000 hands. You'll see that the difference will usually increase. Thus, you can't say that "because I've had more downswings than upswings in the past, I am now more likely to have more upswings".

However, if you divide downsings by upswings it will converge towards 1. They don't "even out", but they become negligible compared to your winnings line assuming you are a winning player.
Take the same graphs as earlier and divide the difference between the winnings and the net-ev line by the amount of the winnings. It should converge towards 0.

NB: I know that the net-EV line doesn't reflect the upswings and downswings, but it works the same way mathematically.

OP, that's a good short explanation if you find that mine is unclear.

That's wrong. The fact that you had a bad run doesn't affect your chances of having a good run later in any way. If that's not what you meant, then your explanation is not clear.

Nope, I was wrong and I retract my statement.

I was thinking that since over 100 flips you expect to get 50 heads that if you got 0 heads over the first 20 you'd expect 50 heads over the next 80. But that's wrong - you expect 40 heads over the next 80.

OP, several posters have given correct answers, I am just going to attempt to articulate it in a slightly different manner.

It evens out in the sense that the amount you deviate from your "true" win rate becomes smaller as a percentage of the whole win rate. It varies proportionately but not absolutely.

This is easy to visualise by looking at a graph of a long term, steady winner (or loser, but the former is less depressing). If you zoom in on any one area of the graph, it's very choppy, with big peaks and troughs. Each of those peaks and troughs might be as much as say 20 stacks. When you look at the graph over 100's of thousands of hands, it looks much smoother, with little peaks and troughs. But they still represent swings of up to 20 stacks, they just look small in terms of the overall picture.