I knew I shouldn't have put those points into charisma.

If you have an overall EV for a strategy against a player pool, all that is accounted for. The law of large numbers applies just as in any series of random trials. If x is your total winnings, e your EV and N the number of hands you play, then lim (N—>infinity) (x/N)/e = 1. Or rearranged x/eN =1.

Note that the difference between your actual winnings and the EV x number of hands may not necessarily get smaller. Just as an example consider a win rate of 3BB/100 or 0.03BB per hand. It would not be absurd after 10000 hands to have an actual winnings amount of 295bb The expected value is 300, so the difference is 5 bb. After 100000 hands you might have a total of 2970 bb winnings. This is a difference from expected (3000 bb) of 30 bb. Seems like the law of large numbers isn’t working.

That isn’t the case though. The convergence is based on ratios not differences. In the first case your winnings are 295/300 = 98.33% of the EV. In the second the winnings are 2970/3000 = 99% of the EV. The point is that while the actual result will converge to the EV, the absolute difference between EV and result will not necessarily converge and may even grow as more hands are played.

The absolute difference may also (and probably will) fluctuate between positive and negative.

I am absolutely sure that the absolute difference will not fluctuate between positive and negative, I think

True. The difference will fluctuate. (Absolute numbers have no sign!) I will blame my brain fart on it being early morning, sort of early anyway.




ETA: I was half way through typing a post explaining how it could happen before I went "doh".

It may fluctuate between positive and negative, but it actually is more likely to remain either positive or negative for extended periods. It seems counterintuitive but a simplified example can help.

Suppose Peter and Paul play a game. They each flip a coin. If the coins match (both heads or both tails) Peter wins 1 point. If the coins are different Paul wins one point. Obviously since it is a fair game, the percentage of wins for each player will tend to equalize over many plays. Instead of that question, suppose we consider a different one — if they play some large number of games, what is the probability that Peter will be in the lead x% of the time during those plays? (Obviously the symmetrical question about Paul would have the same answer).

It is possible to calculate this value for any value of x ranging from zero to 100%. When we do so, an interesting and (IMO) counterintuitive result occurs - you get the maximum probability for either 0 or 100% and the minimum probability at 50%. It is more likely that one player or the other will lead the whole time than it is that they will go back and forth and each will lead about the same amount!

For those interested in the technical details, the calculation can make two assumptions about tie scores - either the player who led last is consistent to still be leading or the player who won to tie the score can be assumed to be the new leader. Either way the result holds. In technical terms the distribution is an arcsine distribution over the interval [0,1], which takes on maxima at the endpoints and a minimum at 1/2.

To make some kind of sense of this consider the first few games. Suppose (without loss of generality) that Peter wins game one. Then depending on your assumption about who the leader is on tie scores, either Peter will always lead after game 2 or either player is equally likely to lead. In the first case, we already have an asymmetry — the player who wins first will automatically lead for two rounds. In the second case, we find that if Peter also wins game 2, we get the same asymmetry - only Peter can lead after 3 games. Loosely speaking, these asymmetries add up over many games, making it more likely that one player or the other will lead most of the time.

Hopefully the parallel with long term poker winnings is obvious. Once you go positive over x hands, you either will go positive over the ensuing x hands, or negative over those hands. If you go negative, you are likely to be close to breakeven. If positive, you become even less likely to go negative (as your total winnings would be even greater after 2x hands than they were after x hands). Unlike the Peter vs Paul game, though a long term winner at poker isnÂ’t a fair game, and the probability of being positive after many hands is much greater than being negative.

Edit — of course the absolute difference, being defined as positive will not fluctuate, but it also is more likely that the signed difference will also not fluctuate, as described above.

I'm absolutely positive that they are approximately exactly the same.

This sentence was the subject to over an hour of discussion at a recent poker table .. amazing .. GL

We are seriously starting to get into “Yogi Berra” territory here (as I age myself)

For those unfamiliar, Yogi was a baseball player back around the 1950s-1960s who was famous for making statements that sounded reasonable at first but would make you just shake your head and go “huh?” when you realized what he said. My favorites were (describing a local restaurant) “Nobody goes there anymore, it’s too crowded” and “You have to go to your friends’ funerals or they won’t go to yours”.

It's true that one player will hold the lead most of the time. It's also true that the probability that one player will lead the whole time is actually 0, and each player will hold the lead an infinite number of times.

I just have to add that if Yogi played poker (and I'd be surprised if he didn't) he would be sure that "80% of the game is half skill".

What will fluctuate infinite times is your relation to the EV line, not to the breakeven line unless your EV is zero. For the same reason that a +EV player's risk of ruin isn't 100%, their chance of hitting the break-even line isn't 100% once they pull ahead.

Was talking about the coin flip game which is 0 EV as described. It's just a random walk.

There are a Few undeniable arguments that can be made that demonstrate that poker is askill based game with an element of some luck involved.

You can not have "professional players" in ANY game that is purely skilled based That is a Fact.

If you dealth the same exact cards to 4 different people facing the same exact situation they now have a choice to fold check call or raise
While 2 people might call 1might fold and 1 might raise THIS NOW leads to a branch in the decision tree which it and of itself
leads to different results. The same thing occurs on the next street where a decision is to be made.
Even IF 2 players made the Same exact decisions on an action their bet amounts on a raise could easily be different amounts of money needed to be
called by the villan. This is would lead to a different amount of profit or loss.

Any reasonable person with half a brain needs to consider the above and its OBVIOUS implications.

What?

No professional chess players.

@ VB

just what I said you cant have professional players in any game that is pure Luck. There ARE professional poker players
when was the last time you heard of a professional roulette player.
And I illustrated about decision tree branches in poker.

Poker is 70% skill 30% luck
anyone who thinks it is the other way around HAS NO IDEA of what they are talking about

But you wrote that you can't have professional players in any game that is pure skill. Perhaps you should read the things you write before posting.

is an obvious Typo. My bad. I dont think there is an edit buttom omce you have already posted ?