assumming......

[RudeboyOi]

youre a break even player
what are the chances
of losing 60 sessions in a row?

[bigpooch]

There is some chance of breaking even exactly. Also, what
if the session length was exactly ONE orbit so the chances
of losing is greater than one half.

A "break-even player" may win much less than half of the
sessions he plays in.

On the other hand, losing 60 sessions in a row is quite a
lot! For example, suppose our hypothetical break-even
player loses in about 70% of his sessions (which may be a
bit high for any players that are "close to break-even").
Even then, (0.7)^60 is a mere 5.08x(10^-10) which is
extremely small.

If a "break-even" player claimed such a massive losing
streak, he may no longer claim to be "break-even": there
could have been changes in the game that make him not only
a loser, but a significant one. Also, many players have a
problem with emotional discipline, so that is often one of
the factors contributing to long losing streaks.

[RudeboyOi]

there was a post in medium stakes forum
about a player that had 60 losing sessions in a row
and someone did the math (.50)^60
assuming he is breakeven
and that just doesnt seem right
ive forgotten most of my statistical knowledge
so was just checking with some people here

[DWarrior]

.5^60 actually seems pretty correct without any additional info. Each session you play, you're 50/50 to be ahead or behind. Maybe the calculation's a bit more complex to account for perfectly break even sessions, which are very rare.

However, what if he's a winning player, but after a bad beat he goes on sick tilt and loses, so he has mostly winning sessions and then a couple of very big losers. What are the odds of him losing 60 in a row then? Or what if he does well when he pays attention, but is a mild loser when he doesn't, and he doesn't pay attention most of the time (basically, what if he's a winning player only 10% of the time or so).

[MtDon]

[ QUOTE ]
youre a break even player
what are the chances
of losing 60 sessions in a row?

[/ QUOTE ]

I don't know off hand how to calculate that.

But, I'm pretty certain that more information is needed to do the calculation.

You need information on the variance of the player's play.

For example, one break even player might average a winning session 1 out of 5 sessions, another 1 out of 3. Obviously, the one with 1 out of 5 winning sessions is more likly to have 60 losing sessions in a row than the player who averages 1 out of 3.

After writing this, I'm not sure if saying that a player is "break even" has anything to do with the mathematical chances the player will lose 60 sessions in a row.

How many winning sessions a player has has as much to do with the games they play in as with how well they play. Heres an extreme example: a player might play 10% of the time in a tough 10/20 Holdem game and 90% of the time in an easy 3/6 holdem game. Be a winning player in the 3/6 game and a losing player in the 10/20 game. He might win 10% of the 10/20 sessions and 90% of the 3/6 sessions. The wins in the 3/6 might make up for the losses in the 10/20 game, so that the player is a "break even" player over all.

-- Don

[SamIAm]

[ QUOTE ]
Each session you play, you're 50/50 to be ahead or behind.

[/ QUOTE ]
Why in the world would this be true? Maybe that's the case for very long sessions, but the OP never said anything about session length. (That's a big problem with the question.) Imagine a session is 1 hand. Chances are high that you'll pay a bit and lose. There's a small chance that you'll win a big pot. Still break-even on average, but way off from 50/50.

The other problem with the OP is there's no discussion of playing style. High variance players (loose aggressive) will have an easier time losing on multiple short sessions than low variance players (tight passive).

[f97tosc]

[ QUOTE ]
Why in the world would this be true? Maybe that's the case for very long sessions, but the OP never said anything about session length. (That's a big problem with the question.) Imagine a session is 1 hand. Chances are high that you'll pay a bit and lose. There's a small chance that you'll win a big pot. Still break-even on average, but way off from 50/50.

[/ QUOTE ]

If we assume that the hands are independent (not entirely true), then by the central limit theorem, the outcome of a session with many hands is normally distributed, and that is a symmetric distribution, so EV=0, would imply 50% to be up or down (disregarding the small possibility of precisely break even).

When using the central limit theorem in practical applications, a common rule of thumb is that "many" means >30. And so for sessions lasting more than about an hour, the proposed answer should be quite good.

[PairTheBoard]

[ QUOTE ]
there was a post in medium stakes forum
about a player that had 60 losing sessions in a row


[/ QUOTE ]

It's about a billion to 1 shot for someone whose percent of losing sessions is 70%. A million to 1 for someone whose percent of losing sessions is about 80%. Neither of these cases sound like breakeven play to me. And they are still prohibitive longshots.

My conclusion is, I don't believe the report.

PairTheBoard

[SamIAm]

[ QUOTE ]
When using the central limit theorem in practical applications, a common rule of thumb is that "many" means >30. And so for sessions lasting more than about an hour, the proposed answer should be quite good.

[/ QUOTE ]
I guess our difference of opinion comes down to this definition of "many". I think it's a pretty bad rule of thumb in general. Let's say a pot is always $100 and I have a 10% chance of winning it and a 10% chance of paying my $10 and losing. I find it hard to believe 30 is enough trials for this to be 50/50 win/lose, but I haven't worked it out. Obviously there's a problem if I had a 1% chance of winning $100 and 99% chance of paying $1.
-Sam

[AaronBrown]

I agree that it takes more than 30 hands to start thinking about the Central Limit Theorem. A tight player might easily fold 25 of those hands preflop.

Let me tackle the question the other way, what is the minimum reasonable session winning percentage for a break-even player? I think it can be significantly less than 50%.

Suppose, for example, in a no-limit game a certain player wins a big pot, averaging $10,000, 0.5% of the time. He is careful, however, to avoid losing big pots anywhere near that often, meaning he usually folds to all-in bets, even with good hands. He also folds most hands preflop.

Over 100 hands, 61% of the time he will get no big hands. Since his EV from these hands is $5,000, he will have an EV of -$5,000 on these sessions. Since he rarely loses big hands most all of these 61% will be losing sessions. On another 30% of sessions he will win one of these hands, making his session EV +$5,000, most of these will be winning sessions but some will be losing, offsetting the winning sessions with no big pot wins. So, overall, it might be reasonable that he loses money 60% of the time, makes a small profit 30% of the time and only breaks even due to the 10% of his sessions in which he wins two or more big pots.

But 100 hands is a short session, and this is a highly skewed distribution of hand outcomes. I played around a bit, and can't come up with reasonable scenarios in which a break even player would make money in fewer than 40% of sessions. Even that's stretching it.