bad luck downswings don't exist?

[beach_bum]

[ QUOTE ]


You would expect about 1.6 out of 20 on average, so your results are just a little lucky, but still in line with expectations. The average win rate for 20,000 hands is 3*200 = 600, and the standard deviation for 20,000 hands is 30*sqrt(200) =~ 424. So being negative after 20,000 hands is 600/424 standard deviations below average, which corresponds to a probability of 1 - NORMSINV(600/420) =~ 7.85%, or about 1.6 in 20.

Repeating this for 100,000 hands would that show being behind after that many hands would be more than 3 standard deviations below average, and the probability comes out to about 1 in 1278. So while rare, this is still possible.



[/ QUOTE ]

Hi Bruce,

I take this to mean that you are saying that a 3BB/100 player with a variance of 30 BB/100 should have a 20K breakeven stretch about 7.85% of the time. Could you tell me what percent of the time a 5BB/100 player with a variance of 50BB/100 should expect to experience a 20K and a 30K breakeven stretch?

Thanks very much.

[BruceZ]

[ QUOTE ]
[ QUOTE ]


You would expect about 1.6 out of 20 on average, so your results are just a little lucky, but still in line with expectations. The average win rate for 20,000 hands is 3*200 = 600, and the standard deviation for 20,000 hands is 30*sqrt(200) =~ 424. So being negative after 20,000 hands is 600/424 standard deviations below average, which corresponds to a probability of 1 - NORMSINV(600/420) =~ 7.85%, or about 1.6 in 20.

Repeating this for 100,000 hands would that show being behind after that many hands would be more than 3 standard deviations below average, and the probability comes out to about 1 in 1278. So while rare, this is still possible.



[/ QUOTE ]

Hi Bruce,

I take this to mean that you are saying that a 3BB/100 player with a variance of 30 BB/100 should have a 20K breakeven stretch about 7.85% of the time.

[/ QUOTE ]

It means that this is the probability that you will break even or lose over your next 20K hands.


[ QUOTE ]
Could you tell me what percent of the time a 5BB/100 player with a variance of 50BB/100 should expect to experience a 20K and a 30K breakeven stretch?

[/ QUOTE ]

For 20K the answer is the same as before since the ratio of SD to WR is still 10. For 30K, the WR is 5*300 = 1500, and the SD is 50*sqrt(300). The probability of breaking even or worse is 1 - NORMSDIST(1500/(50*sqrt(300)) =~ 4.16%.

Note that we use NORMSDIST, not NORMSINV as I had in my last post. This has been corrected.

Also note that the 50 bb/100 is the standard deviation, not the variance.

[beach_bum]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]


You would expect about 1.6 out of 20 on average, so your results are just a little lucky, but still in line with expectations. The average win rate for 20,000 hands is 3*200 = 600, and the standard deviation for 20,000 hands is 30*sqrt(200) =~ 424. So being negative after 20,000 hands is 600/424 standard deviations below average, which corresponds to a probability of 1 - NORMSINV(600/420) =~ 7.85%, or about 1.6 in 20.

Repeating this for 100,000 hands would that show being behind after that many hands would be more than 3 standard deviations below average, and the probability comes out to about 1 in 1278. So while rare, this is still possible.



[/ QUOTE ]

Hi Bruce,

I take this to mean that you are saying that a 3BB/100 player with a variance of 30 BB/100 should have a 20K breakeven stretch about 7.85% of the time.

[/ QUOTE ]

It means that this is the probability that you will break even or lose over your next 20K hands.


[ QUOTE ]
Could you tell me what percent of the time a 5BB/100 player with a variance of 50BB/100 should expect to experience a 20K and a 30K breakeven stretch?

[/ QUOTE ]

For 20K the answer is the same as before since the ratio of SD to WR is still 10. For 30K, the WR is 5*300 = 1500, and the SD is 50*sqrt(300). The probability of breaking even or worse is 1 - NORMSDIST(1500/(50*sqrt(300)) =~ 4.16%.

Note that we use NORMSDIST, not NORMSINV as I had in my last post. This has been corrected.

Also note that the 50 bb/100 is the standard deviation, not the variance.

[/ QUOTE ]

Hi Bruce,

Thanks for the response. Based on your explanation, I am wondering how so many top players who claim to have winrates close to 5BB/100 over hundreds of thousands of hands can claim to have somewhat frequent breakeven stretches of 50K hands or longer. It seems as though they talk about these long breakeven stretches as though they are absolutely to be expected on a regular basis. Yet according to your calculations they should be very rare.

Thanks

[BruceZ]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]


You would expect about 1.6 out of 20 on average, so your results are just a little lucky, but still in line with expectations. The average win rate for 20,000 hands is 3*200 = 600, and the standard deviation for 20,000 hands is 30*sqrt(200) =~ 424. So being negative after 20,000 hands is 600/424 standard deviations below average, which corresponds to a probability of 1 - NORMSINV(600/420) =~ 7.85%, or about 1.6 in 20.

Repeating this for 100,000 hands would that show being behind after that many hands would be more than 3 standard deviations below average, and the probability comes out to about 1 in 1278. So while rare, this is still possible.



[/ QUOTE ]

Hi Bruce,

I take this to mean that you are saying that a 3BB/100 player with a variance of 30 BB/100 should have a 20K breakeven stretch about 7.85% of the time.

[/ QUOTE ]

It means that this is the probability that you will break even or lose over your next 20K hands.


[ QUOTE ]
Could you tell me what percent of the time a 5BB/100 player with a variance of 50BB/100 should expect to experience a 20K and a 30K breakeven stretch?

[/ QUOTE ]

For 20K the answer is the same as before since the ratio of SD to WR is still 10. For 30K, the WR is 5*300 = 1500, and the SD is 50*sqrt(300). The probability of breaking even or worse is 1 - NORMSDIST(1500/(50*sqrt(300)) =~ 4.16%.

Note that we use NORMSDIST, not NORMSINV as I had in my last post. This has been corrected.

Also note that the 50 bb/100 is the standard deviation, not the variance.

[/ QUOTE ]

Hi Bruce,

Thanks for the response. Based on your explanation, I am wondering how so many top players who claim to have winrates close to 5BB/100 over hundreds of thousands of hands can claim to have somewhat frequent breakeven stretches of 50K hands or longer. It seems as though they talk about these long breakeven stretches as though they are absolutely to be expected on a regular basis. Yet according to your calculations they should be very rare.

Thanks

[/ QUOTE ]

How often they occur is really a different issue. Every hand can start a new 20,000 hand downswing, though the overlapping stretches are not independent of each other.

[beach_bum]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]


You would expect about 1.6 out of 20 on average, so your results are just a little lucky, but still in line with expectations. The average win rate for 20,000 hands is 3*200 = 600, and the standard deviation for 20,000 hands is 30*sqrt(200) =~ 424. So being negative after 20,000 hands is 600/424 standard deviations below average, which corresponds to a probability of 1 - NORMSINV(600/420) =~ 7.85%, or about 1.6 in 20.

Repeating this for 100,000 hands would that show being behind after that many hands would be more than 3 standard deviations below average, and the probability comes out to about 1 in 1278. So while rare, this is still possible.



[/ QUOTE ]

Hi Bruce,

I take this to mean that you are saying that a 3BB/100 player with a variance of 30 BB/100 should have a 20K breakeven stretch about 7.85% of the time.

[/ QUOTE ]

It means that this is the probability that you will break even or lose over your next 20K hands.


[ QUOTE ]
Could you tell me what percent of the time a 5BB/100 player with a variance of 50BB/100 should expect to experience a 20K and a 30K breakeven stretch?

[/ QUOTE ]

For 20K the answer is the same as before since the ratio of SD to WR is still 10. For 30K, the WR is 5*300 = 1500, and the SD is 50*sqrt(300). The probability of breaking even or worse is 1 - NORMSDIST(1500/(50*sqrt(300)) =~ 4.16%.

Note that we use NORMSDIST, not NORMSINV as I had in my last post. This has been corrected.

Also note that the 50 bb/100 is the standard deviation, not the variance.

[/ QUOTE ]

Hi Bruce,

Thanks for the response. Based on your explanation, I am wondering how so many top players who claim to have winrates close to 5BB/100 over hundreds of thousands of hands can claim to have somewhat frequent breakeven stretches of 50K hands or longer. It seems as though they talk about these long breakeven stretches as though they are absolutely to be expected on a regular basis. Yet according to your calculations they should be very rare.

Thanks

[/ QUOTE ]

How often they occur is really a different issue. Every hand can start a new 20,000 hand downswing, though the overlapping stretches are not independent of each other.

[/ QUOTE ]


Hi Bruce,

Sorry, but I am a bit confused. Is there a way to determine for a player averaging 5BB/100 who has a standard deviation of 50 BB/100 how often he should be expected to experience a 20K, a 30K, or a 50K breakeven stretch over a sample size of 1,000,000 hands?

Thanks again!

[BruceZ]

[ QUOTE ]
Sorry, but I am a bit confused. Is there a way to determine for a player averaging 5BB/100 who has a standard deviation of 50 BB/100 how often he should be expected to experience a 20K, a 30K, or a 50K breakeven stretch over 1,000,000 hands?

[/ QUOTE ]

This is a hard question that has been asked here before, and I have never seen a closed form solution. I suspect that there isn't one, since even the apparently much simpler problems involving success runs can only be solved recursively or by raising a matrix to a large power. The result can be found by computer simulation.

Since the number of 20K, 30K, and 50K non-overlapping runs in 1 million hands is 50, 33.3, and 20 respectively, we know that the result must be AT LEAST as large as these numbers multiplied by the probabilities derived above. The actual numbers are larger than this, perhaps significantly larger, since a losing stretch can start in one interval and end in another. Anyway, here are the minimum numbers FWIW:

20K: 7.86% * 50 =~ 3.9

30K: 4.16% * 33.3 =~ 1.4

50K: 1.27% * 20 =~ 0.25

EDIT: Changed 20K value to 3.9.

[raze]

[ QUOTE ]
or are all downswings a result of bad play moreso than bad luck?

[/ QUOTE ]

I won't pretend to understand the analysis, but as a 7+bb/100 NL player I have a LOT less downswings than when I was playing 4bb/100 NL and 1.25bb/100 FL.

[beach_bum]

[ QUOTE ]
[ QUOTE ]
Sorry, but I am a bit confused. Is there a way to determine for a player averaging 5BB/100 who has a standard deviation of 50 BB/100 how often he should be expected to experience a 20K, a 30K, or a 50K breakeven stretch over 1,000,000 hands?

[/ QUOTE ]

This is a hard question that has been asked here before, and I have never seen a closed form solution. I suspect that there isn't one, since even the apparently much simpler problems involving success runs can only be solved recursively or by raising a matrix to a large power. The result can be found by computer simulation.

Since the number of 20K, 30K, and 50K non-overlapping runs in 1 million hands is 50, 33.3, and 20 respectively, we know that the result must be AT LEAST as large as these numbers multiplied by the probabilities derived above. The actual numbers are larger than this, perhaps significantly larger, since a losing stretch can start in one interval and end in another. Anyway, here are the minimum numbers FWIW:

20K: 7.86% * 50 =~ 1.6

30K: 4.16% * 33.3 =~ 1.4

50K: 1.27% * 20 =~ 0.25

[/ QUOTE ]


Hi Bruce,

Thanks for the answer. I think the 20K figure should be 3.93, but I may be mistaken. Does this mean that the minimum number of 30K breakeven streaks to be expected by our hypothetical player in 1,000,000 hands would be 1.4 and a 50K breakeven streak would happen a minimum of once every 4,000,000 hands or am I misunderstanding?

Thanks

[BruceZ]

[ QUOTE ]
Thanks for the answer. I think the 20K figure should be 3.93, but I may be mistaken.

[/ QUOTE ]

Right, sorry. I went back and fixed the typo.


[ QUOTE ]
Does this mean that the minimum number of 30K breakeven streaks to be expected by our hypothetical player in 1,000,000 hands would be 1.4 and a 50K breakeven streak would happen a minimum of once every 4,000,000 hands or am I misunderstanding?

[/ QUOTE ]

It means that the average number of such streaks must be larger than this. We could do a computer simulation to get the actual average number.

[Paxinor]

you guys might want to look in here:
http://forumserver.twoplustwo.com/showfl...=0#Post12274178

i calculated downswings that will generaly occur with the correct distribution for a given standard deviation and a given winningrate, assuming the results are independent (=no tilt), and i will look for breakeven stretches in my sample and post them there...