Do you believe in Luck?

[Antiquehunter]

As with all of statistics, that depends entirely on how you phrase things non-mathematically. As I said, all A-K vs 8-5 pre-flop showdowns are the same statistically in heads-up play and full-table play if you ignore all other information available to you. In the former there is virtually no information to ignore. In the latter there might be a great deal of information that requires being ignored.

And - we don't have any of that extra information, and its been raised that this could be a problem to Beth, before. As such, we are looking exclusively at the math. We have no information as to position, or how the betting proceeded. We also don't know the number of players at the table, the number who were involved in the hand, the number who may have dropped behind a raise etc... That said - with concealed cards, when the money is in the middle, heads-up, the math is simple.

Again, it depends on your phrasing. Simply stated, an astute player who repeatedly comes up against a novice will in the long run win more hands than your statistics say he will.

To phrase it another way let us say you analyze the statistics as indicated, which I agree is fine for heads up play. I, by contrast, have observed these players, looked at their stacks, watched the betting, and took note of who bet what from what position. In heads up play, we will reach the same conclusions on the odds. In full-table play, I will fudge the numbers based on inferences.

Do you agree that my numbers will be more accurate in the long run?

Yes, and no. My preference earlier on, was to request that Beth & her husband examine their hypothesis by sitting around the kitchen table and dealing out hands. This would show if indeed, her husband was seeing results that differed dramatically from expectation. This was rejected - something to do with 'game conditions' being required to test the hypothesis. (I don't understand why).

I would agree that as a serious player myself, I'm not interested in this sort of model we're tinkering with here, to assess my own results. When I note hands for later analysis, I DO consider a wider range of factors - my table image, how I believe other players perceive me, how I perceive other players, how other players may believe what I'm thinking about them etc... These factors in addition to all the relevant aspects of the game, and the cards, in precise detail. I also have no doubt that over the long haul, the 'luck' factor will even out - so its of no interest to me to assess whether or not I am 'unlucky'.

If someone says they seem to lose showdowns more than the odds seem to indicate, the questions the player should ask himself is under what conditions is he entering showdowns and were those good choices. We have yet to touch on pot odds, which can make playing a likely-to-lose hand profitable in the long run despite having more lost hands than won.

Yes - but Beth's hypothesis doesn't pretend to examine her husband's skill at poker, and indeed she expressed specifically this isn't up for discussion. I've made a few comments throughout the thread which may serve to prod him a bit, but we have absolutely no information at hand to even begin to look at pot-odds situations, and a stated zero desire to increase the level of record keeping so we could analyze same.

 

[Beth]

More Data

Last week-end (Dec 3) he played poker and had three all-in showdowns. He had three more losses, but in this case, when the cards were shown, he had pretty poor odds to win. They were as follows.

He played pocket queens after the flop, which was 9 high rainbow. His opponent showed pocket aces.

He played an AK suited after a K J 10 flop, with one of his suit. His opponent showed pocket aces.

He played an A9 after an A, 6, 10 and a Q on the turn. His opponent showed Q6.

I added these hands to the dataset. I am using the following probabilities for a loss with wins and ties lumped together.

dh opp Prob of Loss
1 AK 37 Loss 0.3583
2 A6 K3 Loss 0.027
3 AT AK Loss 0.6864
4 AA 99 Loss 0.1966
5 TT 58 Loss 0.1956
6 KK QQ Loss 0.1856
7 KT AT Loss 0.7019
8 JJ A9 Win 0.3167
9 AK QQ Loss 0.5359
10 AA 99 Loss 0.1966
11 QQ AA Loss 0.9162
12 AK AA Loss 0.7141
13 A9 Q6 Loss 0.8181

BTW, can anyone tell me the command to align columns with headings?


I get a probability of 0.0000693 for getting one or zero wins from these 13 hands.

If I truncate the dataset, dropping the best win (#8) and the worst beat (#2), I get a probability of 0.0000695 for getting zero wins out of the remaining eleven hands.

If anybody wants to check my computations, I'd appreciate it.

Also, would any of the poker players here consider collecting data on all of their all-ins for a while and posting them? I could compute the probability of their hands using the same method. That would provide an empirical check on my method of computation. If it turned out that other people also consistently scored such low probabilities, that would imply an error in the method of compution.

 

[jt512]

BTW, can anyone tell me the command to align columns with headings?

[PLAIN][table][B]h1[/B]|[B]h2[/B]
datum1|datum2[/table][/PLAIN]

gives

h1 | h2
datum1|datum2

 

[jt512]

I added these hands to the dataset. I am using the following probabilities for a loss with wins and ties lumped together.

dh opp Prob of Loss
1 AK 37 Loss 0.3583
2 A6 K3 Loss 0.027
3 AT AK Loss 0.6864
4 AA 99 Loss 0.1966
5 TT 58 Loss 0.1956
6 KK QQ Loss 0.1856
7 KT AT Loss 0.7019
8 JJ A9 Win 0.3167
9 AK QQ Loss 0.5359
10 AA 99 Loss 0.1966
11 QQ AA Loss 0.9162
12 AK AA Loss 0.7141
13 A9 Q6 Loss 0.8181

I get a probability of 0.0000693 for getting one or zero wins from these 13 hands.

If anybody wants to check my computations, I'd appreciate it.

Using the above probabilities of loss, I get a 1-tailed p-value of 0.0000357 by using my formula on the previous page. Using the same probabilities, a computer simulation of 20 runs of 1 million hands gives an average probability (± SEM) of winning one hand or less of 3.45 × 10^–5 ± 0.14 × 10^-5, which is in agreement with the calculation.

 

[Beth]

Using the above probabilities of loss, I get a 1-tailed p-value of 0.0000357 by using my formula on the previous page. Using the same probabilities, a computer simulation of 20 runs of 1 million hands gives an average probability (± SEM) of winning one hand or less of 3.45 × 10^–5 ± 0.14 × 10^-5, which is in agreement with the calculation.

Thanks. I appreciate the confirmation that it is, indeed, a small probability. Still not willing to reject the null though.

 

[jt512]

Thanks. I appreciate the confirmation that it is, indeed, a small probability.

Beth, you should be concerned that our calculations are not exactly equal. Being off by nearly a factor of two when doing an exact probability calculation unequivocally means that (at least) one of us is doing the calculation wrong.

Jay

 

[Beth]

Beth, you should be concerned that our calculations are not exactly equal. Being off by nearly a factor of two when doing an exact probability calculation unequivocally means that (at least) one of us is doing the calculation wrong.

Jay

I'm not inclined to spend a couple of hours fussing about an exact match. I do that at work, but this application doesn't require it. I'm just performing the calculations in Excel and I might have made a minor error. In some cases, I used an exact value rather than a finite decimal, which will lead to some minor discepancies. Since my estimate is the higher probability, I'll go with that.

I'm interested in seeing empirical data if any of the poker players here want to record all their all-in showdowns for a similar analysis.

 

[jt512]

I'm not inclined to spend a couple of hours fussing about an exact match. I do that at work, but this application doesn't require it. I'm just performing the calculations in Excel and I might have made a minor error. In some cases, I used an exact value rather than a finite decimal, which will lead to some minor discepancies. Since my estimate is the higher probability, I'll go with that.

We're not calculating estimates in the statistical sense, where differences in assumptions could result in different estimates of the same probability. We're each calculating an exact probability, and therefore should come up with the exact same result up to a rounding error. It doesn't matter that you're doing your calculations in Excel (though I can't imagine why you would). Our answers should agree.

I am flabbergasted that as a "professional statistician" you would be willing to live with this discrepancy.

Jay

 

[Antiquehunter]

I am posting this while after attending a wedding, and I'm unable to post those three hands to a poker calc. But based on reading them, with the post flop play, your hubby is WAY behind when the money is in the middle. In those three situations, he was either basically screwed (QQ vs AA) or based on the flop play, way behind, and went for it.

I guess I COULD post my hands here, but to be honest, I examine my play somewhat differently. I NEVER question my 'luck' factor. If a reliable poker calculator tells me that I'm a 60-40- favorite, I can interpret that to my results. I track my results over the numbers if hands played (online) or hours played (live play). The assessment you suggest you want to track here isn't of interest to me. I could post here one off hands that were typical, or one off hands that were truly bizzarre (like when I beat 4 tens with a Royal flush playing hold'em) but they're not meaningful.

There is no 'luck'.

 

[Beaver Hateman]

Among the millions of people playing poker all over the word, there's bound to be a few for whom the best possibilities keep turning up. Are they lucky? I dunno, they're not dead yet. Herodotus had this one sorted out by 400 BC.

 

[Beth]

We're not calculating estimates in the statistical sense, where differences in assumptions could result in different estimates of the same probability. We're each calculating an exact probability, and therefore should come up with the exact same result up to a rounding error. It doesn't matter that you're doing your calculations in Excel (though I can't imagine why you would). Our answers should agree.

I am flabbergasted that as a "professional statistician" you would be willing to live with this discrepancy.

Jay

A difference of less than 0.00004 could be due to rounding error. I've had discrepancies of the order of .001 disappear with the application of double-precision computations. As this isn't a professional project, but an informal one, I can live with it rather than expend the effort to chase it down.

 

[jt512]

A difference of less than 0.00004 could be due to rounding error. I've had discrepancies of the order of .001 disappear with the application of double-precision computations.

Excel is double precision, and the difference is in the first significant digit, anyway. The difference is not due to rounding error. We're either using different formulas or different data.

 

[Beth]

Excel is double precision, and the difference is in the first significant digit, anyway. The difference is not due to rounding error. We're either using different formulas or different data.

I found an error in my formula. I now get .0000357 too. Thanks.

 

[jt512]

Thanks. I appreciate the confirmation that it is, indeed, a small probability. Still not willing to reject the null though.

So, Beth, if and when you do reject the null hypothesis, what will you conclude? That is, what will your explanation for the results be?

Jay

 

[Beth]

So, Beth, if and when you do reject the null hypothesis, what will you conclude? That is, what will your explanation for the results be?

Jay

I have no explanation for the results if we reject the null. It would imply that random chance events can somehow be affected and we would have to reconsider the basic assumptions we make about reality.

 

[Gods Big Toe]

I have no explanation for the results if we reject the null. It would imply that random chance events can somehow be affected and we would have to reconsider the basic assumptions we make about reality.

I hear you. Just the other day in five card stud poker I got A-Clubs, 10-Clubs, 10-Diamonds, 4-Spades, 2-Diamonds. Do you have any idea the odds of me getting that particular hand? I am four times as likely to be dealt a royal flush than I am to get the exact hand I got.

I am still shaking.

 

[Beth]

More data

My hubby's had another couple of poker nights since the last time I posted, but since he got a few more wins in, it's harder to compute the exact probability. Here is the complete dataset as of Jan 28, 2012:

1 AK 37 Loss 0.3583
2 A6 K3 Loss 0.027
3 AT AK Loss 0.6864
4 AA 99 Loss 0.1966
5 TT 58 Loss 0.1956
6 KK QQ Loss 0.1856
7 KT AT Loss 0.7019
8 JJ A9 Win 0.3167
9 AK QQ Loss 0.5359
10 AA 99 Loss 0.1966
11 QQ AA Loss 0.9162
12 AK AA Loss 0.7141
13 A9 Q6 Loss 0.8181
14 KK 99 Win 0.1899
15 AQ A9 Win 0.2395
16 78 34 Loss 0.7101
17 KT 77 Loss 0.4629
18 99 77 Loss 0.1364
19 T4 T5 Loss 0.1091

The first column is the observation number. The second is my hubby's hands, the third is his opponents. The fourth column is the win(1)/loss(0) outcome. The fifth column is the probability of my hubby losing the hand.

It's taken me a while to post this because programming is not my forte. With the additional complexity in computing the exact probability for the additional wins, I kept putting off the task. I did test my program with a small dataset that I compute the answers for completely, but anyone else wants to do the calculations, feel free to to post your answers and confirm mine or point out errors.

This is what I compute as the probability of getting a run of bad luck as bad or worse than what we've recorded here.

Prob (Wins = 0) 1.32E-10
Prob (Wins = 1) 1.11E-08
Prob (Wins = 2) 3.51E-07
Prob (Wins = 3) 6.28E-06

Prob(Wins <= 3) 0.000006638

 

[jt512]

I get a slightly different answer for P(1). Our other calculations seem to agree up to a rounding error.

P(0) = 1.3217 × 10^-10P(1) = 1.0921 × 10^-8P(2) = 3.5114 × 10^-7P(3) = 6.2752 × 10^-6
P(≤3) = 6.62752 × 10^-6
I had started writing up a Bayesian analysis of your husband's results, but I got sidetracked. Maybe now that the this thread has been revived I'll finish it up.

 

[Senex]

Luck isn't all probability. When I was freshman in college a spectacularly beautiful sophomore girl decided she was frustrated and was just going to arbitrarily pick a guy to flirt with in class. She happened to pick me and although we dated two years and nothing came of it I owe her a great deal. She was so beautiful that the first day I walked her back to her dorm I walked into the steal divider that steal doors lock into when closed. She was a step ahead of me and didn't notice I clocked myself and I was in excellent shape and recovered immediately. There was laughter behind me which she didn't attribute to anything but then some ass came running up to me and asked me if I was alright. (I walked into a stationary steal object because I was distracted by this girl's beauty). I told the guy I was fine and when this girl asked why would a stranger ask if I am all right I claimed ignorance.

I was painfully shy (still may be) and the only way I could date a beautiful woman was if she asked me - and luck would have it that not only a beautiful woman did, but a brilliant and nice one to boot. The complete package. No bulloney I owe her a lot. I had issues she helped me through. I have no issues anymore because of her. We broke up because she was speaking of marriage and I felt too young. I never dated a prettier or smarter woman. My first real girlfriend was the best and that is the story of my life.

I've recently wished to get in touch with her and thank her. I can't find her on the internet. I can find her brother.

No bulloney, I got lucky to have this woman as my girlfriend. I know what it is like to have a woman as beautiful as Cindy Crawford as a girlfriend. It was just plain luck she decided to chat me up.

 

[Beth]

I get a slightly different answer for P(1). Our other calculations seem to agree up to a rounding error.

P(0) = 1.3217 × 10^-10P(1) = 1.0921 × 10^-8P(2) = 3.5114 × 10^-7P(3) = 6.2752 × 10^-6
P(≤3) = 6.62752 × 10^-6
I had started writing up a Bayesian analysis of your husband's results, but I got sidetracked. Maybe now that the this thread has been revived I'll finish it up.

Thanks. I'll computed the values for P(0) and P(1) two ways, but I'll check it again next time. Hubby's out playing tonight, so I may have more hands to post soon.