Do you believe in Luck?

[jt512]

So, all the data you have presented us to date is merely anecdotal. Without a careful log of each and every hand played eliminating confirmation bias is impossible. Have you considered the possibility that your husband is just not that good a poker player?

How is a record of every hand belonging to a well-defined class an anecdote?

 

[Beth]

But my argument is that the one hand specifically should be thrown out because it has such a huge potential to only impact the calculation in one direction. IF the hand had played out as mathematically dictated (ie - your husband didn't get 'unlucky' and his 45:46 edge brought him the pot) then there would be little impact to the calculation. IF he gets 'unlucky' then it has a big impact on the rest of the calculation. Someone more educated in math probably has a principle that suggests why it should be treated separately.

There are a number of situations where that would be appropriate. I don’t think this is one of them. Any time you selectively dump data from your analysis, it needs to be justified. Otherwise, you are opening the door to bias and inviting it in.

Data can be excluded any time the unit that generated the value(s) is determined to be invalid.

When the main concern is accurately characterizing the population mean and variance is not of concern, truncating the dataset by dropping equal numbers of the highest and lowest values may be appropriate. Particularly in situations like surveys where there it is not unusual for people to fill out forms incorrectly as a joke.

In situations where bias in a particular direction is acceptable, the extreme value in the opposite direction of the acceptable bias may be dropped.

When constructing a regression model certain outliers may be very influential and are sometimes dropped in order to create a better model

There are probably a few other justifications for excluding outliers, but I don’t know of any that apply to this situation.

I made a data entry error! I was using 1/5 instead of 4/5 for the one win. Using the correct term, I get a p-value of 0.00022 for all 10 hands. When the post-flop hand is excluded, the p-value is 0.003, which is definitely within the realm of a bad run.

I get p = 0.000142 using Antiquehunter’s probabilities with your computational formula for all ten hands. Dropping only the worst hand is not appropriate. If I drop both the best (the win) and worst (the 1-outer) hands, I still get a p-value of 0.000142. Since these are estimated probabilities, and you may be using slightly different ones, I think both your computation and mine are reasonably assumed to be in the ballpark. What do you get when you drop both the best and worst hands, not just the worst hand?

@JT - The original hypothesis from Beth was that her husband was 'unluckier' than could be expected only in 'all-in' situations. As such, I pointed out that once all the money was all-in (whether it was skill or foolishness that arrived is moot) - one can determine simply the mathematical expectation of that situation.

Yes thanks. That argument was helpful in being able to convince my husband to start recording all of his all-in hands .

At first the test data was only going to include 'race' type of hands. These hands vary only slightly in value (generally 52/48 range). But then we looked at a broader variety of hands. Beth & the player didn't feel that running out hands around the kitchen table was proof enough - hands that came from a 'real' game (either online or live) were seen as necessary to test this 'luck' factor.

Anyways - that is how Beth got to this point. Just to condense the thread for you.

Thanks. That’s a nice summary.

Yes, but you actually have to calculate that expectation. You can't assume that it's 0.5, for reasons I explained above.

For the particular type hand he was initially collecting data on, 50% was within 2% and making that assumption meant the computations can be done as a straightforward binomial probability situation. He knew it wasn’t perfect, but it seemed a reasonable way to determine whether his perception was accurate or biased.

 

[Joey McGee]

Interesting discussion. Just thought anyone interested would love the new Derren Brown special "The Secret of Luck", in regards to this topic. A real heart-warming hour of television actually.

 

[jt512]

I get p = 0.000142 using Antiquehunter’s probabilities with your computational formula for all ten hands.... If I drop both the best (the win) and worst (the 1-outer) hands, I still get a p-value of 0.000142. Since these are estimated probabilities, and you may be using slightly different ones, I think both your computation and mine are reasonably assumed to be in the ballpark. What do you get when you drop both the best and worst hands, not just the worst hand?

Our calculations are not even close, especially for the truncated data (it is suspicious that you get the same result for both). I used the loss probabilities you reported here. The following code block shows the data I used and the results for the original data and a computation without the largest and smallest probability. Results are shown as a vector of three probabilities. The first is P(X = 1), using the formula I posted previously; the second is P(X = 0); the third is the sum of the first two, that is, the one-tailed p-value:

# Correct data, as originally reported by Beth
p.loss <- c(1/3,1/3,1/5,2/3,1/5,2/3,1/5,1/2,4/5,1/50)
[1] 2.125432e-04 3.160494e-06 2.157037e-04

# Above data, minus the largest and smallest
p.loss <- c(1/3,1/3,1/5,2/3,1/5,2/3,1/5,1/2)
[1] 0.0035555556 0.0001975309 0.0037530864

If you used different data in your calculations, if you could post or PM me the exact data you used, I'll do the calculation.

For the particular type hand he was initially collecting data on, 50% was within 2% and making that assumption meant the computations can be done as a straightforward binomial probability situation. He knew it wasn’t perfect, but it seemed a reasonable way to determine whether his perception was accurate or biased.

Just saying that it is a reasonable approximation with no clear justification, does not make it so. I have already shown that this assumption is questionable. To repeat, if the actual probabilities for the individual hands can run from 0.48 to 0.52, then the one-tailed p-value for 22 wins out of 57 hands could range from 0.029 to 0.098. Not only does this range differ by a factor of more than 3, at α = 0.05, you would reject the null in the first case, but not in the second.

Jay

 

[JoeB]

Luck is far and away the best player in this upcoming NFL Draft class, I think he.... oh.... nevermind.

 

[Beth]

Interesting discussion. Just thought anyone interested would love the new Derren Brown special "The Secret of Luck", in regards to this topic. A real heart-warming hour of television actually.

This is very cool! We're about halfway through watching it. I'm hoping we can watch the rest later this evening.

Our calculations are not even close, especially for the truncated data (it is suspicious that you get the same result for both).

I'll have to disagree. The relative difference is large, but in terms of assessing the overall probability, they are both below 0.001. I figured you were likely using different probabilities, but they weren't that far off. The changes were minor.

Anyway, as I was going over the probabilities I was using, I discovered an error in my computations similar to the one you had made. I accidently switched the win/loss probability on the game he won. I now have p = 0.00006282. The truncated result didn't change since my error was on one of the dropped games.

If you used different data in your calculations, if you could post or PM me the exact data you used, I'll do the calculation.

I'm using the following probabilities:

Mark Other Win/Lose Prob Loss
1 AK 37 0 0.38
2 A6 K3 0 0.02173913
3 AT AK 0 0.7
4 AA 99 0 0.2
5 TT 58 0 0.2
6 KK QQ 0 0.2
7 KT AT 0 0.666666667
8 JJ A9 1 0.3
9 AK QQ 0 0.5
10 AA 99 0 0.2

I wish I knew how to align the columns. The first column is my hubby's hands, the second is his opponents. The third column is the win(1)/loss(0) outcome. The fourth column is the probability of my hubby losing the hand.

Just saying that it is a reasonable approximation with no clear justification, does not make it so. I have already shown that this assumption is questionable. To repeat, if the actual probabilities for the individual hands can run from 0.48 to 0.52, then the one-tailed p-value for 22 wins out of 57 hands could range from 0.029 to 0.098. Not only does this range differ by a factor of more than 3, at α = 0.05, you would reject the null in the first case, but not in the second.

Jay

Actually, I do consider that a reasonable approximation. This is not a formal experiment. This is a rather light-hearted informal experiment to try and prove a point. Kind of like a wager but without any stakes. A p-value of 0.029 is not sufficient for either of us to reject the null. A p-value of 0.098 is unacceptably low to simply accept the null. So he keeps collecting data.

 

[Antiquehunter]

Here are poker calculator results for the hands, Beth. Where pairs are involved, I assumed no suits match for all 4 cards. Where connecting cards are involved, I assumed they are suited. Where two different sets of connecting cards are involved, I assumed that the two players hands are of different suits. Basically - I made it as favorable/unfavorable as possible by keeping the maximum number of flush 'outs' as possible for both players. I have used the format AxAx to indicate suit, by the small letter h,s,c,d for hearts, spades etc... The third value is the chance of a split pot. First value is odds for first player's hand to win.

AsKs 3h7h - 63.67 / 35.83 / 0.50
As6s Kh3h with community cards as described above - 97.73 / 2.27
AsTs AhKh - 27.15 / 68.64 / 4.21 (Looks like you had this one reversed in your calculation Beth)
AsAh 9d9c - 80.05 / 19.66 / 0.28
TsTh 8d5d - 79.96 / 19.59 / 0.45
KsKh QdQc - 81.05 / 18.56 / 0.38
KsTs AhTh - 27.94 / 70.19 / 1.87 (Looks like you had this one reversed too)
JsJh Ad9d - 67.98 / 31.67 / 0.35 (And this one was reversed in your calc)
AhKh QsQc - 46.02 / 53.59 / 0.39
AsAh 9d9c - 80.05 / 19.66 / 0.28

 

[Beth]

Here are poker calculator results for the hands, Beth. Where pairs are involved, I assumed no suits match for all 4 cards. Where connecting cards are involved, I assumed they are suited. Where two different sets of connecting cards are involved, I assumed that the two players hands are of different suits. Basically - I made it as favorable/unfavorable as possible by keeping the maximum number of flush 'outs' as possible for both players. I have used the format AxAx to indicate suit, by the small letter h,s,c,d for hearts, spades etc... The third value is the chance of a split pot. First value is odds for first player's hand to win.

AsKs 3h7h - 63.67 / 35.83 / 0.50
As6s Kh3h with community cards as described above - 97.73 / 2.27
AsTs AhKh - 27.15 / 68.64 / 4.21 (Looks like you had this one reversed in your calculation Beth)
AsAh 9d9c - 80.05 / 19.66 / 0.28
TsTh 8d5d - 79.96 / 19.59 / 0.45
KsKh QdQc - 81.05 / 18.56 / 0.38
KsTs AhTh - 27.94 / 70.19 / 1.87 (Looks like you had this one reversed too)
JsJh Ad9d - 67.98 / 31.67 / 0.35 (And this one was reversed in your calc)
AhKh QsQc - 46.02 / 53.59 / 0.39
AsAh 9d9c - 80.05 / 19.66 / 0.28

Thanks. With those adjustments - I put the third column of numbers (prob. of loss) into my excel spreadsheet and I get p-values of .00006594 for the set of all ten hands and 0.00005857 for the truncated set of eight hands.

 

[jt512]

Here are poker calculator results for the hands, Beth. Where pairs are involved, I assumed no suits match for all 4 cards. Where connecting cards are involved, I assumed they are suited. Where two different sets of connecting cards are involved, I assumed that the two players hands are of different suits. Basically - I made it as favorable/unfavorable as possible by keeping the maximum number of flush 'outs' as possible for both players. I have used the format AxAx to indicate suit, by the small letter h,s,c,d for hearts, spades etc... The third value is the chance of a split pot. First value is odds for first player's hand to win.

AsKs 3h7h - 63.67 / 35.83 / 0.50
As6s Kh3h with community cards as described above - 97.73 / 2.27
AsTs AhKh - 27.15 / 68.64 / 4.21 (Looks like you had this one reversed in your calculation Beth)
AsAh 9d9c - 80.05 / 19.66 / 0.28
TsTh 8d5d - 79.96 / 19.59 / 0.45
KsKh QdQc - 81.05 / 18.56 / 0.38
KsTs AhTh - 27.94 / 70.19 / 1.87 (Looks like you had this one reversed too)
JsJh Ad9d - 67.98 / 31.67 / 0.35 (And this one was reversed in your calc)
AhKh QsQc - 46.02 / 53.59 / 0.39
AsAh 9d9c - 80.05 / 19.66 / 0.28

One-tailed p-value = 0.000074 for the full set of 10 hands; 0.059 without the second hand. So, the loss of the second hand highly influences the p-value.

To compute these p-values I calculated the probability of winning exactly one hand P(X = 1) by the formula I gave earlier, using 1 – (p_i / 100), where the p_i's are the first-column percentages above. Then I added P(X = 0).

P.S. Any chance we can all say "probability" when we mean probability, and "odds" when we mean odds?

 

[Antiquehunter]

One-tailed p-value = 0.000074 for the full set of 10 hands; 0.059 without the second hand. So, the loss of the second hand highly influences the p-value.
To compute these p-values I calculated the probability of winning exactly one hand P(X = 1) by the formula I gave earlier, using 1 – (p_i / 100), where the p_i's are the first-column percentages above. Then I added P(X = 0).

P.S. Any chance we can all say "probability" when we mean probability, and "odds" when we mean odds?

This is my earlier point. You have a very rare circumstance which is skewing the results. Its like if you bought 10 lotto tickets, and one of them happened to be a big winner. That would look like you're REALLY lucky over the short term. Its an atypical result. Its also a result where IF the player had happened to win the highly expected hand, it would have little impact to the final calculation. By including it in the set, it can only have one possible effect - that of skewing the data.

 

[Beth]

One-tailed p-value = 0.000074 for the full set of 10 hands;

When values get this low, the number of decimal places typed into EXCEL can make that sort of difference in the results we are getting. Both values can be described as < 0.00001. As far as I'm concerned, this validates the probabilities I was getting as a generally appropriate estimate. Thank you.

0.059 without the second hand. So, the loss of the second hand highly influences the p-value.

Certainly it does, but it's not reasonable to simply drop observations without justification. Truncation on both sides is permissible, but dropping off the worst datapoint with no more justification than because it is an outlier is not. What do you get when you drop off that hand AND the one he won?

P.S. Any chance we can all say "probability" when we mean probability, and "odds" when we mean odds?

Probably not. Sorry, I'll try, but I'm always getting similar terms like probability and odds switched around. I generally try to stick with only referring to probabilities not odds, but I'll likely still make mistakes along those lines. Feel free to call me out about such errors.

 

[Beth]

This is my earlier point. You have a very rare circumstance which is skewing the results. Its like if you bought 10 lotto tickets, and one of them happened to be a big winner. That would look like you're REALLY lucky over the short term. Its an atypical result. Its also a result where IF the player had happened to win the highly expected hand, it would have little impact to the final calculation. By including it in the set, it can only have one possible effect - that of skewing the data.

Dropping just that one observation will have the effect of skewing the data in the opposite direction. That's why truncation, when it is applied, is applied to both sides of the dataset. If you drop the worst hand from the analysis, you also need to drop the best hand out of the analysis in order to balance things out and avoid the data being skewed in either direction.

 

[Antiquehunter]

Sure - drop hand #2 and hand #3 then in line with that principle and see what you get.

 

[jt512]

One-tailed p-value = 0.000074 for the full set of 10 hands...

When values get this low, the number of decimal places typed into EXCEL can make that sort of difference in the results we are getting.

The difference between our calculated values is probably not due to rounding error in Excel, but rather because I used the 1 – P(Win)'s in my calculations, whereas you used the P(Loss)'s. Due to ties, they're not equal. I suspect that if you were to redo your calculations using the 1 – P(Win)'s, you'll get the same results I did to several significant digits.

Both values can be described as < 0.00001.

< 0.0001

Thee loss of the second hand highly influences the p-value.

Certainly it does, but it's not reasonable to simply drop observations without justification.

I am not suggesting that you delete the data point per se; but, rather, compute p-values with and without the data point in order to elucidate its influence on the p-value.

Truncation on both sides is permissible, but dropping off the worst datapoint with no more justification than because it is an outlier is not.

Truncation "on both sides" seems rather silly to me.

ETA: Maybe I take that last comment back. You're effectively neutralizing outliers by using a trimmed mean. Fair enough.

Jay

 

[Antiquehunter]

Just a general comment - I see that of 10 all-in hands, only 1 apparently took place post-flop. Most skilled players prefer to minimize the pre-flop all-in move, as there are too many fluctuations - at least I would say this is a rule in tournaments where the stacks are deep, or in cash games. So I don't know what structure the tournaments are that your husband plays, but if the stacks are deep enough and the rounds long enough, I would say as an indicator of stronger overall play, I would like to see the ratio of pre-flop all-ins to later-game all-ins move away from 9:1.

 

[Gods Big Toe]

I think there is something major being overlooked here. The odds are being calculated based on heads-up play. If there were other players in the hands before it went heads up, that plays a major factor in how it plays out. In an extreme example look at A-K vs 8-5 in a hand that started with 10 players. If other players had A-LowCard and K-LowCard such that all of the other aces and kings were mucked, then at that point A-K stands no chance of getting pairing. So while the heads-up odds may say it is X, if we could see the mucked cards, we\'d know it was Y. The odds calculators are correct in that both players are affected equally when the much cards are unknown.

What makes hold \'em so interesting is that we can infer from the betting what cards might have been mucked. It all depends on the style of play for each player as well as their stack size. For example, some players might bet A-8 pre-flop but not call any raises. They might call raises with A-J but choose not to go against an all-in. Somebody in late position with a big stack might infer from the betting that there were lots of face cards dealt and figure that somebody going all-in is banking on face cards coming on the flop. That person might then choose to gamble with 8-5 suited, especially if a lot of money was already in the pot.

To look at A-K vs 8-5 as if it was purely a heads-up match when it wasn\'t *dealt* as a heads up match is to ignore the finer points of hold \'em. Instead of trying to calculate how bad his luck is, your husband should be looking to see if his opponents are better at inferring what was mucked and thus making more prudent bets than he.

 

[Antiquehunter]

So long as the player doesn't KNOW any of the other cards (via exposed cards etc...) then the calculations should be fine for the purposes of Beth's query.

Yes, one could do a rough Bayesian analysis of the likelihoods of what cards may be held by other players, but this isn't really relevant to the conversation.

 

[Gods Big Toe]

So long as the player doesn\'t KNOW any of the other cards (via exposed cards etc...) then the calculations should be fine for the purposes of Beth\'s query.

Yes, one could do a rough Bayesian analysis of the likelihoods of what cards may be held by other players, but this isn\'t really relevant to the conversation.

It is actually at the heart of the conversation. The assumption here is that a showdown of A-K vs 8-5 is a random event and that all such showdowns have the same odds. That is incorrect. Any showdown is the result of decisions made by some number of players. In a 10 hand game, you will find A-K and 8-5 dealt far more often than you will find A-K and 8-5 in a pre-flop showdown. There are *reasons* why some end up in showdowns while others don\'t.

You can\'t ignore those reasons, look at the odds based purely on chance, then compute a p-value that has real meaning. Suppose in a 10 player game that everyone mucks all the way around to the small blind. That tells you something about the number of face cards still in the deck. Playing A-K is a much stronger proposition than going through a few rounds of betting pre-flop until you\'re left with a showdown.

If you could get the details on actual hands of poker played, you would find that A-K vs 8-5 in a pre-flop showdown is *not* going to match up statistically with the calculated odds because, as I said, it\'s not a random situation.

 

[Antiquehunter]

No matter HOW AK vs 85 ended up getting all-in (whether good decisions or not were made leading to that point) the math is the same once the money is in the middle.

Your point that some of the A's and/or K's may be dead, or that a good player may be able to deduce this based on preflop betting patterns is true - however it would never be a KNOWN fact. At best, a skilled player could make the appropriate deduction and choose a different course of action - however the mathematics if he makes an all-in move & is called, would be unaffected.

I don't know if you've read the whole thread from the beginning, but you'll note earlier on I made a number of suggestions to Beth on how this scenario could be examined. Some were accepted, some were rejected.

However what we're discussing at this point is not about Beth's husband's skill (or lack) at poker, but a somewhat contrived way of examining if for some reason he is 'unlucky' at all-in bets made in a live game situation. Based on the way the data is being gathered, and based on the fairly loose hypothesis, I think we're doing a reasonable job here, and also, the results aren't indicating anything to get excited about.

 

[Gods Big Toe]

No matter HOW AK vs 85 ended up getting all-in (whether good decisions or not were made leading to that point) the math is the same once the money is in the middle.

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.

Your point that some of the A\'s and/or K\'s may be dead, or that a good player may be able to deduce this based on preflop betting patterns is true - however it would never be a KNOWN fact. At best, a skilled player could make the appropriate deduction and choose a different course of action - however the mathematics if he makes an all-in move & is called, would be unaffected.

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?

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.