# Measuring Luck Using the Binomial Distribution

My husband claims to have insanely bad luck at cards.  So, we’ve developed a scheme to compute his luck.  He usually plays Texas Hold ’em.  Whenever he has an all in hand with cards yet to come, he records the cards and outcome.

We did this for several months last year.  I started a thread about it at the James Randi forum (JREF Thread) which was helpful in deciding what hands could be included. I was persuaded that at this point in the game, no amount of skill can affect the outcome. It’s pure probability theory to compute the odds at that point.  I compute this probability for each hand using an on-line calculator.

Last year his statistics were running in line with his complaints – i.e. his computer percentage of winning hands was considerably less than the expected value.  It was a statistically significant deviation.

He put down his cards for an all-in showdown with an average probability of 0.60 to win or tie.  His computed percentage win (no ties) was 0.40.   The prob. of this occurring, using a binomial computation for the 81 hands is 0.0014.

This is computed using the EXCEL function BINOMDIST.  It was designed specifically for this situation.  Wins are “successes” in BINOMDIST parlance while the number of hands is the ‘trials’.  The ‘probability of s’ is our average probability of win computed from the individual hands played – 0.60.  Last,  BINOMDIST has an input labeled ‘CUMULATIVE’.

This is asking whether you want it to compute the probability of getting exactly 35 wins in 81 hands with an average prob. of 0.60 or the probability of getting 0 wins or 1 win or 2 wins or…. 35 wins.  The latter choice is a cumulative sum, which is actually what we want.  So set cumulative equal to 1.

BINOMDIST(35,81,0.6,1)=.0017

Wait, you say, that’s not the 0.0014 you claimed.  What’s up with that?

This is rounding error due to the actual numbers I’m using are average prob. of win = 0.603246914 which gives a more probability of 0.00136078.

But that small difference isn’t much.  The major point is that he had documented an unusually bad string of luck at cards over the course of 6 months of data collection.  I wasn’t quite sure what to do next and ceased working on this project when our lives became quite chaotic last spring. Mark’s mom was diagnosed with terminal cancer and passed away a few brief months later.  But Mark began recording his all-in showdown hands once again at the beginning of this year.

So far, I’ve input the probability for 50 hands.   His average probability of a win is 0.495488.  Out of 50 hands, he has had 3 ties, so I’m working out the computations for wins and for wins and ties combined:

He had 22 wins, giving him a percent win of .44,  not too far below the expectation of wins.  His probability of having 22 or fewer wins is BINOMDIST(22,50, 0.495488, 1)=0.2603, which indicates his actual wins are not statistically significantly different from his expected wins.

If we analyze including ties, he has 25 successes with an average probability of success of 0.566353.  (Average prob of tie is 0.070865) His probability of having 25 or fewer wins and ties is BINOMDIST(25,50, 0.566353, 1)=0.2102, which is about the same as previous.

Now, technically, these binomial computations are not exact because we are using an average value for the prob. of winning in our binomial computation.  However, a more exact computation should result in a smaller rather than a larger probability. If we can conclude, as we did for the 2012 data, that his results are statistically significantly lower than expected by random chance, then that conclusion will not change.  I think it is unlikely to be so much lower that it would change our conclusion for 2013 hands.

Another option for including ties is to use a normal distribution and compute the prob. of getting an empirical mean as low or lower that our computed value.

Setting up the following distribution:

wins = 1, ties = 1/2 and losses = 0,

We compute the expected value (or mean of the distribution) as follows:

1*0.495488 (prob of a win) + 0.5*070865 (prob of a tie) + 0*0.43637 (;prob of a loss) = 0.53092.

For the mean of 50 tries with an expected mean of 0.53092 compared with an actual mean of  0.489583.   We get a prob. of 0.2344 which lies between the two binomial distribution computations.  I plan to go into more detail on how to use EXCEL to compute a normal probability in a later post.

50 hands isn’t really enough to make a firm conclusion yet, but he has data on a few hundred more.  As I work through the recorded hands, I’ll update the results here.

If anyone wants to record their own poker hands and compute their luck this way, I’d love to hear from you.

I’m attaching my excel file if anyone wants to check our these computations:  Poker 2013