Sports Statistics Logic Puzzle

[Kim Lee]

There are handicapper picks on the "Other Games" page, and we can disagree about whether they are any good. I would like to separate disagreement about assumptions from disagreements about logic. So here is an instructive puzzle. You can use intuition, math, computer simulation, or just guess.

First, a warm-up. What is the probability a coin flipper hits 65% on 100 flips? What is the probability a coin flipper hits 55% on 900 flips?

Now suppose there is one true 55% handicapper mixed with nineteen coin flippers. You get the record from one person at random. What is your advantage (at market lines -110) on their next pick if their record is 65% on 100 flips? What is your advantage on their next pick if their record is 55% on 900 flips?

Finally, what is the lesson here?

[Copernicus]

Without doing the math, the lesson is to be the bookie, not the bettor. It only takes 52.5% to show a slight profit (assuming ties are a push), not 55%, but the number of handicappers who can even make that nut over the long haul is miniscule.

The lesson for bookies of course, is that daddy-bookie was right when he told you not to risk the whole bankroll on one game. Lay-off and balance your action in case there are a lot of those 65%ers on the wrong side. Then you dont have to worry about the heavy math.

[BruceZ]

What is the probability a coin flipper hits 65% on 100 flips?

We're really going to want at least 65% here. We can use the Excel function BINOMDIST to compute 1 minus sum of the probabilities of 0 to 64:

1 - sum{n = 0 to 64} C(100,n)*0.5^n * 0.5^(100-n)

= 1 - BINOMDIST(64,100,0.5,TRUE) = 0.1759%

Call this P(65|f) for probability of 65% given a flipper.


What is the probability a coin flipper hits 55% on 900 flips?

Again, we want at least 55% or 495, which is 1 minus the sum of the probabilities of 0 to 494:

1 - sum{n = 0 to 494} C(900,n)*0.5^n * 0.5^(100-n)

= 1 - BINOMDIST(494,900,0.5,TRUE) = 0.1495%

Call this P(55|f).

We will also need P(65|h) and P(55|h) for the probabilities of a true handicapper getting 65% of 100 and 55% of 900 respectively.

P(65|h) = 1 - BINOMDIST(64,100,0.55,TRUE) = 2.724%

P(55|h) = 1 - BINOMDIST(494,900,0.55,TRUE) = 51.38%


Now suppose there is one true 55% handicapper mixed with nineteen coin flippers. You get the record from one person at random. What is your advantage (at market lines -110) on their next pick if their record is 65% on 100 flips?

From Bayes' rule:

P(h|65) = P(65|h)*(1/20) / [ 19/20*P(65|f) + 1/20*P(65|h) ]

= P(65|h) / [ 19*P(65|f) + P(65|h) ]

= 2.724 / (19*0.1759 + 2.724) = 45% probability that he is a true 55% handicapper. There is a 45% chance that he will win 55% of the time, and a 55% chance that he will flip a coin and win 50% of the time. There is a 45% chance that he will lose 45% of the time, and a 55% chance that he will flip a coin and lose 50% of the time. If we bet $110 and win, we only win $100, but if we lose, we lose the whole $110. From this we can compute the EV, and dividing this by the $110 bet, we get the edge or advantage. In this case, we are at a slight disadvantage of

[ (0.45*0.55+0.55*0.5)*100 +(0.45*0.45+0.55*0.5)*-110) ] / 110 = 0.25%.


What is your advantage on their next pick if their record is 55% on 900 flips?

P(h|55) = P(55|h)*(1/20) / [ 19/20*P(55|f) + 1/20*P(55|h) ]

= P(55|h) / [ 19*P(55|f) + P(55|h) ]

= 51.38 / (19*0.1495 + 51.38) = 95% probability that he is a true 55% handicapper. You are at an advantage of

[ (0.95*0.55 + 0.05*0.5)*100 + (0.95*0.45 + 0.05*0.5)*-110) ] / 110 = 4.5%.


Finally, what is the lesson here?

When trying to identify a successful handicapper for the purpose of getting advice, the length of his record is much more important than the size of his edge. It is dangerous to try and identify a successful handicapper after a relatively small number of games, even as many as 100 in this example, even if he seems to have an outstanding record like 65%. Listening to that person may be no better, and may actually be slightly worse, than flipping a coin yourself. On the other hand, if you take advice from someone who has a more modest record of 55%, but who has sustained that record for a long period time like 900 games, you can have a very nice edge, and there will be a much higher probability, like 95%, that you have correctly identified a successful handicapper.

[bigpooch]

Just thinking that you should word the problem a bit more
carefully: instead of 55%,... should have 55%+ (55% or
better). Then for the 65%+ in 100 or 55% in 900 by
normal approximation, phi(1.5) = 0.9332 so the likelihood
is 0.0668 for a coinflipper. On the other hand, for the
"good" handicapper which has a 55% edge, phi(1.0)=0.8413
giving a chance of 0.1587 in the case of hitting 65+/100
and of 0.5 of hitting 495+/900 (it's clearly more of
course!). Now if there is a group of 20 handicappers
and only one of them has a record of >=65 picks of 100,
and if their choices are independent (unlikely, after all
most people are swayed by others' opinions!) how likely
is this "lucky" guy "good" ? Probabilities can be
computed then approximately (for you nits out there, you
can get out your computer!):

(0.1587)(0.9332**19) = 0.04267 (the "good" guy is lucky)

(0.8413)C(19,1)(0.0668)(0.9332**18) = 0.3076
(a not so good guy got lucky!)

(shouldn't have so many decimal places!)

Then the chances are but 0.1218 ! It's easily
more than 1/9 that this only guy was lucky!

In the case of more than one "lucky" picker,
say 2 of them, even if you were to choose the
picker with the better record after 100 trials,
it's no surprise you could still be choosing
someone who is simply lucky and not good!

What about 495/900: now the numbers will change to
0.1344 and 0.1828 and again it is still more likely
that the sole 55%+ picker was just lucky and not so
good! Clearly, now if you were to bet, you would
still be an underdog: 42.4% you would be getting
picks from the "good" handicapper but the rest of
the time you are at the mercy of the vigorish!

Suppose we try 2500 trials with the same criterion
of 55%. Then the numbers will be 0.444 and 0.0527.
This is kind of a nice number of trials as the last
number is close to 0.05. Also, it's much more
likely that the only 55% picker is really "good".
There is about an 89% chance that the sole lucky
handicapper is actually good! Even then, about
11% of the time you are just paying the vig.

On the other hand, how would anyone know that at
least one of any group of 20 handicappers was
"good"? They could all be not so good!

The lesson to be learned is: even if among a large
pool of handicappers, if the number of trials is
not really big for each of them, there is no
assurance that you will be able to beat the vig
unless you had a large sample size.

[Zeno]

[ QUOTE ]
Finally, what is the lesson here?

[/ QUOTE ]

Time outdistances all.

-Zeno

[bigpooch]

If among the 20 handicappers, more than two are lucky
and you pick one of them randomly, you will be clearly
a dog: at most 1/3 of the time you have the "good"
handicapper with an EV of +$5.5 per $100 bet but any
time you have a not so good handicapper, you are raked
at -$5 per $100. In the case of exactly two "lucky"
pickers, if one of them is the "good" handicapper,
then half of the time you are at -$5 and the other
half of the time at +$5.50 for every $100 so you average
+$0.25 (coincidentally the same positive EV guaranteed
in the two envelopes switching problem!). Of course
the problem is that both of these two lucky pickers
are not so good! That probability just swamps whatever
gains you get in the case one of them was "good".

That's why in my earlier post the analysis just focussed
on the case when there was exactly one lucky picker.

The conclusion is don't bet! You will have a negative
EV for any of the two cases (65%+ at 100 picks or 55%+
at 900 picks) discussed. I didn't bother to compute
this EV because you don't have an advantage (misleading
question!) because from a practical point you would not
make the bet! For the 2500 trial case, I wonder if any
2+2er could determine the positive EV in this case for
a $100 bet (why do I have to do all the work?).

The rest of the analysis I had placed in an earlier post.

[BruceZ]

Then for the 65%+ in 100 or 55% in 900 by
normal approximation, phi(1.5) = 0.9332 so the likelihood
is 0.0668 for a coinflipper. On the other hand, for the
"good" handicapper which has a 55% edge, phi(1.0)=0.8413

For the normal approximation, it's 3 standard deviations, and 2 standard deviations, instead of 1.5 and 1.0. SD for binomial distribution = sqrt(npq) = sqrt(100*0.5*0.5) = 5, so 65 - 50 = 15 is 3 SDs. sqrt(900*0.5*0.5) = 15, so 495 - 450 = 45 is also 3 SDs. Then sqrt(100*0.55*0.45) = 5, so 65 - 55 = 10 = 2 SDs. So it's much less likely to get these results, and we have an edge betting on the 900 trial guy as I've shown above by Bayes' theorem.

The normal approximation works pretty well here since p is close to 1/2, and n is fairly large. 3 SDs is 0.0013%, and the exact answers from the binomial distribution is 0.00176% and 0.00149%. Did you know that you can make the normal approximation more accurate by using fractions instead of integers? You just use >=65.5 instead of 65. It's called correction for continuity. It fits the normal distribution better to the histogram. In any case, you can use the binomial tail distribution in Excel for the exact answer.


(0.1587)(0.9332**19) = 0.04267 (the "good" guy is lucky)

(0.8413)C(19,1)(0.0668)(0.9332**18) = 0.3076
(a not so good guy got lucky!)

You really want Bayes' rule here, not binomial. We are picking one at random and seeing these results, not that 1 out of 20 was lucky.

I might as well fix one of my typos while I'm in here. It doesn't change any of the numbers:

What is the probability a coin flipper hits 55% on 900 flips?

Again, we want at least 55% or 495, which is 1 minus the sum of the probabilities of 0 to 494:

1 - sum{n = 0 to 494} C(900,n)*0.5^n * 0.5^(100-n)

Of course it should be 900-n, that was a cut and paste.

[bigpooch]

Quite right! Thanks for the correction; should remember
that the variance for a Bernoulli r.v. is p(1-p)=1/4.

[CCass]

I read all of the responses, and now I have a headache!!!

All of the fancy number crunching just proves what I have known for years....be the bookie, not the bettor. [img]/images/graemlins/grin.gif[/img]

[BruceZ]

Since it is so difficult to get 65 out of 100 by random chance (0.1759%) you might be tempted to think that if you found someone with that record then he must be a truly successful handicapper. The reason this isn't true is because there are so many more people trying who are not truly successful, that it becomes almost even money that a successful one will have gotten that way by random chance, even though this is improbable for any single individual. This is what Bayes' theorem tells us. This is also tells us that we should develop our own systems, rather then relying on other's systems which may not have been tested over a sufficiently long time. Conversely, after 900 plays, it is not probable that any of the pretenders will still be maintaining even a 55% success rate, so achieving that would distinguish the truly successful handicapper.