Probability of Binomial Variable ≥ Another Binomial Variable

[SirTristan]

If two binomially distributed variables are generated as paired events, how often will the variable with p=X be greater than the variable with p=Y? Also what is the "equity" if ties are counted as .5 for each?

For instance in Excel I generated 10,000 numbers with p=.8 and 10,000 with p=.6. The first set of numbers was greater 3,173 times, they were equal 5,642 times, and the second set was greater 1,185 times. So p=.8 was greater than p=.6 31.73% of the time. Counting ties as equal the total equity for the first set was (3173+5642/2)/10000=.5994.

Repeating this for p=.7 and p=.4, the first was greater 4,204 times, they were equal 4,610 times, and the second set was greater 1,186 times. p=.7 was greater than p=.4 42.04% of the time, and the "equity" for the first variable was (4204+4610/2)/10000=.6509.

 

[micromass]

Hi SirTristan!

What you are looking for is

P\{X\geq Y\}=P\{X-Y\geq 0\}

Thus you must know the distribution of X-Y. Sadly, I do not know any nice formula for this. However, if X~B(n,p) and Y=B(m,q) and n and m is large, then we can appy the Central Limit Theorem.

Indeed, if n is large, then X~N(np,np(1-p)) and if m is large then Y~N(mq,mq(1-q)). Thus X-Y~N(np+mq,np(1-p)+mq(1-q)). Thus if Z is standard normal, then you need to calculate

P\left\{Z\geq \frac{-np-mq}{\sqrt{np(1-p)+mq(1-q)}}\right\}

which can be easily done by using some kind of table...

 

[bpet]

Looks like you mean Bernoulli variables (Binomial with n=1). For this case it's easy to set up a 2x2 table, e.g. with P[X=1]=p and P[Y=1]=q you have P[X=0,Y=1]=(1-p)q etc and thus P[X>Y]=p(1-q) and P[X=Y]=pq+(1-p)(1-q) which should match reasonably closely the percentages you found by Monte Carlo simulation. You may like to try a Chi square test to see if the observations are close enough to the predictions.

 

[SirTristan]

Thank you very much guys :)

bpet, yes those numbers seem to match the simulations quite precisely. More simple math than I expected :) Here's what those formulas give:

P	Q	X>Y	X=Y	Equity
0.8	0.6	0.32	0.56	0.6
0.7	0.4	0.42	0.46	0.65

That's almost exactly the simulation numbers.

I'm having trouble with micromass's formula though - perhaps I'm doing something wrong? Since n=m=1, here's what I get for the numerator [-p-q] and the denominator [sqrt(p*(1-p)+q*(1-q))], the Z score, and the probability of being higher than that Z score:

P	Q	Num	Den	Z	Probability
0.8	0.6	-1.4	.6325	-2.2136	.9866
0.7	0.4	-1.1	.6708	-1.6398	.9495
0.6	0.8	-1.4	.6325	-2.2136	.9866

Perhaps I'm misapplying the formula, because from how I'm gathering it, when P is less than Q gives the same result as when P is higher than Q. Shouldn't it be that X-Y is distributed with a mean of np-mq rather than np+mq? And the numerator should be -(np-mq) rather than -np-mq? Using that numerator gives me:

P	Q	Num	Den	Z	Probability
0.8	0.6	-0.2	.6325	-.3162	.6241
0.7	0.4	-0.3	.6708	-.4472	.6726
0.6	0.8	0.2	.6325	.3162	.3759

These numbers make more sense to me, although I think they're a bit less accurate relative to the simulation.

 

[micromass]

I'm sorry SirTristan, you are correct! The numerator should indeed be -(np-mq).

Also, the formula I gave you will only approximate the real probability for large n and m. If you pick n=m=1, then this will be highly inaccurate, as your example shows!

Maybe you could try the same thing for n,m>20 or so, you'll see that the formula approximates your simulation quite closely!