Yeah, you got it. You just needed to take it a little further to prove it.

When the lowest bit for a and b is 0, then the formula is equal, which happens, as you say, 25% (1/4) of the time.

When the lowest bits for a and b are 01, the formula is equal. This happens 1/16 of the time (there are 16 possible combinations of the lowest 2 bits of 2 variables, and only this particular combo will give you a problem).

So let's say the lowest two bits of both variables are 11. If the next bit is 0 (making the lowest bits of each variable 011), then the formula is equal 1/64 of the time (64 combinations of the lowest 3 bits of 2 variables).

The math majors in the audience will look for a pattern and realize that, to figure out the total probability that the equation will be equal, they just need to sum up the series (1 / (4 ** x)), where x goes from 1 to the number of bits in yer integer. A Perl one-liner that shows the convergence of the series to 1/3 is shown below. 

perl -e "$x=0;for $y (1 .. 25) { $x+=(1/4**$y));print $x.chr(10)}"

0.25
0.3125
0.328125
0.33203125
0.3330078125
0.333251953125
0.33331298828125
0.333328247070313
0.333332061767578
0.333333015441895
0.333333253860474
0.333333313465118
0.33333332836628
0.33333333209157
0.333333333022892
0.333333333255723
0.333333333313931
0.333333333328483
0.333333333332121
0.33333333333303
0.333333333333258
0.333333333333314
0.333333333333329
0.333333333333332
0.333333333333333

[download]
Edit: changed from 2 ** (2 * x) to 4 ** x, which seems to me to be a bit clearer.

In reply to Re: Re: This may be more of a math question by jima
in thread This may be more of a math question by Fingo

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