Well, it's simply true - we need first to look at the lowest bit of a and b. If they are both zero, then, obviously, `a^b == (a^b) + (1^1) == (a+1)^(b+1)`.This occurs 25% of the time.
The next (interesting) case would be that both lowest bits of a and b are 1, and the second lowest bit of both, a and b, is 0. Then again, `a^b == (a+1)^(b+1)`.
I'm not sure, if this method of counting will arrive at 33% overall, as there are some other possibilities (like a=5, b=6) which I didn't count, but 25% is close enough to 33% to mark this method of generating random numbers as dangerous.
`perl -MHTTP::Daemon -MHTTP::Response -MLWP::Simple -e ' ; # The
$d = new HTTP::Daemon and fork and getprint $d->url and exit;#spider
($c = $d->accept())->get_request(); $c->send_response( new #in the
HTTP::Response(200,$_,$_,qq(Just another Perl hacker\n))); ' # web
`
| [reply] [d/l] |

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
`
**Edit:** changed from 2 ** (2 * x) to 4 ** x, which seems to me to be a bit clearer. | [reply] [d/l] |

Just in case...
Perhaps you misunderstand the `^` operator? That's the XOR operator, although many people mistakenly assume it's the exponent operator. Exponents are done with the `**` operator. See perlop.
**Update:** Ah, slow typing will be the death of me someday...
buckaduck | [reply] [d/l] [select] |

Note also that ^ is XOR in perl, not exponentiation (which would be ** ). | [reply] |

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