How do I prove mathematically that one gets less variance if one plays the board three times instead of two times when going all-in preflop in Texas No-Limit Poker with AK against QQ?

The  probability of AK beating QQ is ~43%, assuming we don't know the other  48 cards and all 4 suits are represented between the two hands. You could simply say that, since we know this probability distribution with certainty, the more hands played, the more likely the distribution observed will be the asymptotic distribution. has the mathematical proof nicely laid out above. In "layman's terms", perhaps the best explanation is the more hands you play, the more likely it is you'll see the "true" probability. I.e., every additional hand you play, you increase the chances that AK will have won 43% of the time and QQ will have won 57% of the time, in total. Formally, if each time you play the board is an element of your sample, as the sample grows, its distribution approaches the true distribution. I believe that comes directly from how we define probability.

So far, the other answers have assumed that the hands are independent. They are not, and this is not just esoteric. People know that if you hit an out in the first board, you are less likely to hit in the second board. This gives no net advantage or disadvantage to the drawing hand, but it does reduce the variance more than if the deck were reshuffled between deals. It is difficult to say the exact value that running it twice reduces the variance over running it once. There is still a simple explanation for why running it 3 times reduces the variance compared with running it 2 times, and running it n+1 times reduces the variance compared with running it n times. When you average identical gambles, the variance goes down, even if these are correlated. Dealing it 3 times is a 3-way average of dealing it twice. Deal out the board 3 times face down. Call these boards A, B, and C. To deal it only twice, pick board A, B, or C to discard. Call the results D(A), D(B), and D(C) from discarding board A, B, or C, respectively. Dealing it 3 times uses all 3, and gives you the average (D(A)+D(B)+D(C))/3. If boards A, B, and C agree, then it doesn't matter if you pick one to discard. If 2/3 favor one hand and 1/3 favor the other, then what has lower variance, taking the average, or picking one board to discard, with a 1/3 chance to cause a scoop? Choosing the average has lower variance than picking one, and the law of total variance says that we can write the variance as a sum of the variance of dealing the 3 boards, plus the variance from deciding whether to use the average or pick a random board to discard. By the way, QQ vs AK is not that close to 50%. QQ vs AKo, averaged over the suits, is close to a 4:3 favorite (56.8% equity). This doesn't affect the argument above which works regardless of the equities. If we look at AKs vs 22, which is much closer to 50%, what difference does that make? It would mean that splitting the pots when you deal it twice seems fair. If you ignore the rare hands where AKs and 22 both play something like a straight on the board and split, then you won't get such a fair result by dealing it 3 times. I think this is the root of the question. There is almost a 50% chance to get a fair result when we deal it twice, but by dealing it 3 times we almost guarantee an unfair result where one side gets at least 2/3 of the pot. So, in what sense does dealing it 3 times reduce variance? An answer is that the magnitude of the unfairness matters, not just its probability. Dealing it 3 times reduces the chance of a scoop to about 1/4 from 1/2. Variance measures the average value of the square of the difference between the result and the expected value. This square is huge when one side scoops, and decreasing the chance of a scoop from 1/2 to 1/4 drops the variance even though there is almost no chance that the result will be very close to 1/2.

I don't know LaTeX, so the math will look ugly, but anyway, a slightly more general result than Hagop's: We are interested in the variance of the random variable X whose value is the amount you stand to win. If the number of times you win is denoted by the random variable Y, the total pot is r, and the number of runs is n, then X=r*(Y/n) (for example, if if you ran it 10 times, you won 3 times out of those, and the pot is 100, you get 100*(3/10)). So we are looking for Var[X]=Var[r*(Y/n)]=((r/n)^2)*Var[Y]. Y is binomially distributed, so if the probability of you winning on a single run is p, and the number of runs is n: Var[Y]=np(1-p). We get: Var[X] = (p*(1-p)*r^2)/n.  (You can check it with Hagop's results for p=1/2, n=1 and 2). r and p are constant, so the larger n is, the lower the variance.