In a game of yahtzee what is the probability of getting 2 pair?

What are the odds against 26 consecutive hands without so much as a pair in a five-player game of Texas hold 'em?

The probability of no player ever getting a pair with 5 players of Texas Hold 'Em in 26 hands is 1.4937 x 10-8, but see the note at the bottom. A five-player game of Texas Hold 'Em has 15 cards in play on each hand; the two down cards for each player, and the 5 common up cards. The probability of one of the down cards matching the other down card or one of the five up cards is 6 in 52 or 0.11538. The probability of one of the up cards matching one of the four other up cards is 4 in 52 or 0.076923. Invert these probabilities, and you get the probability of not matching: 46 in 52 for the down card, and 12 in 13 for the up card. Multiply these two together, and you get the probability of 552 in 676 or 0.81657 that one player does not have a pair. To determine the probability that no player has a pair, include the four other down card probabilities in the calculation. The up card probability is common to all players, so you only count that once. You get a probability of 2471555712 in 4942652416 or 0.50005 that no one has a pair in one game. To determine the probability that no player has a pair in 26 games, simply raise that to the 26th power, giving a probability of 1.4937 x 10-8 (0.000000014937) that no one gets a pair in 26 games. Note that this does not include the probability of someone not getting a flush or straight - it only gives the probability of no one ever getting a pair, and the result is so low as to be practically impossible.