For those unfamiliar with the rules, in Pick ’Em Poker the player is dealt two cards, plus the choice of one of two more. The game then gives the player two more cards to complete a five-card poker hand. The question at hand is what is the probability of having at least a pair of nines on the deal. Let’s call the players initial two cards that he must keep the "pocket," and the other two cards the "field." This could be accomplished the following ways:
Four of a kind: 13 combinations
High (9-A) three of a kind: 1,152 combinations
Low (2-8) three of a kind with the singleton in the field: 672 combinations
Two high pairs: 540 combinations
One high pair, one low pair, with at least one high card in the pocket: 1,260 combinations
High pair, with at least one in the pocket: 31,680 combinations
The sum of those combinations is 35,317. The total number of ways to pick 4 cards out of 52 is combin(52,4)=270,725. So the probability of getting a pair of nines or better on the deal is 31,680/270,725 = 13.05%. The probability of not getting a pair of nines or better is 100%-13.05%=86.95%. The probability of going 66 hands without a pair of nines or better is (1-(31,680)/270,725))66 = 0.00009848, or 1 in 10,155. That could have just been ordinary bad luck, and doesn’t rise to the level to make a convincing case of foul play. A bigger sample size is warranted to make a better case.