Probability My Opponent Has a Specific Scrabble Letter Mid-Game How does one calculate the probability that an opponent would have a specific letter mid-game?
For example: my opponent would have to have a T in his rack in order to hit the Triple Word Score off the word I am considering playing. There are 50 tiles still in the bank (so, that doesn't include the 7 tiles currently on each of our racks.) There are 7 Ts total in the game (I'm actually playing Words With Friends), 2 Ts have been used on the board and I have 1 T in my rack, leaving 4 Ts unaccounted for. So how would I approach calculating the odds that my opponent currently has 1 or more of the remaining 4 Ts on his rack?
Thank you!!!
 A: I don't think you can get a definite probability out of that without some very strong assumptions about your opponent's playing style.
The other answers basically assume that the tiles you can't see are distributed randomly between the bank and your opponent's rack. This assumption doesn't seem to be warranted here -- which tiles your opponent holds will depend to some degree on which ones he prefers to play.
Suppose, for example, that your opponent follows a strict strategy of holding on to any T tile he gets until he can hit that particular triple-word field with it. (I grant you that's a somewhat strange strategy, but it's possible within the rules of the game). In that case, if any of the Ts currently on the board were played by your opponent, he'll have another one ready with probability 1.
Even if you don't remember who played what, an opponent who tends to hold on to his T's will be more likely to have one than an opponent who doesn't like holding T's and attempts to play them as soon as he can.
And then there's the question of how many chances to play a T your opponent has had (and recognized) recently, which your description of the state of the game doesn't say anything about at all ...
A: There are 57 tiles unbeknownst to you, seven on your opponent's bench and 50 in the boneyard.  There are 4 "loose" Ts.
$$P(\hbox{oppponent has > 0 Ts}) = 1 - P(\hbox{ opponent has no Ts}).$$
The opponent has 7 tiles. There are ${57\choose 7}$ possible holding of tiles for him and ${53\choose 7}$ that are T-less. Can you do the rest?
A: There are $50+7=57$ letters whose place is unknown to you at the moment (they might be in the bank, they might be in his rack), out of which $4$ are "interesting" and you know he has $7$ letters in his rack.
The binomial coefficient $\binom{57}{7}$ tells you the total number of possible racks he can have (you "choose" $7$ out of $57$ tiles). Out of these, $\binom{53}{7}$ are good for you; they don't put any of the "interesting" letters in his rack. The ratio of these two quantities is the probability that he has no "interesting" letter at all in the rack; and if you subtract it from $1$, you'll get the probability he has at least one.
Putting it together and evaluating yields $$\frac{\binom{53}{7}}{\binom{57}{7}}=\frac{53.52.51.50.49.48.47}{57.56.55.54.53.52.51}=\frac{50.49.48.47}{57.56.55.54}\approx 58\%$$
So, he has about $42\%$ chance of having at least one T in the rack.
