so I was recently involved in a probability debate with a friend involving Scrabble: I had been selected to be Player 1. It was my turn to play (first play of the game) and I had a q and was looking for a word that would have a U in it. I was contemplating skipping my turn in the hopes that my adversary would play a U. However, he told me that, from my perspective, it would be about a 4 * 4/93 chance that he would play a word with a U in it, since 4 was the average length of a scrabble word and there are 4/93 possible tiles in the bag.
I disagreed, but could not come up with a valid counter, aside from some vague musings about dependent probabilities.
Later on, I did some digging. It looks like you'd have to figure out the probability that both that my friend (Player 2) has a U and will play it. I did some math and it seems like one approach could be to use a hypergeometric solution, like that posited here: Scrabble Probability to figure out the probability that he has a u.
I followed this approach to figure out that the probability that my friend had at least one U (given that I don't have one), which came out to about 25.54%. I then multiplied this by the (number of words that are seven letters or less and have at least one u in them in the Scrabble dictionary)/(total number of words that are seven letters or less in the dictionary) to arrive at at about a 4.21% probability.
My main question is where did I err? Would appreciate any and all insights!
There are $86$ tiles left that you don't have, so the chance your opponent does not have a $U$ is $\frac {82 \choose 7}{86 \choose 7}\approx 0.7075$. This ignores the fact that two of the tiles are blank and could be used as $U$s.
Assessing the chance your opponent plays a $U$ assuming he has one is very difficult. The word is definitely not chosen at random from the seven letter or less words of the dictionary, which is what your calculation does. If your opponent has no other vowels the chance the $U$ will be played is (almost) $1$. If your opponent has other vowels he is probably more likely to play them than the $U$ as the other vowels are easier to use. If your opponent will play the highest value word he finds it is likely to be long as there is no constraint from other played letters. I would then think it more likely than not the $U$ would get played but wouldn't want to be more precise than that. This gives a probability of at least $15\%$ you will see a $U$