Geometric Distribution problem where p accumulates at each round my friends and I are hosting a Beerio Kart tournament and we are having trouble calculating a probability pertaining to it.
So in Mario Kart on the switch there are 56 courses and we are going to play a certain number of times. Each time we play, the course we play on is chosen at random. Before this happens, I get to guess which course it will be at a 1/56 probability. If I guess correctly, I win a prize. So say I guess rainbow road, and I am wrong. Then, the second time we play I get to guess again. There are still 56 courses, so say I guess moo moo meadows. This time, however, if either moo moo meadows OR rainbow road is selected, I win. So my odds of winning at this round are 2/56. The next round, I get to guess again, giving me 3/56 odds. What is the expected number of times until I get the guess correct?
I calculated P(getting it correct at round n) as P(correctly guessing the course) * P(not guessing the course before round n). I think this is pretty much the geometric distribution, with the twist that each round p grows. So my formula is: n/56(1-p(n-1))(1-p(n-2))…(1-p(1)). I calculated each p(n) for 1-56 with a spreadsheet. Then I did SUM(n*P(n)) for n=1-56 to get the expected value. I am confused though because my p(n)s sum to over 1, and my expected value was like 86, so I am guessing I definitely did something wrong. Thanks in advance for any help you might be able to offer!
 A: Your formula is not quite right. Let's put $K=56$. And let's build up a tree of the possibilities.

Each branch to the left corresponds to a success, each branch to the right to a failure, except for the last branch, because after $K$ attempts, you will have proposed all the courses and are thus certain to win. At each branching, I have the conditional probability to land in that branch from the previous node.
At step one, we have a success with probability $p_1=1/K$ and failure with probability $(K-1)/K$. At step two, our chances of guessing correctly are now $2/K$. Hence, following the tree we have
$$p_2 = \frac{2}{K} \frac{K-1}{K}$$
At the next step, we'll have
$$p_3 = \frac{3}{K} \frac{K-1}{K} \frac{K-2}{K}$$
The general case will be for the probability to succeed at the nth round (with $n<K$)
$$p_n = \frac{n}{K} \frac{K-1}{K} \cdots\frac{K-n+1}{K} = \frac{n(K-1)!}{(K-n)!K^n}$$
At the Kth round, we'll succeed with certainty one, but the probability of having to play until the Kth round will be
$$p_K=\frac{(K-1)!}{K^{K-1}}$$
You can verify that all the probabilities sum to 1. (They should by construction with a probability tree).
I've computed the expected value for the case $K=56$ and I arrive at $9.05905$.
