Geometric Distribution problem where p accumulates at each round for Beerio Kart! 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!

Assuming the game chooses races uniformly at random with replacement.
$E[X]=\sum_{n=0}^{n_{max}}P(X>n)=\sum_{n=0}^{56}\prod_{i=0}^n(1-\frac{i}{56})\approx9.06$ guesses, where $n$ is the guess number [1].
In R:
sum(cumprod(1-(0:56)/56))
#> [1] 9.05905

