Say you have a cake that is cut into 100 pieces.
One of the pieces has a hidden prize.
100 people take turns, each taking a piece in sequence hoping to find the prize.
Note: I slightly modified (see italics) the question below to provide greater clarity; I hope that is allowed. If not I will revert it.
Q: Which turn (C) yields the (highest probability/optimal risk) of finding the prize? factoring in failed attempts from previous players
Please express your solution in a formula and provide an explanation behind your reasoning.
1st person has C1 = 1/100 chances of finding the prize.
IF the first person does not find it, then the 2nd person has (and correct me if I'm wrong)
C2 = 1/99 - (chance of the first person did not find it)
^ C2 chooses 1 out of the remaining 99 pieces of cake ^
If the first 98 people did not find the prize then the 99th person has:
C99 = 1/2 - (cumulative chance the first 98 people did not find it)
To sum up:
The 1st person has the highest chance of not having someone else finding the prize, but the largest pool to choose from.
the 100th person has the smallest pool to choose from however takes the largest risk of someone finding the prize before them.
Note: I do not know the answer to this question. I am interested in your logical approach.
Thank you and have fun!
Food for thought: The Monty Hall problem
what would happen if you took into consideration the knowledge that the previous people did not find the prize?
Can we predict a turn that will yield a high reward vs efficient risk, given the knowledge that the people before you had not found the prize. The more people that fail to find the winning piece, the higher the chance that the next person will find it, however, let too many people take a piece before you and chances are you will lose because someone else will find it.