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.

For example:

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.

  • 2
    $\begingroup$ If your C2 has a subtraction, it is certainly incorrect since the chance that the first person didn't find the piece is significantly greater than 1/99, making C2 less than 0. An "and" typically indicates multiplication in probabilities. $\endgroup$ Commented Jul 10, 2018 at 14:57
  • $\begingroup$ I agree with Ian. Could you recheck that part again, please :) $\endgroup$
    – ABcDexter
    Commented Jul 10, 2018 at 15:02
  • $\begingroup$ Related: math.stackexchange.com/questions/96331/… $\endgroup$ Commented Jul 10, 2018 at 15:08
  • 3
    $\begingroup$ Monty Hall doesn't really come into play here. Monty Hall was interesting because the door to reveal was not chosen at random; it was specifically chosen not to reveal the prize. The cake here will be selected at random. $\endgroup$ Commented Jul 10, 2018 at 21:01
  • 7
    $\begingroup$ This is not a puzzle but a statistics question. $\endgroup$
    – user29705
    Commented Jul 11, 2018 at 0:24

6 Answers 6


What you're missing here is the chance of playing at all, given that the game ends when someone finds the prize. (or, chance of finding a prize goes to 0, which is the same thing)

Person 1 has a 100/100 chance of playing, and a 1/100 chance of winning.
Person 2 has a 99/100 chance of playing, and a 1/99 chance of winning.
Person 3 has a 98/100 chance of playing and a 1/98 chance of winning.
Person 100 has a 1/100 chance of playing and a 100/100 chance of winning.
(Fractions left unreduced for clarity)

When you multiply these out, you get 1/100 chance of playing AND winning.

In other words, playing sequentially doesn't change the odds from playing simultaneously.

  • 4
    $\begingroup$ While your reply was 12 seconds later than athin's, it better explains why everyone has the same chance. $\endgroup$ Commented Jul 10, 2018 at 16:28
  • $\begingroup$ While I accept your answer for the insight provided, allow me to extend my question to: 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 wield a high reward vs efficient risk, or is the knowledge of previous losers negligible since every choice is predefined with equal chances since before 1st turn? $\endgroup$ Commented Jul 11, 2018 at 12:58
  • $\begingroup$ @TheCakeisaLie... this wouldn't change the math in any way. The only thing that would increase your odds if you know X amount of people didn't find the reward and you are offered to switch your piece. $\endgroup$
    – Ontamu
    Commented Jul 11, 2018 at 13:14
  • $\begingroup$ The answer still holds. Assume 3 people failed already. Starting the puzzle over again gives a similar solution, 97/97 * 1/97, 96/97 * 2/96 ... $\endgroup$ Commented Jul 11, 2018 at 13:16
  • $\begingroup$ @IliaL but that is exactly my question( clarified in Edit ) Since I can choose X amount of people, that means I am offered to switch every time I let someone take another piece before me. ChrisCudmore does it though? since (100/100)*(1/100) not equal to (97/97)*(1/97) etc... $\endgroup$ Commented Jul 11, 2018 at 13:40


They are same.

First person: $1/100$.
Second person: $99/100 \times 1/99 = 1/100$.
Third person: $99/100 \times 98/99 \times 1/98 = 1/100$.
Last person: $99/100 \times 98/99 \times 97/98 \times ... \times 1/1 = 1/100$

  • 1
    $\begingroup$ Sniped! I type too slow $\endgroup$ Commented Jul 10, 2018 at 15:05
  • 1
    $\begingroup$ Welp sorry, I also tried to type this quickly on my mobile and hoping noone sniped it >< $\endgroup$
    – athin
    Commented Jul 10, 2018 at 15:07

Other people have already given correct answers, but I wanted to suggest a different way of thinking about the question that involves less calculation:

First of all, let the people pick their pieces of cake but not check them for prizes. One of those hundred pieces of cake contains the prize. It's equally likely to be any one of them. (Assuming that there's no visible clue to which one contains the prize, which of course would complicate things.) Therefore, the winner is equally likely to be any of the hundred people. It makes no difference in what order they check their pieces of cake: all that matters is which person's piece has the prize.



no one knows anything about the piece of cake they pick, it's a random choice. The problem then becomes this: Randomly distribute 100 pieces of cake to 100 individuals - who is most likely to have the special piece?

It should be clear that

everybody has the same chance of having the special piece. It doesn't matter if everyone chooses their piece and checks it sequentially, or if everyone gets their piece and checks simultaneously.


I will answer your edited question. In case that you have prior knowledge that the ones before you didn't find the price then the better turn is...the last. It's a little underwhelming but basically you have a 1*100/n% chance in getting the price with any number left of cake pieces (being "n" the number of pieces) so 100/1 is 100%.

  • $\begingroup$ Well the last piece has 100% chance yes, If you manage to get there, however I do not accept it as an answer since the accumulative chance of the prize being found before that is much higher, at least to my understanding.. $\endgroup$ Commented Jul 11, 2018 at 13:45
  • $\begingroup$ @TheCakeisaLie... Strictly speaking, if none of previous 99 participants got the prize, you are sure to get it as a hundreth one. That makes probability equal 1. However, the probability to get there is 1/100, so the overall probabillity to win as the last to take your piece is ...1/100, same as for every one. $\endgroup$
    – CiaPan
    Commented Jul 11, 2018 at 23:54

Put a number from 1 through 100 under each piece of cake (suppose the number 1 is the one with the prize). The order of choosing pieces generates a sequence of numbers. There are 100! possible outcomes (permutations of the 100-term sequence), and each number appears exactly the same number of times at any chosen position. Hence the probability of hitting the number 1 is the same, whichever position you take in a row of consumers.


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