You have been captured for the crime of making a rude comment on Puzzling Stack Exchange and sent to Puzzlcatraz, a special prison for perfectly logical logicians. The warden is fair and decides to play a game with you, as is standard for this prison. Failure to win the game results in death.

The Rules:

1: You are in a group of 10 prisoners. Each prisoner has a number on their head. These numbers are RANDOM, from 1 to 100.

2: To win the game, you must guess if your number is even or odd. If you are correct, you will immediately be freed from prison. If you are incorrect, the warden will throw you into a pit of sharp objects, killing you.

3: You have a 50/50 chance of survival if you simply guess, but the warden, who is a terrible person, will kill all prisoners if he finds out you are guessing. Thus, you must make a strategy.

4: If less than 6 prisoners survive, then all prisoners will die because the warden assumes you are guessing.

5: Prisoners can see other prisoners numbers.

What is the optimal strategy to free as many prisoners as you can?

  • 2
    $\begingroup$ With the number of variants of this puzzle already on the site, I'd be surprised if this isn't a duplicate. $\endgroup$
    – F1Krazy
    Commented Jun 6, 2023 at 15:20
  • 1
    $\begingroup$ I tried to make it not a duplicate but I won’t be surprised if it is. $\endgroup$ Commented Jun 6, 2023 at 15:21
  • 5
    $\begingroup$ I assume we're dealing with the standard "Communicate before to make a plan, no communication once numbers are on" restriction? As written, you could just all tell each other your numbers and get 100% $\endgroup$
    – StephenTG
    Commented Jun 6, 2023 at 15:59
  • $\begingroup$ I feel there are some bits of information missing. 1. Are the numbers exclusive or can the same number appear multiple times in the group? 2. How are the answers communicated? Is the only thing you are allowed to say "I think my number is odd/even?`" or are there other forms of answers you can make? 3. Can anyone else guess, or is it only me? $\endgroup$
    – MichaelK
    Commented Jun 6, 2023 at 16:10
  • 3
    $\begingroup$ I don't understand the question. Why can not I simply ask others to tell me if my number is even or odd? $\endgroup$
    – tsh
    Commented Jun 7, 2023 at 8:24

2 Answers 2


The first person can't do better than 50%, as they have no prior information to act on, so the ceiling on survival is 9.5 prisoners.

In fact, we can reach that ceiling using the following strategy

First prisoner guesses the odd-even status of the numbers he sees, the rest of the prisoners can use that information plus the 8 shared seen numbers to determine the parity of their own number.

  • $\begingroup$ Good job! I’m surprised this wasn’t marked as a duplicate. $\endgroup$ Commented Jun 7, 2023 at 13:56

Sconibulus gives the correct answer assuming prisoners guess sequentially. If they are required to guess simultaneously then an upper bound on the expected number to survive is

$5$, since each prisoner has exactly $50\%$ chance of being correct (we can pair up configurations where the other numbers are the same but that prisoner's number goes from $k\leftrightarrow 101-k$).

This can be achieved by

each prisoner assuming the total of all numbers is even. Now they all guess correctly (and all survive) $50\%$ of the time.


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