Brils M.

asked • 10/21/23

Probability question

Hello, my question is about probabilities from a game called "Deceit". Here are the premises.



  1. 6 players, 2 unknow infected players among them.
  2. Each one has therefore 1/3 chance of being infected (2/6).
  3. We isolate two random people from the group.
  4. From the isolated group of two, ONLY one of the two is guaranteed to be infected.
  5. Therefore, the players of the isolated group have 1/2 chance of being infected, and the other group of 4, have 1/4 of being infected, since there is still an unknow infected player among them.
  6. After, we bring back the isolated group of 2 to the group of four.


The question is : What are the odds of being infected for EACH and SINGLE player after putting them

back together?


Thank you for your answers!

Patrick F.

tutor
Maybe I misunderstand the question, but unless there are new infections, there is no reason that the probability would change from 1/3.
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10/21/23

Brils M.

I thought that maybe since we know that from the isolated group one of them is infected, their chance of being infected would be higher than the rest [at least more than 1/6]. And the four remaining people would have less chance of being an infected individually.
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10/21/23

Patrick F.

tutor
OK, so you put the isolated two back together with the four. If they both have a tee shirt with a big I on it then we know the probability they are infected is 1/2. If they return and all 6 are indistinguishable then the probability that any one individual is infected is 1/3. Depends on what you assumed is known.
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10/22/23

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