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In each round of the board game "The Resistance" three players are randomly and secretly chosen to be spies while the rest of the players are part of the resistance. The spies are then made aware of each other while the players do not know the identity of the rest.

Imagine that 8 persons are playing the game and that three rounds have been completed. A new round has just begun. Among the players are Ted and Bob. Bob thinks that Ted is a spy and tries to convince the other players: "I think that Ted is a spy", he says. Ted replies: "I have been a spy in the last three rounds. It is very unlikely that I am a spy four times in a row. It is more likely that I am good in this round. If you let $X$ denote the number of times I am a spy in the four rounds, then X will be binomially distributed. The expected value of $X$ is then $3/8\cdot 4=1.5$. So in two of the rounds I should be good and in two of the rounds I should be a spy. Thus I should be good in this round." Bob then says: "Your status in each round of the game is independent of your status in the other rounds, so your reasoning is wrong. The probability of you being a spy in this round is $3/8$." Which player is correct?

FredrikAa
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  • Are the roles assigned independently of how were they assigned previously? If so, you have your answer... – Tim Feb 04 '16 at 11:33
  • Yes, @Tim, so Bob is correct, but exactly where is Ted wrong? – FredrikAa Feb 04 '16 at 11:47
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    If the spies are chosen randomly and independently of previous rounds, this is (in effect) the Gambler's fallacy. Versions of the fallacy are dealt with in numerous questions on site. One such example is [here](http://stats.stackexchange.com/questions/136870/does-10-heads-in-a-row-increase-the-chance-of-the-next-toss-being-a-tail/) – Glen_b Feb 04 '16 at 13:01

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If roles are assigned independently of how were they assigned in previous turns, than the probability is the same each turn. Check Wikipedia articles on independence and conditional probability. Recall that it two events $A$ and $B$ are independent then

$$ \begin{align} P(A \cap B) &= P(A) P(B) \\ \Leftrightarrow P(A|B) &= P(A) \\ \Leftrightarrow P(B|A) &= P(B) \end{align} $$

The situation that you are describing is a conditional probability

$$ P(\text{currently have role of spy} | \text{previously had role of spy}) = P(\text{currently have role of spy}) $$

If the events are independent than knowing previous role tells us nothing about current role. This is pretty abstract, so let me give you a simple real-life example. Imagine two people who interdependently of each other throw two different coins (assume that it is fair). Does the result of the first throw influence anyhow the second one?

The problem that you are describing is closely connected to Gambler's fallacy that is a kind of magical thinking that

if something happens more frequently than normal during some period, it will happen less frequently in the future, or that, if something happens less frequently than normal during some period, it will happen more frequently in the future (presumably as a means of balancing nature).

or hot-hands fallacy, i.e. belief in sports or gambling that previous success may somehow increase the future probability of success leading to "successful series". If you are interested in reading more about this kind of fallacies you can check Thinking, fast and slow book by Daniel Kahneman or his paper written with Amos Tverksy titled Judgment under Uncertainty: Heuristics and Biases.

Tim
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