So this is part of a larger question on this game theory problem set, but I'm really just having troubles with this one part—apologies, it's probably really elementary. We have a defendant on trial in front of four judges. He is either guilty or innocent. Each judge receives a "signal" about whether he is guilty or innocent, and that signal is right with probability 0.9, ie. if he is guilty then each judge has a 0.9 chance of receiving a "guilty signal" and a 0.1 chance of receiving a "not guilty" signal. I need to find the probability that, given that the defendant is not guilty, exactly three judges receive the guilty signal.

So at first I thought it was easy enough: $0.1\times 0.1\times 0.1 \times 0.9$, but then I realized that this is only the probability that, if we were to order them Judge 1, 2, 3, 4, Judges 1, 2, and 3 receive the guilty signal and Judge 4 receives the not guilty signal. Is that correct? So, my question is, do I need to somehow incorporate all the possible orderings of the judges into this calculation?

Thanks for your help, and again, sorry if this is very easy.


Yes, you need to incorporate the fact that there are different possibilities for which judge receive the guilty symbol (or equivalently, if you prefer, which judge receives the "not guilty" signal). Since there are four such ways, you end up with $$0.9 \times 0.1 \times 0.1 \times 0.1+0.1 \times 0.9 \times 0.1 \times 0.1+0.1 \times 0.1 \times 0.9 \times 0.1+0.1 \times 0.1 \times 0.1 \times 0.9=4(0.1 \times 0.1 \times 0.1 \times 0.9)=0.0036$$

By the way, this is an example of what is called a binomial random variable. If you have $n$ independent trials, each with probability $p$ of a given outcome, then the probability of having exactly $k$ trials with that outcome is $\binom{n}{k}p^k(1-p)^{n-k}$, where the symbol $\binom{n}{k}$ represents the number of ways to pick $k$ items out of a set of $n$ such items (in our case we were picking which three judges got the "not guilty" signal).

  • $\begingroup$ ahhhh, thank you, I knew that this was a binomial random variable but somehow I forgot to apply my knowledge from one class to another class. Thank you very much $\endgroup$ – crf Apr 15 '12 at 18:59

Your intuition is right. The answer $(0.1)^3 (0.9)$ is not correct. There are $4$ possible choices for which judge votes correctly. So we need to multiply $(0.1)^3(0.9)$ by $4$.

Caveat: The number $(0.1)^3(0.9)$ that forms part of the intended answer is of dubious validity. That is too weak: it is real-world dead wrong. For it is based on the assumption of independence. So it assumes essentially that whether a judge receives a guilty "signal" (fidgeting? skin pigmentation?) is independent of whether another judge receives a guilty signal. You can readily assess how true to life that is!

But back to theory. In general, suppose that an experiment is repeated independently $n$ times, and that each time the probability of "success" is $p$, and the probability of failure is $1-p$. Then the probability there will be exactly $k$ successes is $$\binom{n}{k}p^k(1-p)^{n-k}.$$ Here $\binom{n}{k}$ is the number of ways of choosing $k$ objects from a collection of $n$ objects. For details, search for the Binomial Distribution.

One formula for $\binom{n}{k}$ is $$\binom{n}{k}=\frac{n!}{k!(n-k)!}.$$ Although the notation $\binom{n}{k}$ is the one most commonly used by mathematicians, there are many other notations, such as $C(n,k)$, $C_k^n$, ${}_nC_k$, others.

  • $\begingroup$ Thank you very much. I've just realized I think that I'd done the previous part incorrectly then, which is to calculate the probability that all four judges receive the guilty signal given that the defendant is not guilty... I said $0.1^4$, but I should be multiplying that by 4 as well shouldn't I. $\endgroup$ – crf Apr 15 '12 at 18:52
  • $\begingroup$ The $(0.1)^4$ would be right, I will add general material. $\endgroup$ – André Nicolas Apr 15 '12 at 18:57
  • $\begingroup$ The $(0.1)^4$ is right because there is only one way to pick which four judges get that "not guilty$ signal (must be all of them). $\endgroup$ – Brett Frankel Apr 15 '12 at 18:58

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