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So a friend asked me that:

A cop is going after a criminal with 2 police dogs. At some point the road diverges. The cop knows that the probability that a dog will go in the right way after the criminal is $p$ independently with other dogs choice.

The cop decides as follows: If both dogs have chosen the same way, he will go that way. If each dog has chosen a differnet path, he will choose the path to go through, randomally.

Now the question is: Is this method is better then letting one of the dogs choose the path?

It feels like I can use Bernoulli distribution to build the equation I need but I can't figure out how.

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    $\begingroup$ Equivalent: $1\cdot p^2+\frac12\cdot2p(1-p)+0\cdot(1-p)^2=p$. $\endgroup$
    – Did
    Nov 16 '14 at 9:50
  • $\begingroup$ can you please expand? explain? I have no idea why that's correct. $\endgroup$ Nov 16 '14 at 9:55
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    $\begingroup$ The probability that one dog will get it right is $p$. The probability that you will get it right given the two dog case is the sum of the probability that they both get it right: $p^2$, plus the probability that you choose (randomly) the correct path $\frac{1}{2}$ after the dogs disagreed (which can happen in two ways) $p(1-p)+(1-p)p$. Now sum all of this. $\endgroup$
    – alonso s
    Nov 16 '14 at 10:03
  • $\begingroup$ @alonsos Thanks. $\endgroup$
    – Did
    Nov 16 '14 at 10:05
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If you let one of the dogs choose the probability of finding the criminal is p. If you follow the second method, probability can be written as :

P(Finding Criminal)=P(Finding Criminal| Both dogs go the right way).P(Both dogs go the right way) + P(Finding Criminal| one dog goes the right way).P(one dog goes the right way) + P(Finding Criminal | Both dogs go the wrong way).P(Both dogs go the wrong way).

This gives $p^2 +\frac{1}{2}.2.p(1-p) + 0.(1-p)^2 =p $ Thus both the methods are equally likely to lead to the criminal.

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If the cop decides to follow Dog $1$ when there is a split betweeen the dogs, this has the same effect in this problem as always following Dog $1$. Similarly with Dog $2$. So both methods are as good as each other.

More generally, having an even number of dogs, following the majority, and a random decision when they divide equally, is no better than doing the same with one fewer dog. In fact having one fewer might be faster as you never need to stop and take a random decision.

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