# How to use Bernoulli distribution

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.

• Equivalent: $1\cdot p^2+\frac12\cdot2p(1-p)+0\cdot(1-p)^2=p$.
– Did
Nov 16 '14 at 9:50
• can you please expand? explain? I have no idea why that's correct. Nov 16 '14 at 9:55
• 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. Nov 16 '14 at 10:03
• @alonsos Thanks.
– Did
Nov 16 '14 at 10:05

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.
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.