4. (1b, 7-9) A hunter has two dogs. One day, getting lost in the forest, he came to a fork in the road. The hunter knows that each dog will choose the way home with probability $p$. He decided to release the dogs one after the other. If both choose the same road, he will follow them; if they split, the hunter will choose a road by flipping a coin. Will this method increase the hunter's chances of choosing the road home, compared to if he had only one dog?

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# Solution.

The probability of choosing the correct path with one dog is $p$. The probability of choosing the correct path, if acting in the specified way, with two dogs is:

$$
\tilde{p}=p \cdot p+p(1-p) \cdot \frac{1}{2}+(1-p) p \cdot \frac{1}{2}=p \cdot p+2 \cdot \frac{1}{2} p(1-p)=p
$$

where $p \cdot p$ is the probability that both dogs choose the correct path, and $p(1-p)$ and $(1-p) p$ are the probabilities that only one of the dogs chooses the correct path. We obtain that the probabilities in both cases are equal $\tilde{p}=p$.

Answer: will not increase.