Probability of finding a prize in the box? There are $7$ concealed boxes, and $6$ of the boxes are empty while $1$ of the boxes contains a nugget of gold. You are required to select two boxes out of the $7$, and after selecting two boxes the host will open $3$ of the remaining $5$ boxes, and he only opens boxes that are empty! After he opens $3$ of the $5$ remaining boxes, $2$ out of the $5$ boxes remaining closed alongside the $2$ you had originally chosen. You are then given two choices:


*

*You can open both of the two boxes you had originally selected.

*You can open only $1$ of the other $2$ boxes that remained from the $5$ other boxes where $3$ were opened by the host and shown to be empty.


What is the probability of winning using either strategy $1$ or strategy $2$?
 A: Looking at the probabilities:
$$
\mathrm{P}( \text{Nugget in 1 of the choosen boxes}) = \frac{ 2 }{ 7 }
$$
$$
\mathrm{P}( \text{Nugget in 1 of the other boxes} ) = \frac{ 5 }{ 7 }
$$
$$
\mathrm{P}( \text{Nugget in 1 of the remaining other boxes} ) = \frac{ 5 }{ 7 }
$$
$$
\mathrm{P}( \text{Nugget in the one of the remaining boxes you choose} ) = \frac{ 1 }{ 2 } \frac{ 5 }{ 7 }
$$
Therefore, strategy 2 is better with probability to win $\frac{5}{14}$, where strategy 1 has probability $\frac{4}{14}$.
A: Define the events


*

*$G=$ the nugget of Gold was in one of the two boxes that you choosed in the beginning and

*$W_i=$ you win when you apply strategy $i$, for $i=1,2$.


Now, by the law of total probability $$P(W_1)=P(W_1|G)P(G)+P(W_1|G')P(G')=1\cdot\frac{\dbinom{1}{1}\dbinom{6}{1}}{\dbinom{7}{2}}+0=\frac{2}{7}=\frac{4}{14}$$ and $$P(W_2)=P(W_2|G)P(G)+P(W_2|G')P(G')=0+\cdot\frac{1}{2}\frac{\dbinom{1}{0}\dbinom{6}{2}}{\dbinom{7}{2}}+0=\frac{5}{14}$$ Since $$P(W_2)>P(W_1)$$ we conclude that there is a higher probability to win using the second strategy.
