"The Mario Party Problem" My roommate and I were trying to figure this one out last night after a heated game of Mario Party:


*

*This is a minigame in Mario Party that pits 3 players on a team against 1 solo player.

*The game consists of 4 cannons, each with a corresponding fuse

*The camera pans away from the cannons, and each of the 3 players chooses one cannon to hide in.

*Note that any number players can choose to hide in the same cannon.

*Once the 3 players are hidden, then the solo player chooses 3 cannons to set off. If there were players that were in a cannon that gets set off, then those players are revealed.

*After choosing the 3 cannons, if even a single player from the team of 3 has not been revealed, then the team of 3 wins. Otherwise, the solo player wins.


Here's an example video of the game.
What we were trying to figure out: Assuming the hiding locations of the three players are independent, and both the hiding locations and cannon selection are completely random, what is the probability that the solo player wins, versus the team of three?
 A: Without loss of generality we may assume that the solo player chooses cannons $1$, $2$, and $3$. She wins unless there is one or more player in cannon $4$. 
There are $4^3$ equally likely ways that the three team players can choose their cannons. Of these, $3^3$ do not involve cannon $4$. So the probability the solo player wins is $\frac{3^3}{4^3}$.
Equivalently, the probability Team Player A avoids cannon $4$ is $\frac{3}{4}$. So the probability they all avoid cannon $4$ is $\left(\frac{3}{4}\right)^3$.  
A: As André Nicolas stated, we can assume that the revealing player chooses cannon $1, 2$ and $3$.
Assuming that the three players choose their location all uniformly distributed (which is quite sure not the case!) they have a probability of $\frac{1}{4}$ for every player to choose $4$. So they have a probability of $\frac{3}{4}$ not to do so.
The single player wins, if nobody chooses canon $4$. As they are independent you can simply multiply:
$$P(\text{"Single player wins"}) = (\frac{3}{4})^3 = \frac{27}{64} = 0.421875 < 0.5$$ 
and 
$$P(\text{"Group wins"}) = 1 - P(\text{"Single player wins"}) = 0.578125 > 0.5$$ 
This means you want to be in the group, if you want to win.
But when you take a look at the wiki article Rock-paper-scissors you can see that there are strategies for such games because players do not play truely random.
(And - although this might not be important for only integers from $1$ to $4$: People don't choose numbers uniformly distributed. Just ask your friends to say a random number between $1$ and $10$. Most people will not say $1$ or $10$ at first. There was website for that, but I forgot the URL :-/ )
