Magic Box Game

average of 1,2,3,4 = 2.5 so expected value of 1 throw is 2.5 pts. So the expected value of 4 throws is 2.5*4= 10 points .

Relevant wiki: Central Limit Theorem

By putting $4$ identical balls into $4$ different boxes, the total combinations will equal $\binom{4 + 4 -1}{4} = \binom{7}{4} = 35$ ways according to Stars & Bars Equation , and we know that the lowest possible points equal to $4$ while the highest being $16$ .

By using partition, we will see that there is only one way to earn $4$ points: $1+1+1+1 = 4$ .

And there are $5$ ways to earn $10$ points: $1+1+4+4 = 1+2+3+4 = 2+2+3+3 = 1+3+3+3 = 2+2+2+4$ .

Now by going for all partitions, we will obtain expected value = $\dfrac{(4+ 5 + 15 + 16)\times 1 + (6+14)\times 2 + (7+13)\times 3 + (8+9+11+12)\times 4 + 10\times 5}{35} = \dfrac{350}{10} = 10$ .

Alternatively, by using Central Limit Theorem , the distribution of the point combination will form a bell-like curve of normal distribution as the following:

From this distribution graph, we can depict the expected value or average = $\dfrac{4+5+6+...+14+15+16}{13} = 10$ , which is the mean $\mu$ in the normal distribution.

Therefore, the expected score of this game is $\boxed{10}$ points.

This sum can be solved by linearity of expectation. We have four independent random variables $X_{1}, X_{2}, X_{3}$ and $X_{4}$ , denoting the reward from throwing ball 1, ball 2 ball 3 and ball 4 respectively. We want $E[ X_{1} + X_{2} + X_{3} + X_{4}]$ , which is same as sum of expectations of $X_{i}$ by linearity. Hence the answer is $4 \times 2.5 = 10$

1 + 2 + 3 + 4 = 10. Since you have to throw all four, well . . . That sounds like a pretty boring game show!

Welcome back to Ten in the bank! where we give you ten thousand no matter what!

[This solution is to be edited]

The minimum sum of these numbers is 4 (1 4) and the maximum could be 16 (4 4). However the probability of each of these sums occuring is not distributed equally ( hence is not $\frac{1}{13}$ .