Gamble Machine

I solved this problem by creating a scheme of the possible outcomes, see the scheme at the end of this solution. For now, we ignore the last two columns. A "0" in a column $i\in\{1,\ldots, 5\}$ means that in that possible situation $C_i$ does not pass along the signal. Else, it is "1" and it does pass along the signal. As an example, the row $(1, 0, 1, 1, 0)$ is not possible because $C_2$ did not pass along the signal to $C_4$ , which means that $C_4$ could not pass along the signal to the light bulb following it.

Now we have to calculate for each of those possible scenarios the chance it will happen. In the scheme, I have put this under the label $\bold{P}$ . To safe the most calculations, I will go over two of these possibilities, namely $(0, 0, 0, 0, 0)$ and $(1, 1, 1, 0, 1)$ . The other cases will be similar. For the first case, notice that $C_1$ leads towards all the other contraptions, which makes the other four contraptions not working as well. Because the chance that $C_1 = 0$ is equal to $50\%$ , we see that the chance on the first scenario happening is equal to $\frac{1}{2}=\frac{16}{32}$ . For the latter case, we see that $C_1, C_2, C_3$ and $C_5$ are on. Because $C_1$ leads into $C_2$ and $C_3$ , and furthermore $C_2$ leads into $C_4$ and $C_5$ , we notice that every contraption is reached by the signal, which makes chance on this scenario happening equal to $(\frac{1}{2})^5 = \frac{1}{32}$ .

Lastly, we have to calculate how many light bulbs are turned on in every case, in the scheme this is put under the label $\bold{\# L}$ . This is easy, because it is not hard to observe that the number of light bulbs turned on is equal to the sum of the terms in the corresponding row; columns 3, 4 and 5.

Calculating the expected gain of the player is now simply calculating the product of the last column and the number of coins corresponding to the number of light bulbs turned on. We then add them together minus the coin he put in. We get $0 \cdot \frac{16+4+1}{32} + (1+1) \cdot \frac{4+1+1+1}{32} + (2+1) \cdot \frac{1+1+1}{32} + (3+1) \cdot \frac{1}{32} - 1= -\frac{5}{32}$ , so the expected loss is $\frac{5}{32}$ , resulting in the answer $5 + 32 = \boxed{37}$ .

Here the scheme I was talking about:

For each $i$ , let the indicator variable $C_i$ denote the event that the $i$ th bulb is ON and $C_i^c \equiv 1-C_i$ . Hence, the number of light bulbs $N$ which are on may be expressed as $N= C_1C_3 + C_1C_2C_4+ C_1C_2C_5.$ Using linearity of expectation and independence, the expected number of bulbs which are ON may be computed as:
$\mathbb{E}(N)= \frac{1}{2}\cdot \frac{1}{2}+ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}+ \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}= \frac{1}{2}.$ Next we will compute the probability of the event $F$ that all bulbs are off. From the circuit diagram, this event can be expressed as $F = C_1^c + C_1 C_3^c (C_2^c+C_2C_4^cC_5^c).$ Taking expectation of both sides and using linearity of expectation and independence, we have $\mathbb{P}(F)= \frac{1}{2} + \frac{1}{2}\cdot \frac{1}{2} (\frac{1}{2}+\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2})= \frac{21}{32}.$ Note that the player loses one coin when the event $F$ takes place. Thus, the expected loss of the player is given by $\frac{21}{32}-\frac{1}{2}= \frac{5}{32}.$