The question I saw was: In a game, a man wins 100 points if he gets 5 or 6 on a throw of a fair die and lose 50 points for getting any other number. If he decides to throw the die either till he gets a five or six or to a maximum of 3 throws, what is his expected gain/loss?
On the solutions provided to me, they made 4 cases of outcomes, which are: LLL, LLW, LW, W where L denotes getting any of 1,2,3 or 4 and W denotes getting 5 or 6. The cases respectively got -150, 0, +50 and +100 points. The probability of L was taken to be $\frac46$ and probability of W was taken as $\frac26$ which makes sense. Using a probability distribution table, they found the expected value by adding the products of the probability of each case and their respective points, which came out to be zero.
| -150 points | 0 points | +50 points | +100 points | |
|---|---|---|---|---|
| Probablity | $(\frac{4}{6})^3$ | $(\frac{4}{6})^2×(\frac{2}{6})$ | $(\frac{4}{6})×(\frac{2}{6})$ | $(\frac{2}{6})$ |
$Expected$$value$ = $(-150)×(\frac{4}{6})^3$+$(0)×(\frac{4}{6})^2(\frac{2}{6})$+$(50)×(\frac{4}{6})(\frac{2}{6})$+$(100)×(\frac{2}{6})$$= 0$
My approach to the problem was slightly different and I'm unable to understand what I'm doing wrong. I made the same cases, i.e. LLL, LLW, LW and W.LLL has 4×4×4 = 64 total outcomes (that look like: 1,1,1 and 1,2,4 and so on)
Similarly, LLW has 4×4×2 = 32 outcomes, LW has 4×2 = 8 outcomes and W has 2 outcomes. The sum of all the outcomes would be the total outcomes = 64+32+8+2=106
So, I thought that the probability of "LLL" happening was $\frac{64}{106}$, "LLW" happening was $\frac{32}{106}$, "LW" happening was $\frac8{106}$ and "W" happening was $\frac2{106}$. This is different from $(\frac{4}{6})^3$ and $(\frac{4}{6})^2(\frac{2}{6})$ etc. used in the actual solution. And therefore my answer isn't right. How?