Average time of hitting the bonus on arcade game In some coin dropper games, you can hit a bonus by lighting up all of the bonus slots.  For an example of such a coin game, look at https://www.youtube.com/watch?v=gPLTL0IlvTI&t=11s
Here, there are 6 different slots that a coin can go into, and if the coin matches up with the travelling LED light, then that will "close" the slot.  When all slots are closed, the bonus jackpot is hit.
When there are no slots closed, the probability that the coin will match up with the travelling LED light is simply 1/6. Once a slot is closed, the odds change.
With no slots initially closed, how many coins does it take on average to close all of the slots and hit the jackpot?
 A: Let us first compute the expected number of coins needed to close one slot. (In this solution, I will use "expected," which here means the same as "average.") We know the probability of doing this is $\frac{1}{6}.$ Let us use a state argument, with $x_{1}$ being the expected number of coins to close one slot. If the first coin succeeds (with probability $\frac{1}{6}$), then this would take up $1$ coin. Otherwise, we expect to need another $x_{1}$ coins. So we write
$$x_{1} = \frac{1}{6} + \frac{5}{6}\left(1 + x_{1}\right)$$
$$\frac{1}{6}x_{1} = 1$$
$$x_{1} = 6.$$
So we expect to need $6$ coins to get $1$ win.
Now we extend this argument to $x_{2}, x_{3},$ and so on. For $x_{n},$ $n - 1$ slots are already filled. The probability of success at stage $n$ is $\frac{7 - n}{6} \times \frac{1}{6} = \frac{7 - n}{36},$ because we first need to land on an unfilled slot, then we need it to be on the LED light. For example, for $n = 2,$ we have $2 - 1 = 1$ initial light and the probability of success is $\frac{5}{36}.$
Now we apply the state argument. Notice that a fail is a fail - it doesn't matter whether you landed on a closed slot or a non-lighted slot. So we have
$$x_{n} = \frac{7 - n}{36} + \frac{29 + n}{36}\left(1 + x_{n}\right)$$
$$x_{n} = \frac{36}{7 - n}.$$
This should make some intuitive sense as well.
The expected number of coins to win the jackpot is equal to the sum of the individual expected amounts. So, we see that
$$\sum_{n = 1}^{6} x_{n}$$
$$= \sum_{n = 1}^{6} \frac{36}{7 - n}$$
$$= 6 + \frac{36}{5} + 9 + 12 + 18 + 36$$
$$= \boxed{\frac{441}{5}}.$$
This is $88.2,$ or almost $90$ tokens. That's a lot of tokens, man!
