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Let's say we have a slot machine, with $3$ reels. For simplicity let's only consider one line. We have symbols $A,B,C,D$ in each reel. $N_A$ denotes how many times that symbol appears in each reel. Same for $N_B, N_C, N_D$.

There's also the symbol $J$, the joker, appearing $N_J$ times per reel. It counts as any symbol you want (except when $JJJ$)(Let $N=N_A+N_d+N_C+N_D+N_J$)

When a joker appears (in the only line we are considering) that reel doesn't spin for the rest of the games. We play $10$ games.

What's the expected number of times you'll have $AAA$?

If we wouldn't have the rule where the joker stays for the rest of the game, it'd be simple: $10\times(\frac{N_A+N_J}{N})^3$

I also found that on the $n^{th}$ game the probability of having a $J$ in a particular reel is $1-(1-\frac{N_J}{N})^n$.

Couldn't go further than that.

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Let $P_j(j,n)$ be the probability of j jokers in n spins.

$P_j(j,n)={3 \choose j}\left[1-\left(\frac{N-N_J}{N}\right)^n\right]^j\left(\frac{N-N_J}{N}\right)^{n(3-j)},\: n > 0$

$P_j(j,0) = 1$ for $j=0$, and $0$ otherwise.

Let $P_k$ = P(completing AAA spinning k wheels)

$P_1 = \frac{N_A}{N}$

$P_2 = \left(\frac{N_A}{N}\right)^2 + 2\left(\frac{N_A}{N}\right)\left(\frac{N_J}{N}\right)$

$P_3 = \left(\frac{N_A}{N}\right)^3 + 3\left(\frac{N_J}{N}\right)\left(\frac{N_A}{N}\right)^2+3\left(\frac{N_A}{N}\right)\left(\frac{N_J}{N}\right)^2$

$E(number \: of \: AAA) =$ $$\sum_{n=1}^{10}\bigg[P_J(0,n-1)P_3+P_J(1,n-1)P_2+[P_J(2,n-1)+P_J(3,n-1)]P_1\bigg]$$

I actually simulated this (because apparently I have nothing better to do), and the results agreed with the calculated values. For example, for 10 symbols each of A-D and 3 jokers, the average number of AAA is about 0.90.

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