Now, let’s determine the probability of getting all 3 numbers after four spins of our roulette wheel.
First, we calculate that there are 34 (or 81) results after 4 spins.
Notice that in this case, it is much more difficult to determine and to list these 81 results than it was when we were listing just 27 numbers.
| 1 1 1 1 | 1 1 1 2 | 1 1 1 3 | 1 1 2 1 | 1 1 2 2 | 1 1 2 3 | 1 1 3 1 | 1 1 3 2 | 1 1 3 3 |
| 1 2 1 1 | 1 2 1 2 | 1 2 1 3 | 1 2 2 1 | 1 2 2 2 | 1 2 2 3 | 1 2 3 1 | 1 2 3 2 | 1 2 3 3 |
| 1 3 1 1 | 1 3 1 2 | 1 3 1 3 | 1 3 2 1 | 1 3 2 2 | 1 3 2 3 | 1 3 3 1 | 1 3 3 2 | 1 3 3 3 |
| 2 1 1 1 | 2 1 1 2 | 2 1 1 3 | 2 1 2 1 | 2 1 2 2 | 2 1 2 3 | 2 1 3 1 | 2 1 3 2 | 2 1 3 3 |
| 2 2 1 1 | 2 2 1 2 | 2 2 1 3 | 2 2 2 1 | 2 2 2 2 | 2 2 2 3 | 2 2 3 1 | 2 2 3 2 | 2 2 3 3 |
| 2 3 1 1 | 2 3 1 2 | 2 3 1 3 | 2 3 2 1 | 2 3 2 2 | 2 3 2 3 | 2 3 3 1 | 2 3 3 2 | 2 3 3 3 |
| 3 1 1 1 | 3 1 1 2 | 3 1 1 3 | 3 1 2 1 | 3 1 2 2 | 3 1 2 3 | 3 1 3 1 | 3 1 3 2 | 3 1 3 3 |
| 3 2 1 1 | 3 2 1 2 | 3 2 1 3 | 3 2 2 1 | 3 2 2 2 | 3 2 2 3 | 3 2 3 1 | 3 2 3 2 | 3 2 3 3 |
| 3 3 1 1 | 3 3 1 2 | 3 3 1 3 | 3 3 2 1 | 3 3 2 2 | 3 3 2 3 | 3 3 3 1 | 3 3 3 2 | 3 3 3 3 |
That list of 81 numbers contains all the results of 4 spins of a three-numbered roulette wheel.
Next, we must search that list to see how many of those 81 numbers contain “1 2 3 ” in any order.
It turns out that there are 36 such occurrences:
| 1 1 2 3 | 1 1 3 2 | 1 2 1 3 | 1 2 2 3 | 1 2 3 1 | 1 2 3 2 | 1 2 3 3 | 1 3 1 2 | 1 3 2 1 |
| 1 3 2 2 | 1 3 2 3 | 1 3 3 2 | 2 1 1 3 | 2 1 2 3 | 2 1 3 1 | 2 1 3 2 | 2 1 3 3 | 2 2 1 3 |
| 2 2 3 1 | 2 3 1 1 | 2 3 1 2 | 2 3 1 3 | 2 3 2 1 | 2 3 3 1 | 3 1 1 2 | 3 1 2 1 | 3 1 2 2 |
| 3 1 2 3 | 3 1 3 2 | 3 2 1 1 | 3 2 1 2 | 3 2 1 3 | 3 2 2 1 | 3 2 3 1 | 3 3 1 2 | 3 3 2 1 |
We see there are 36 ways out of 81 that we can get all 3 numbers after four spins of the wheel.
36 ÷ 81 = .44444444…
By now, you are probably thinking that there must be an easier way to do these calculations … and there is.
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