Figuring out how to calculate the probability of different multi-trial outcomes given the single trial probability of success isn’t new math. And it really doesn’t matter what someone’s day job is getting the calculation correct. There’s only one right way to do it.

Let me see if I can teach you how to do it, since your physics professor acquaintance apparently failed to do so.

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Click to learn how to calculate probability for problems like this

### Counting outcomes and calculating probability

When trying to figure out the probability of multiple discrete outcomes, it helps to list out what can happen, and then apply the percentages.

We can start with looking at one feeding of one 1*, and look at the possible outcomes:

Failure (98% probable)

Success (2% probable)

This seems easy enough, right? No one would debate this. 2% of the time we will succeed in leveling the special, and 98% of the time we will fail.

But what happens if we do another training?

Well, now, there are 4 total possible outcomes of those two trainings:

Failure, Failure (no special levels gained)

Failure, Success (one special level gained)

Success, Failure (one special level gained)

Success, Success (two special levels gained)

In order to combine probabilities through multiple trials we multiply.

So the probability of that first case–that we failed once (98% chance), then failed a second time (98% chance)–is 98% * 98% = 96.04%

Using this rule, we can figure out the probabilities of all four outcomes:

Failure, Failure: 98% * 98% = 96.04%

Failure, Success: 98% * 2% = 1.96%

Success, Failure: 2% * 98% = 1.96%

Success, Success: 2% * 2% = 0.04%

Let’s look at this, now, in terms of the number of successes:

0 (the first case)

1 (the second and third case)

2 (the fourth case)

The only way to get 0 successes is our first case, and it happens 96.04% of the time.

Getting 1 success could happen either from our second or third cases, so the probability is 1.96%+ 1.96% = 3.92%.

The only way to get 2 successes is our fourth case, and it happens with a probability of 0.04%.

So after two feedings, our possible special level increases are:

0: 96.04%

1: 3.92%

2: 0.04%

We can extend this to a third feeding, for which there are now 8 possible outcomes across all three feedings:

Failure, Failure, Failure

Failure, Failure, Success

Failure, Success, Failure

Failure, Success, Success

Success, Failure, Failure

Success, Failure, Success

Success, Success, Failure

Success, Success, Success

The probabilities of these are calculated the same way as for the previous example. So three failures in a row is: 98% * 98% * 98% = 94.1992%

The outcomes have the following pribabilities:

Failure, Failure, Failure: 94.1992%

Failure, Failure, Success: 1.9208%

Failure, Success, Failure: 1.9208%

Failure, Success, Success: 0.0392%

Success, Failure, Failure: 1.9208%

Success, Failure, Success: 0.0392%

Success, Success, Failure: 0.0392%

Success, Success, Success: 0.0008%

The possible number of special levels from these 3 feedings is, then:

0 (case 1)

1 (cases 2, 3 and 5)

2 (cases 4, 6 and 7)

3 (case 8)

Summing up the probabilities of those different cases gives us the probabilities of those outcomes:

0: 94.1992%

1: 5.7624%

2: 0.1176%

3: 0.0008%

### Doing the calculation fast

Looking at the above example, we can see that there are two parts to getting the probability of a certain number of successes:

- The probability of any particular case with that number of outcomes happening
- The number of cases in which that number of outcomes happens

For part 1, the quick calculation is:

P(success)^(# of successes) * P(failure)^(# of failures)

So looking at Failure, Success, Failure, the probability would be: 2%^1 * 98%^2 = 1.9208%

Part 2 is a combinatorial problem. There are three possible places that one success can happen in three trials:

First position: Success, Failure, Failure

Second Position: Failure, Success, Failure

Third Position: Failure, Failure, Success

The question we’re needing to answer is: how many ways can I choose one position out of three options?

The general form of this equation is:

(# of positions)! / ((# of positions - # of successes)! * (# of successes)!)

Where the ! means “factorial,” or “multiply all the numbers from 1 to that number.”

0! = 1 (this is by definition)

1! = 1

2! = 1 * 2 = 2

3! = 1 * 2 * 3 = 6

4! = 1 * 2 * 3 * 4 = 24

5! = 1 * 2 * 3 * 4 * 5 = 120

And so on.

In this case, we have: 3! / ((3-1)! * 1!)

Or 6 / (2 * 1) = 3

So you can see the math works right.

The short-hand notation for this is C(n, k), where n is the total number of places, and k is the number of successes.

So for 3 feedings, the calculation of outcome probabilities is, using these quick methods, are:

0: 98%^3 * 2%^0 * C(3,0) = 94.1992%

1: 98%^2 * 2%^1 * C(3,1) = 5.7624%

2: 98%^1 * 2%^2 * C(3,2) = 0.1176%

3: 98%^0 * 2%^3 * C(3,3) = 0.0008%

### Calculating the average number of successes

We calculate the average number of successes by multiplying the number of successes in a given outcome by the probability of that outcome, and summing.

So, for our 3 1* feeders one at a time case, the average number of successes is:

0 * 94.1992% + 1 * 5.7624% + 2 * 0.1176% + 3 * 0.0008% = 0.06 successes on average

Indeed, the rule for distributions like this is that the average is (# of trials) * (probability of success on one trial)

3 * 2% = 0.06

This method can be used to calculate the probabilities of various numbers of outcomes from our two ways of training:

We can see that the average number of successes is exactly the same no matter which method we use:

Looking at the table, you can see that the probability of getting 0 successes from the 1-at-a-time method is actually worse than for the 10-at-a-time method.

But in exchange, you pick up a chance of getting 2, 3, 4, or more successes (although 4 to 10 successes are almost a zero probability).