# Hero Down-B Probabilities

Tweet

Hero just came up, and with it a whole host of RNG. For now, here’s some napkin math for his Down-B, which brings up a menu of 4 randomly selected spells.

So, what if you just want one spell? What’s the probability of getting that specific spell? How many times do I have to roll in expectation? How many times if I *really* want to be confident?

## Getting a specific spell #

Something to remember in this case is that we don’t care about the **order** of the spells; as long as it’s in the selected list, we good. With that, it just becomes a problem of counting.

We’ll fix one of the spots to be the spell we want. As far as I can tell, the most “updated” count is **21** spells; 17 showed off, 4 on the menu but not explicitly displayed.

There are \( 20 \choose 3\) ways to draw the remaining 3.

There are \( 21 \choose 4 \) ways to draw 4 spells from a list of 21 without replacement.

So \(P(spell) = \frac{1 \cdot 20 \choose 3}{ 21 \choose 4} = \frac{4}{21} = 0.19 \)

## How many times on average? #

This is a geometric distribution, so the expected value is \(\frac{1}{0.19} = 5.25 \). On average, you’ll have to open the menu 5.25 times before you get a specific spell.

## How many times in confidence? #

We can use a Chebyshev bound to get some upper limit, though there’s probably better bounds out there. Let \(H = Geom(0.19) \).

\( Pr(|H - 5.25| > k\sigma) \leq 1/k^2, k = 1.025 \)

\( Pr(H - 5.25| > 4.85) \leq 0.95 \)

So, if you open the menu 10 times, you will get it *at least* 95% of the time. Note that this is an upper bound (and not a very tight one at that).

*Contact me at stu2b@statsmash.io or @stu2b50 on Twitter*