User talk:84.124.149.7

Your math is not correct. You are correct that (1-0.2)^5=0.33, but That does not mean anything. To find the average of a probability distribution, you must sum up the probability of each possibility multiplied by the value of that possibility.

For example, if you wanted the average value a rolling a die, you would do:

(1/6 * 1) +

(1/6 * 2) +

(1/6 * 3) +

(1/6 * 4) +

(1/6 * 5) +

(1/6 * 6) = 3.5

The case of the ender eye is more complex. We need to find the probability of each possibility. There is a 1/5 chance of it breaking on each throw, so if you throw it once, there is a 20% of it breaking. On throw two, the probability of it breaking would be (1-0.2) × 0.2 = 16%. This is because there is an 80% chance of it not breaking on throw one than a 20% chance of it breaking on throw 2. On throw 3 it would be (1-0.2)*(1-0.2)*0.2 = 12.8%.

So the probability (P) of throwing exactly any number of throws (n) would be:

P(n) = (1 - 0.2)(n-1) × 0.2; Where n is a positive integer.

So to find the average value of n, we take the sum from 1 -> ∞ of P(n) × n. This is exactly equal to 5.

Side note: I wrote a program that simulated throwing an eye, and recording the number of throws before it broke. Doing that for a few billion trials, my result was consistent with the math above.

71.212.98.101 05:32, 30 March 2020 (UTC)