A probability of 0.67 does not result in an average of 1. It results in an average of 0.67. If 100,000 players each try 5 20% moonshots then 67% of them can be expected to be successful.

Just in case anybody cares, the 67% figure is indeed the number of players who do not experience failure. 32.768% is the number expected to get 0 moons from 5 20% shots; 67.232% receive one

*or more* moons. So, in addition to the 32.768% who receive 0 moons, 40.96% receive 1 moon for .4096, 20.48% receive 2 moons for .4096, 5.12% receive 3 moons for .1536, 0.64% receive 4 moons for .0256, and 0.032% receive 5 moons for .0016. 0 + .4096 + .4096 + .1536 + .0256 + .0016 = 1.000. That's really why the expected number of moons from 5 20% shots is 1.

But, of course, that's not really how getting (or destroying?) moons works, you can't do

*n* shots (regardless of how many moons you get) and then quit for good. You go until you succeed once at the planet. The expected number of shots required for a given probability is calculated in a different way, with an infinite geometric series. (Hence, the "geometric distribution.") The probability of getting a moon on the first 20% shot is 20%, so for the expected value that contributes 0.2. The probability of getting it on exactly the second shot (from the perspective of the beginning) is 80% * 20% = 16%, multiplied by 2 because this is the second shot for 0.32 more. The third, 80% * 80% * 20% = .128 * 3 = .384. Fourth, .8

^{3} * .2 * 4 = .4096. And so forth, on to infinity. If you sum all those infinite terms, each one

*n* * 0.2 * 0.8

^{(n - 1)}, you get... 5.