3 ms·
Here's my heuristic. I wonder if it can be made rigorous. Write P_n for the probability of throwing at least n sixes with 6n dice. 1. P_0 = 1 2. P_n -> 1/2
by tome 10y ago
Here's my heuristic. I wonder if it can be made rigorous.
Write P_n for the probability of throwing at least n sixes with 6n dice.
1. P_0 = 1
2. P_n -> 1/2 as n -> infinity, by the central limit theorem.
3. I can't see any reason that P_n shouldn't be monononic.
Thus I expect that P_1 >= P_2 >= P_3.
- Chinjut 10y agoI had originally heuristically reasoned in a similar but opposite way, to the wrong answer. To wit: 1. P_{1/6} = 1/6 [when rolling one die, you have a 1/6 chance of getting more than 1/6 many sixes] 2. P_n -> 1/2 as n -> infinity, by the central limit theorem 3. I couldn't see any reason that P_n shouldn't be strictly monotonic. Thus I expected that P_1 < P_2 < P_3
- thomasahle 10y agoIf you calculated P_1 = 0.665 I suppose you would guess P_2 would be between that and P_infty = 1/2, thus smaller than P_1.