3 ms·
You're pretty far off man. Assuming we're uniformly sampling the whole space, I thought of it as five independent games of choosing each digit, then "winning" e
by BreakfastBob 10y ago
You're pretty far off man. Assuming we're uniformly sampling the whole space, I thought of it as five independent games of choosing each digit, then "winning" each time by not getting a six. So the probability of 5 wins i.e. A number without a six is (9/10)^4 * (5/6). But that counts from 0 to 69,999. If we count up to one less in the tens of thousands place (5/5) we get 0 to 59,999. Which is off by one in both directions. So we can "slide" the window up but 0 was number without a 6 but 60,000 isn't so we have to discount another choice. Leaving us with (9/10)^4 * (5/5) * (59999/60000) = 0.656089065 or 39365 out of 60000.
- deleted 10y ago[deleted]