I definitely like the estimation method. I’m totally convinced that the answer is higher than 1/1000 as you describe. The bit about dividing that by the number of ways you can roll three ones on five dice sounds sketchy—I can’t tell for sure that that’s sensible. But it does sound intuitively right. There are five ways to roll 4 1′s (simplifying it to ten sided dice is a great move for my intuition), ten ways to roll 3 1s, and 1 way to roll 5 ones; so that’s 16. That would be 1.6%, which is different than GPT4′s .86%. So I think that does get into the ballpark, like you said, but it’s not exactly right. Anyway, we’re into the details. I think you’re right about the order of magnitude, and that’s good enough for a Fermi estimate.
Yes, that’s the main place I’m still uncertain, the ten combinations of three 1’s have to be statistically independent which I’m having trouble visualizing; if you rolled six die, the chance that either three pre-selected specific die would be 1’s or the other three die would all be 1’s could just be added together.
But since you have five die, and you are asking whether three of them will be 1’s, or another overlapping set will be 1’s, you have to somehow get these to be statistically independent. Part of that is actually what I left out (that GPT told me, so not sure but sounds sensible), you take the chance that the other two leftover die will both not be 1’s; there’s a 9⁄10 chance that each will not be a 1, so .81 chance that both will not be ones, and you actually have to multiply this .81 by the 1/1000 for each set of three 1’s. So that slightly lowes that part of the estimate to (1/100010).81=.81%
So you have excluded the extra 1’s from the sets of three 1’s but then you have to do the same calculation for the sets of four 1’s and the one set of five 1’s. The set a five 1’s is actually very easy, there’s a 1⁄10 chance that each will land on one, so all of them together is 10^5=1/100,000, adding only .001% to the final calculation, and the four 1’s are also about a factor of 5 less likely then three 1’s because you have to roll another 1 to get four 1’s. So you have to roll four 1’s and one not-1, or (1/10,000).95=.045%
.81+.001+.045=.856%
Still not 100% sure because I suck at combinatorials but this seems pretty likely to be correct. Mainly going off that 1⁄1,000 intuition for any three sets of 1’s and that being repeated ~10 times because there are five die, and the rest sounds sensible
I definitely like the estimation method. I’m totally convinced that the answer is higher than 1/1000 as you describe. The bit about dividing that by the number of ways you can roll three ones on five dice sounds sketchy—I can’t tell for sure that that’s sensible. But it does sound intuitively right. There are five ways to roll 4 1′s (simplifying it to ten sided dice is a great move for my intuition), ten ways to roll 3 1s, and 1 way to roll 5 ones; so that’s 16. That would be 1.6%, which is different than GPT4′s .86%. So I think that does get into the ballpark, like you said, but it’s not exactly right. Anyway, we’re into the details. I think you’re right about the order of magnitude, and that’s good enough for a Fermi estimate.
Yes, that’s the main place I’m still uncertain, the ten combinations of three 1’s have to be statistically independent which I’m having trouble visualizing; if you rolled six die, the chance that either three pre-selected specific die would be 1’s or the other three die would all be 1’s could just be added together.
But since you have five die, and you are asking whether three of them will be 1’s, or another overlapping set will be 1’s, you have to somehow get these to be statistically independent. Part of that is actually what I left out (that GPT told me, so not sure but sounds sensible), you take the chance that the other two leftover die will both not be 1’s; there’s a 9⁄10 chance that each will not be a 1, so .81 chance that both will not be ones, and you actually have to multiply this .81 by the 1/1000 for each set of three 1’s. So that slightly lowes that part of the estimate to (1/100010).81=.81%
So you have excluded the extra 1’s from the sets of three 1’s but then you have to do the same calculation for the sets of four 1’s and the one set of five 1’s. The set a five 1’s is actually very easy, there’s a 1⁄10 chance that each will land on one, so all of them together is 10^5=1/100,000, adding only .001% to the final calculation, and the four 1’s are also about a factor of 5 less likely then three 1’s because you have to roll another 1 to get four 1’s. So you have to roll four 1’s and one not-1, or (1/10,000).95=.045%
.81+.001+.045=.856%
Still not 100% sure because I suck at combinatorials but this seems pretty likely to be correct. Mainly going off that 1⁄1,000 intuition for any three sets of 1’s and that being repeated ~10 times because there are five die, and the rest sounds sensible