Closet survey #1

What do you believe that most people on this site don't?

I'm especially looking for things that you wouldn't even mention if someone wasn't explicitly asking for them. Stuff you're not even comfortable writing under your own name. Making a one-shot account here is very easy, go ahead and do that if you don't want to tarnish your image.

I think a big problem with a "community" dedicated to being less wrong is that it will make people more concerned about APPEARING less wrong. The biggest part of my intellectual journey so far has been the acquisition of new and startling knowledge, and that knowledge doesn't seem likely to turn up here in the conditions that currently exist.

So please, tell me the crazy things you're otherwise afraid to say. I want to know them, because they might be true.

What do you believe that most people on this site don't?

So please, tell me the crazy things you're otherwise afraid to say. I want to know them, because they might be true.

I have no scientific explanation for the event.

Yes, you do: all four dice were weighted. You did your math assuming only one of them was weighted, but if they all were then the event you saw wasn't unlikely at all. Assume that a weighted die rolls the side that it favors with probability p, each of the sides adjacent to it with probability (1-p)/4, and never rolls the side opposite the favored side. How strongly weighted do the dice have to be (that is, what should p be) for 26 consecutive victories for the defender are assured?

The defender automatically wins on a 5 or 6, which come up with probability p + (1-p)/4. If the defender rolls a 2, then for the defender to win, each of the attacker's dice must either be a 1 (which it is with probability p) or a 2 (with probability (1-p)/4), so the defender wins in this case with probability (p+(1-p)/4)^3. The cases where the defender rolls a 3 or 4 are similar. Summing all the cases, we get that the defender wins with probability

p + (1-p)/4 + (1-p)/4 * ((p+(1-p)(3/4))^3 + (p+(1-p)(2/4))^3 + (p+(1-p)(1/4))^3)

Which simplifies to

(1/64)(-9p^4-6p^3+54p+25)

To win 26 times in a row with 50% probability, the defender would have to win each battle with probability 0.974. To win 26 times in a row with 95% probability, the defender would have to win each battle with probability 0.998.

(1/64)(-9p^4-6p^3+54p+25) > .974 --> p > .841

(1/64)(-9p^4-6p^3+54p+25) > .998 --> p > .958

In other words, to produce the event you saw with 50% reliability would require weighted dice that worked 84% of the time. To produce the event you saw with 95% reliability would require weighted dice that worked 96% of the time. I'm unable to find any good statistics on the reliability of weighted dice, but 84% sounds about right.

I'm unable to find any good statistics on the reliability of weighted dice, but 84% sounds about right.

here is a set of loaded dice for sale that are advertised to roll a seven (6 on one, 1 on the other, I think) 80% of the time.

1Fly17y

"all four dice were weighted" I used three reddish, semi-transparent plastic dice with white dots (as I always did). My opponent used standard opaque, plastic ivory dice with black dots. I noticed nothing unusual about the dice and by the end of the run I was examining dice, cups, methods of rolling closely. "Assume that a weighted die rolls the side that it favors with probability p, each of the sides adjacent to it with probability (1-p)/4, and never rolls the side opposite the favored side." This assumption does not match my recollection of the dice rolls. As I stated previously, I rolled 6's, 5's, 4's, 3's, 2's, and 1's. I also never rolled a 1,1,1 which should happen frequently if my dice were heavily weighed to roll 1's. Nor do I remember rolling large numbers of 1's. Your probability model for a trick die also fails to match my observations of my opponents die rolls. E.g., in your model my opponent would be expected to roll similar numbers of 5's, 4's, 3', and 2's. However, he only rolled a 2 once and he rolled far more 5's than 3's. Besides with your probability model for trick dice, I would have easily noticed if my opponent rolled a 6 84% of the time and I never rolled a 6 at all. PS You used 26 in the above calculation. I had 26 armies and in Risk the attacker must have at least 4 armies to roll three attack dice. So the 3vs1 dice scenario only happened 23 times.

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Closet survey #1

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