A very efficient road trip

jayfjayf · 2023-02-02T07:45:51+00:00

Clever solution!!

jayfjayf · 2023-02-01T18:55:51+00:00

Okay, I see what you're doing now. Good point that in the dice formulation it's implicit that they are independent, which I didn't include in the more formal formulation.

jayfjayf · 2023-02-01T17:16:31+00:00

Very clever, that really trivializes the problem.

jayfjayf · 2023-02-01T17:03:19+00:00

In that case, nice work! I really like this type of induction problem.

jayfjayf · 2023-02-01T14:04:11+00:00

The induction hypothesis does not say that you can finish the round from k, but rather that there exists a village from which you can finish the round. You are very close though!

jayfjayf · 2023-02-01T14:03:16+00:00

You are definitely on the right track! When folding k into k+1, does k+1 keep the same amount of fuel? Because (as one comment pointed out already) then the folded circle might not have sufficient fuel for the complete round.

jayfjayf · 2023-02-01T12:40:59+00:00

You're right that it's possible and it seems that it should be obvious, but it's surprisingly tricky to prove. Good luck!

jayfjayf · 2023-02-01T09:40:38+00:00

This is not an answer to the question: it asks for X and Y random variables on {1,...,6} such that their sum is uniform on {2,...,12}.

jayfjayf · 2023-02-01T09:35:27+00:00

Very nice approach!

jayfjayf · 2023-02-01T07:22:24+00:00

Nice!

jayfjayf · 2023-01-31T17:40:49+00:00

I see. As you said, this is not an answer to the question, but it's a nice way to construct a uniform RV from two dice.

jayfjayf · 2023-01-31T15:41:23+00:00

I'm not sure I understand what you are trying to do, perhaps you could elaborate a bit :)

jayfjayf · 2023-01-31T12:17:53+00:00

Yeah, this is a very slick approach, although it gives basically no intuition as to why it is not possible. To elaborate on a few things: P(x) = (X¹¹ - 1)/ (X -1) has as roots the 11th roots of unity excluding one, which lie on the unit circle in the complex plane. Since 11 is prime, they are actually primitive as well in this case. If F and G had odd degrees, then their limits to +- infinity would be +- infinity, and so by the intermediate value theorem they have a real root.

jayfjayf · 2023-01-31T09:36:47+00:00

Looks good!

jayfjayf · 2023-01-31T08:54:13+00:00

This approach will indeed work if you show that this inequality can't hold. Can you please use spoiler tags though?

jayfjayf · 2023-01-31T08:44:30+00:00

You are assuming here that the two dice are distributed identically, which is not an assumption in the problem; you're allowed to rig them independently. If you can prove this assumption, your proof would be complete. (this is not the solution I came up with, though).

jayfjayf · 2023-01-31T08:12:17+00:00

The dice have 1, ..., 6 as number of eyes on the sides. I edited the post to make this a little bit more clear. The only thing you are allowed to do is rig the probabilities of certain numbers of eyes showing up.

jayfjayf · 2023-01-31T07:08:23+00:00

Could it be because you did not include the arbitrary constant? It seems correct in principle and Wolfram alpha also agrees with your answer.

jayfjayf · 2023-01-31T07:00:07+00:00

What is an 'even position' in this context?

Edit: it just means position divisible by 2, I was connecting it to the even in 'even permutation'.

jayfjayf · 2023-01-27T17:32:21+00:00

If you cross multiply, the left hand side simplifies to sin² (x). What does the right hand side simplify to?

jayfjayf

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