# Solution 69: States, Expected Values, Ants, and an Octahedron

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**Published on Dec 6, 2018**- Let's follow the ants until at least two of them meet on the same vertex.

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Master mystery4 months agoWill this method even work if we need to find expected number of moves until all 3 ants land on the same vertex. Is it that only the calculated probabilities will change but the rest will remain the same??

Chris X5 months agoNice video, but I got a little confused with the mu1, mu2... I liked the problem though and I just solved it here usclip.net/video/mFZZPRal5fw/video.html using just programming (so not exact solution). I hope you liked it. I added this video's url as a reference to the problem :)

DerivativeOfSenpi6 months ago^{+1}Nice

Michael Empeigne6 months agowhat is the sum of the interior angles of a octohedron ?

Datta Subrahmanyam6 months ago^{+1}Can you please show step by step solution for 'Monty hall's problem ' for calculating probability of Win ?

adandap6 months ago^{+2}@Datta Subrahmanyam Well, no. The change of door is very much *not* about independence of events because the host cannot choose at random - his choice is restricted in 2/3 of cases. So the a posteriori probability is not 1/2, it's 2/3. You might also like to read about 'the principle of restricted choice' in contract bridge, which depends on the same mathematical reasoning. en.wikipedia.org/wiki/Principle_of_restricted_choice

Datta Subrahmanyam6 months ago@adandap here it looks like a bernoulli's estimation of 3 independent events & with in each independent event, there are dependent events so the posteriori probability becomes 1/2 instead of 2/3, in 2nd iteration (after revealing one of the three options)

adandap6 months ago^{+2}@Datta Subrahmanyam I think that is the explanation. The a priori probability is 1/3 and the a posteriori probablity is 2/3. (Or 1/00 and 99/100 in the example at the end.) What more is there to it? Some problems just aren't that sophisticated...

Datta Subrahmanyam6 months ago@adandap well thank you. I'm expecting a different explaination rather than just naive explaination.

adandap6 months agoLots of places to find this done. Here's the numberphile version: usclip.net/video/4Lb-6rxZxx0/video.html

matyourin6 months ago^{+2}Complicated way to approach it... I solved it using a Markov process and calculating the Eigenvalues of the transformation matrix...seemed to be an easier approach to me.

matyourin6 months ago^{+2}As far as I know, historically, it was a similar problem that was given by a western university professor when everybody solved it in a more "counting orientied way" when Markov as an exchange student or something of that kind came up with his approach which was rather unfamiliar outside of Russian mathematic schools. So history sort of repeated itself here ;)

adandap6 months ago@LetsSolveMathProblems I don't think we were serious that Markov was easier. I think this was a case of using a sledgehammer to crack a nut since I happened to have one handy! It has the merit of being applicable to more complex problems where the combinatorics are harder to work, but in this case your counting argument had the great merit of being accessible to all.

LetsSolveMathProblems6 months ago^{+2}I agree that a Markov process would facilitate the problem solving process for questions similar to the one in the video. Unfortunately, I have not had much experience with a Markov process while learning linear algebra, and, quite frankly, I would not been able to produce a cogent explanation using the technique. The one advantage of the solution in the video, I suppose, is that it allows for a more diverse audience--only basic counting methods are needed to understand the given solution.

adandap6 months agoSince I did it that way too, I can only agree. :)