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Assume that the number of doors which conceal a prize each lead down a multitude of different paths. Imagine the Monty hall problem extended. version, where each door has a another set of n doors behind it.The initial probability of chosing the correct path on the first try should be 1/n with n representing the amount of finial doors which each path leads to. enter image description here

My question is, assuming that the host reveals a incorrect path, where will it be most optimal to switch (or does it matter); meaning at which level should one switch at? Please include variables like if the host was to reavel a path in the domain [domain=level 1 doors: A,B,C,...,N] of your original guessed path, versus a seperate domain.

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We start with $n$ doors, and each door has $n$ doors behind it? The answer will be the same—you're not really changing the problem very much; you're just jiggling some parameters around a little bit. Now not switching wins when you select the right door and then select the right door, which happens with probability $1/n^2$. Switching wins when you select the wrong door, change to the right door, and then select the right second door, which happens with probability $\frac{(n-1)}{n}\times\frac{1}{n-1}\times\frac{1}{n}>1/n^2$. Have I interpreted your question correctly? – crf Jun 18 '14 at 5:13
I allready know that adding the doors behind the doors do not change the extended version of the problem. My question ask if it makes a difference at what depth within the doors the contestant choosees to switch at.I will try to add a diagram that may help explian what I am asking. – Numoru Jun 18 '14 at 14:43

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