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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: The Darn Old Cocked Hat - the sequel 1
From: Tom Sult
Date: 2013 Mar 17, 14:13 -0500
This explanation is a commonly held view but I think it is in error. These are really two independent events. The first was a choice of three. The second was a nonrandom elimination  of one of the choices. This makes the second event an independent choice of one out of two. Therefore keeping your current curtain or changing to the other curtain results in the same 50-50 probability.

If the move from the three curtain choice to the two curtain choice was a random event then the probabilities may well change. In the event that it were a random choice occasionally the world cruise would have been revealed and eliminated. This would've placed the likelihood of winning at zero. But because that is never an option in this scenario they are actually two independent throws of the "dice".

Tom Sult
Sent from my iPhone

On Mar 17, 2013, at 13:17, Don Seltzer <timoneer{at}GMAIL.COM> wrote:

Geoffrey replies:
>But surely Don, once Monty has revealed that one of the doors does not hide the prize, there are only two doors left and we don't know behind which door the prize sits. So, regardless of which door was chosen first, the probability that the prize is behind each of the remaining two doors is then just 1/2, isn't it?

If this is not the case, why not?
>

If Monty were making his pick blindly, I would agree that the odds had become 50/50. But his pick is not random and there is always an empty curtain for him to deliberately choose. Nothing he has done has altered the 1/3 probability of my initial choice being correct. But Monty has eliminated one of the alternate choices.

My first pick remains at 1/3 chance of being correct, which means the remaining curtain must have a 2/3 chance of being the big prize.

Don Seltzer

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