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There are actually two puzzles and they appeared in the 'Skeptical Adversaria' (the newsletter of ASKE, the Association for Skeptical Enquiry), January 2006.
In the last issue of this journal I promised readers another variation on the Monty Hall problem. By way of compensation to readers who may be familiar with this or who are simply not interested in this kind of problem, I include a completely different problem involving no maths at all.
Once again I am indebted to ASKE member Jan Nienhuys, this time for drawing my attention to the following variation. As usual, any errors are entirely due to me.
In the last issue I expressed the puzzle in the form of the familiar 3 shells game of chance. I show you 3 shells, under one of which is a gold sovereign. You are to guess which shell (the ‘target shell’) covers the sovereign, and if you guess correctly, the sovereign is yours. In the standard Monty Hall puzzle, the rules dictate that I must always know which is the target shell and once you have made your choice, before you look under the shell, I must reveal to you one of the two remaining shells that does not cover the sovereign. Having done this, I must then offer you the chance of changing your mind, that is sticking with your original choice or switching to the shell that neither you nor I have indicated. What do you do – stick or switch?
Well, we’ve already been through all that, so let’s look at the variation. It’s actually simpler than the usual version.
As before, I know exactly where the sovereign is but before you make your choice I draw to one side two shells at random and I have then to point to one of them under which I know for sure that there is no sovereign. Now, what should be your strategy when playing this game and why?
Incidentally, why have I stipulated that the two shells should be chosen at random?
Having solved that, demonstrate what the general formula is when there are n shells (n>2) and each time I randomly draw aside m of them (where 1<m<n) and I again indicate one of the m shells that I know does not cover the sovereign.
You are standing in a field with, let’s say, 100 other people. Connecting this field with the adjacent one is a narrow style. Everyone is to proceed over the style into the adjacent field. As each person is climbing over the style either a red or a green hat is placed on his or her head. As they enter the second field, people must immediately divide off into those with red hats and those with green. No one can see his or her own hat (directly or through a reflecting surface) and no one, either in the group or observing, must inform other people of the colour of their hat.
What simple instruction could you give to the group on how they can perform this task?
Once the whole task is completed, is it guaranteed that everyone will know the colour of his or her hat?
The solutions to the puzzles are as follows.
The answer is that you always choose the other shell that I have drawn to one side.
As with the original problem, there are several ways of demonstrating why this is so but the one I favour (as I did last time) is as follows. By drawing aside 2 shells and pointing to one that I know is empty I am doing the same as saying to you, ‘You can choose both of these (hence you now have a 2/3 chance of winning as opposed to 1/3) but don’t bother looking under this one because I know it’s empty’ (and you still have a 2/3 chance of winning if you choose the other).
Why should the two shells I set aside be chosen at random? Because I could deliberately choose the two empty shells!
Now if you have n shells and I randomly draw aside m (where 1<m<n), then the initial probability of your choosing the target shell is 1/n, whether you choose one of the m shells or one of the remaining n–m shells. The probability that it is amongst the m shells is of course m/n. I am now obliged to indicate one of the m shells that I know to be empty. The probability that one of the remaining m–1 shells is the target shell is now m/n(m-1).
However, the probability that it is one of the shells from the n–m group remains at 1/n. (Because I was restricted to the m group of shells when I uncovered an empty shell means I convey more information about that group than the other group.)
Now, which gives the greater probability, 1/n or the formula above? Clearly the latter, since m/(m–1)>1. So in the long run you will increase your chance of winning by always choosing from the remaining m–1 shells.
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