The Monty Hall problem is a well known brain teaser. Here is a variant.
You're on a game show, and you're given the choice of three boxes. Two boxes each contain $1000; the third contains $4000.
You pick one box and the host picks another. The host removes this box, which contains $1000, and adds $1000 to your box. He then offers you the choice to switch to the remaining box.
Do you switch, or stick with your original box?
This is assuming that (as in the classic Monty Hall problem) the host knows what amounts are in which box, and that he will always remove a box containing $1000.
2/3 of the time, my first choice will have been a $1000 box. If I switch, I get $4000 instead of $2000. 1/3 of the time, my first choice will have been the $4000 box, in which case if I switch, I get only $1000 instead of $5000.
If I always switch, my expected average winnings over repeated games would be $3000 per game (4000*2/3 + 1000*1/3).
If I never switch, my average winnings would be... also $3000 (2000*2/3 + 5000*1/3).
So there would seem to be no optimal strategy between switching and not switching -- but this assumes playing the game many times over. If I had only one shot, as on an actual game show, I would likely stick with the classic Monty Hall strategy and switch boxes.
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Posted by Jyqm
on 2015-06-17 09:42:01 |
$1K is better than a goat, at any rate!
