The Monty Hall paradox is a fun problem to think about. It is named after the "Let's Make a Deal" TV show.
With a lot of fanfare about prizes, what you do for a living, etc., you are presented with three colorful doors, behind one is a prize, there are no prizes behind the other two (except maybe something like a skunk...). You pick a door, then the host opens one of the two remaining doors to show you what is behind it--it, of course, is a non-winning door. There are only two closed doors remaining now, and one of them is a winner. You are asked if you want to stay with the door you chose or switch to the other door. You make your decision and the remaining doors are opened to see if you won the prize. ...cut to commercial...
Ignoring all the stagecraft, in the end you are faced with two doors, one winning and one loosing, so it seems like you have a 50/50 chance of picking the right door. So either strategy, switching or staying, should be equivalent...
I was playing with Small Basic some more to try to illustrate this. In the first example the "stay" strategy is illustrated.
The chosen door is colored in yellow, another, non-winning door is opened, then the prize containing door is revealed.
And here is the second, switch, strategy.
After you choose a door and the other door is opened you always switch to the remaining unopened door.
After a while you may see that the stay strategy only wins about 1/3 of the time and the switch strategy wins 2/3 of the time. So you should always switch and the probabilities are not 50/50.
How can one of three doors have a 2/3 chance of winning...? It seems like it should be 1/2, or maybe 1/3...? This puzzle conflicts with our intuitive way of thinking about the problem. When I first learned about it I told several people that I worked with (at a print shop) about it and they said I was absolutely wrong, that the chances were 50/50 and that I didn't understand statistics...
This is also an excellent example of a situation where history (data) and prior information matters, which is a cornerstone of modern Bayesian statistics (and here). If there were only two doors to start with the probability would indeed be 50/50, but the history of arriving at the two doors changes the probabilities.
To make it more intuitive imagine a more extreme situation. You win an opportunity to participate in a contest at a newly built US Post Office. There are 1,000 mailboxes along a wall in the building. In one of them is a check for $1,000,000 and you have to pick which one. You pick box number 207. Then all the other boxes are opened to show that they are empty except for box 207, that you chose, and box 853. Would you stick with box 207, that you chose as one out of 1,000 boxes, or switch to the only remaining box that could possibly contain the check? In other words, there is probably nothing special about box 207 that you chose haphazardly, but there seems to be something special about box 853 because it was the only other box not opened.
Tuesday, July 24, 2012
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1 comment:
Switching obviously helps.
Suppose you have to choose the Ace of Spades from a pack of cards.
After you choose your card, the dealer then goes through the pack, takes one card out, and show you the rest of the cards – none of which are the Ace of Spades.
Should you switch your selection to the card the dealer removed from the pack before he showed you the 50 cards that were not the Ace of Spades?
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