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Friday, March 2, 2012

Friday Probability Conundrum

 First, here's a review of a new book that may be of interest to puzzle-solvers:


And now, on to a Friday blog puzzle: I adapted this one from another recent Car Talk puzzler, and I like it a lot as another Marilyn vos Savant-type simple (...but not quite so simple) probability dilemma. In fact I've re-written it into a 'Monty Hall'-like framework:

Monty Hall shows you 3 cards, one that is the color red on both sides, one that is green on both sides, and one that is red on 1 side and green on the other side. He puts all 3 into a black bag and shuffles them around. A blindfolded assistant randomly pulls out one card from the bag and lays it flat on the table so only the top side is observable. That side is red. You as a contestant can only win a prize if you accurately guess the color of the card's opposite unseen side. Should you guess red, green, or are your chances equal whichever you guess?

Answer will be given in the comments section…

1 comment:

"Shecky Riemann" said...

ANSWER: You should guess red, which is twice as likely as green. One way to think of it is to realize that you are really dealing, not with 3 cards, but with 6 faces (3 red and 3 green). With a red face showing you know the card CANNOT be the green/green card -- thus 2 green faces are eliminated from possibility of being on other side -- what is left as possibilities are 2 red faces and 1 green. (To understand the probabilities one must realize that the card on the table could be the red/red card flipped in EITHER of 2 ways.)