Hey all,
I was hoping that some mathematically/statistically inclined forum member might be kind enough to comment on the following "problem'"which has had me going 'round in circles for a while:
Here's the Setup:
Imagine you're a contestant on Let's Make a Deal. Monty Hall calls you up to the stage and explains the game to you. He tells you there are three doors. Door #1. Door #2. And Door #3. Behind two of the doors (he won't tell you which of course) are worthless gag prizes, but behind the third is a valuable prize (for this group, we'll imagine it's an unlimited one-on-one Q&A with Robert Pirsig as he sails you down the Hudson River on the Arête).
Monty asks you choose a door... you pick door #2. Monty says, "Well it's a good thing you didn't pick door#1." Door#1 opens and you see one of the gag prizes revealed (let's say... a goat in a wheelbarrow).
Now you're down to your chosen door (door#2) and the remaining door (door#3). Monty says, "I'll give you $100 to switch to door#3.... I'll give to $200...$300...etc...etc."
Here's the Question:
Does switching doors improve your odds of winning?
Here's the Possibilities:
1. Switching won't help. It's a 50/50 chance. Door #2 or Door #3. Switching don't mean diddley.
2. Switching will help. You started with 3 doors. 2 bad and 1 good. Odds are, you picked a bad one to begin with. So odds are, if you switch, you're switching to a good one.
Can anyone crack this one for me? Does switching improve the odds?
thanks,
rick
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