Quote:
Originally posted by SilverTab
indeed
|
See, this is where i think the problem is. You say "The host reveals B or C"
How can two seperate events be counted as one?
Nobody has adequately explained to me why that is.
I've gone through every possible combination of events in this problem and i wind up with a 50% chance of winning.
Outcome number 1:
Prize is in Box A.
Contestant chooses Box A.
Box C is removed.
Stays = Win
Switches = Lose
Outcome number 2:
Prize is in Box A.
Contestant chooses Box A.
Box B is removed.
Stays = Win
Switches = Lose
Outcome number 3:
Prize is in Box A.
Contestant chooses Box B.
Box C is removed.
Stays = Loses
Switches = Win
Outcome number 4:
Prize is in Box A.
Contestant chooses Box C.
Box B is removed.
Stays = Loses
Switches = Win
Nobody has explained to me why the first two outcomes together are somehow only as likely as one of the other outcomes?
Equinox kept posting the results from that C program that someone had written. I can't read programming for shit. My suspicion is that the person has also lumped the first 2 outcomes into a 1 in 3 chance instead of a 2 in 4 chance.
In the end, there are 4 possible outcomes regardless of what box is initially chosen. And out of those 4 outcomes, 2 will win when you stay, and 2 will win when you switch.