I think I've posted the Monty Hall Problem here before, but it goes like this. You've got a game show where there are three doors to choose from. Behind one door is the prize, behind the other two is a goat. Once the player selects a door, the host (who knows which door has the prize) will then reveal the goat behind one of the unselected doors. The player is then allowed to pick the remaining unselected door, or stick with the first selection. For the best chance to win should the player make the switch, stick with the originally selected door, or does it make any difference?
Answer: The Player should switch doors, which will double the chances of winning.
I think I've posted the Monty Hall Problem here before, but it goes like this.
ACK! I think this is one of those things that, no matter how many ways it is explained to me, I will never understand!
Wait.
Double? It goes from 1/3 to 1/2, doesn't it?
For the best chance to win should the player make the switch, stick with the originally selected door, or does it make any difference?
I knew the answer, but only because Charlie explained it on Numb3rs back when I still watched it.
If I could save today's XKCD with its mouse-over text, I would.
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Double wait -- Gud, check my whitefont in my post above. I realized looking back, no one would know it's there.
Jesse:
no, it goes from 1/3 to 2/3.
Oh right, I guess that makes sense.
slowest natter day ever, or me being too bored?
