I love you all.

]]>“The game show host as you to pick one door without opening. After opening the door, the game show host selects a door and opens it showing a goat.”

I was able to get the idea that you meant ‘the game show asks you to pick one door without opening it’ not, the mistyped “as you” or just “opening” (‘opening it’).

That aside, the other confusing instance to me is then the next sentence implies that my pick of a door was opened. Then you state after my pick of door is opened, the game show host selects a door themselves and opens that one.

According to the way you phrased it, it looks like two doors have been opened already. So if two doors are opened, I would know what’s behind them already, and if its two goats, I would change my pick.

“This problem causes endless confusion because its answer depends on how it is phrased.” I agree. ðŸ™‚ Am I missing something here?

]]>In the four door case … the prior probability of winning is 1/4 and the prior probability of losing is 3/4 … conditioned on the probability that the host selectively shows a goat after you arbitrarily pick a door, the probability of winning by switching is 3/8 and the probability of losing by switching is 5/8.

The fact that the prior probability of not switching and the conditional probability of switching add up in the three door case is coincidental. Well, technically not quite coincidental .. in the three door case, you will be forcing the game show host’s hand in 2 out of 3 case (if your door hides a goat). This leaves 1 out of 3 which equals your initial probability.

In the four door case the host is never forced.

ðŸ™‚

I certainly wouldn’t want a car. Not much used to me. A new pickup would be nice, but probably not worth the extra $ it would take to insure and license it.

So yeah, gimme the goat. As far as probability…. my interest waned about 2 lines in ðŸ™‚

]]>In the first part, you never directly state that switching is preferred with a 2/3 probability (but you do hint at it).

Your final example is wrong though. Simply, the total probability does not add up to one. If P(switch)=3/8 and P(not switch)=1/4, then P(switch)+P(not switch) = 5/8. Where are the phantom 3/8 of probability hiding?

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