Imagine that the set of a game show has three closed doors. Behind one of these doors is a car; behind the other two are goats. The contestant does not know where the car is, but the game show host does.
The contestant picks a door and the game show host opens one of the remaining doors, one he knows doesn’t hide the car. If the contestant has already chosen the correct door, the game show host is equally likely to open either of the two remaining doors.
After the game show host has shown a goat behind the door that he opens, the contestant is always given the option to switch doors. Are the chances for winning (the car, not the goat) better to stay, switch or it doesn’t matter if you switch or stay it’s a %50 chance of getting it right or wrong? Also, what is the probability for winning with staying (or switching)?
I’m sure some of you already know the answer and have seen this before, but I just recently ran into. I’ll post the answer later if no one else does first.
The Answer is the question is 2/3 if you always swap since originally (2/3 of the time your initial guess was wrong) and the host can’t remove the option you took, or the winning option. I.E. they have no option. Therefore u can safely bet 2/3 of the time that the winning option is the other door.
This is a neat question that almost no-one gets even after you explain it to them. Its almost a revelation when u first realise the answer…
Sometimes you have to simulate it for them with 3 cups and let them be the host…
I have a rather trivial riddle for you
What is greater than God,
More evil than the devil,
The poor have it,
The rich need it,
and if you eat it you will die.
It’s even clearer with more cups (like 10, ie. empty 8 and give them the choice).
[quote=“Tyc00n”]I have a rather trivial riddle for you
What is greater than God,
More evil than the devil,
The poor have it,
The rich need it,
and if you eat it you will die.[/quote]
Really really dirty socks? Nothing with a side of arsenic? X? The new 2 Pac CD? Man, I’ve heard this one before, but the answer has faded. I’m sure a google search would find the answer, but I’ll sleep on it.
The host has to chose one with a goat. This increases your chances of winning a car to 50/50 no matter whether you switch or not. Or am I missing something?
I guess, but it’s sounds funny. Can you drive a car you don’t have (drive nothing)? Or kick nothing (you can do a kicking motion, but with nothing to kick it’s not kicking)?
Maybe if the last line was changed to: and if you have it to eat, you will die.
Edit: I had: and if you have it eat you will die. That was a slight mistake.
I ate nothing. I did nothing. I want nothing etc. Sounds o.k. to me. It is just a question of holding on to the meaning of words a little less tightly I think. Second language students hang on too tight but I am suprised that teachers do too. I mean a big part of my day is spent explaining that most words are homonyms and that people use figurative speech all the time etc.
Take 3 cards (e.g. 2 red, one black) and take 100 samples where you do not change your initital choice and another 100 samples where you do change your choice. Then see which yields the better results.
I’ve seen this puzzle before, and while understanding the logic, it still drives me a bit nuts because it is counterintuitive.
For the doubters, yes, computer programs have been written to simulate it, and do confirm the results.
The reason it works is known as ‘conditional probability’. By opening the door, the host has shown you where the car ISN’T. That represents valuable information which changes (or rather, updates) the odds. If the host shows you the goat first, and THEN you pick a door from the remaining two, the odds are indeed 50/50. As jeff points out, the host does indeed have to know where the car is - that’s the key to the puzzle which changes the odds. If he didn’t, he might accidentally show you the car
Perhaps a way of thinking of this puzzle which makes more ‘sense’ without going into the logic is this:
You have a low chance that you picked right the first time so it’s probably one of the other 2. So if you know which of the other two it probably is, then go for that.
It’s false to think that just because there’s two choices it must be 50/50.
Here’s another one:
I hold up a pack of cards facing me, and ask you to pick the Ace of Spades. You pick a card and put it face down on the table without looking. I can see the cards. I put all the other cards but one face up on the table. None of them are the Ace of Spades. Do you really think it’s just as likely that you picked the Ace of Spades out of 52 cards as it is that I left that as the one card in my hand? Of course it’s not 50/50. Same with the doors.
