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Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Do any of you have any noteworthy experiences where knowledge of math helped you in an unusual way?
https://en.wikipedia.org/wiki/Monty_Hall_problem
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Get Those Brain Cells Working: The Monty Hall Problem
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Can anyone remember when the times were not hard, and money not scarce?
(Score: 4, Insightful) by MrGuy on Friday July 29 2016, @05:35PM
Here's one way to understand WHY the 2/3 answer is correct.
Take a slightly simplified version of the game, with the following rules. There are three doors. One always contains a car, two are always losers. The player is allowed to pick one door, and the host has the other two doors. At the start of the game, neither the player nor the host know where the prize is. Immediately before the prize is revealed, the host MUST offer the player the option to trade the player's current door for BOTH of the other doors (i.e. the host has no discretion on whether or not to offer the trade).
First iteration of the game - the player picks the door, and the host does nothing else. No doors are opened, nothing is said, nothing. The player picks a door, and is immediately offered the chance to trade with the host for the host's two doors.
I'd hope in this instance we'd all agree the odds from staying are 1 in 3, and from switching are 2 in 3 - you can trade one door for two.
Second iteration. The player picks their door.
Before offering the switch, the host says to the player "Neither of us know where the prize is. However, given we both know there's only one prize, that means at least one of my two doors must be a loser. So, if you trade with me, you're guaranteeing yourself that you're going to get at least one losing door as part of the bargain."
Does this change the odds of winning from the first iteration? It shouldn't. It's still 1 door vs. 2 doors. The host's statement doesn't give you any information you didn't have in iteration 1. Sure, having two doors means having at least 1 losing door. But having all three doors means having TWO losing doors. The game isn't about losing doors - it's having the best chance to have the WINNING door.
Odds of staying - 1/3. Odds of switching - 2/3.
Third Iteration. Player picks a door. Host gives the same speech as before. Then, before the player is allowed to switch, the host goes backstage and sees where the prize is. The host says nothing additional to the player after doing this. Then the switch is offered, as required by the rules.
Does THIS change the odds? It shouldn't - the host is required to offer the switch, so it doesn't really matter if he knows where the prize is or not. The host hasn't given you any new information.
Odds of staying - 1/3. Odds of switching - 2/3.
Fourth Iteration. Player picks a door. Host makes the same comment that he must have at least one losing door. Then the host goes backstage. This time, when he returns, he says to the player "I was right - I just checked, and I definitely have one door that's a losing door."
Does this change anything? Again, it shouldn't - the host just confirmed what you ALREADY know - that one of his doors is a losing door. We've known this since iteration 2. The fact that the host has seen the prize doesn't change anything.
Odds of staying - 1/3. Odds of switching - 2/3.
Fifth Iteration. This is where the trick happens. After the host returns from backstage, the host tells the player "I told you I have at least one losing door." He points to one of his two doors and says "That door is a losing door."
This is where people get hung up, because this is the first time the host has conveyed any information we didn't know in iteration 1. We know know one SPECIFIC door is a losing door.
The trick, however, is that this is actually NO DIFFERENT from Iteration 4. In Iteration 4, the host found out which of his doors were losing doors. Since we've always known the host had one losing door, we've always known he'd have at least one losing door. The host has at least 1, possibly 2, losing doors. Does the host telling us WHICH door is a loser give us MATERIAL information?
Again, it really shouldn't - he's confirming what we already knew. There's at least one door the host has which is a loser.
The host MIGHT have 1 losing door and the prize, or might have 2 losing doors, but in both cases, once the host knows where the prize is, he can definitely point to a losing door.
Odds remain the same - stay, 1/3. Switch, 2/3.
Iteration six. After the host tells you which door is a losing door, he says "Look, I'll prove it to you!" and opens the door he knows is a losing door, revealing there is no prize.
Does this change things from Iteration 5?
Again, it shouldn't. The host already told us that door was a loser. Why does it matter that he "proved" it by opening it?
Odds remain stay 1/3, switch 2/3.
The odds never change, even though we've opened a door. How can that be? There are two keys to this remaining true, both of which apply to the actual game.
First, the offer to switch is mandated by rule. It's not a discretionary action by the host. If the host looked at the doors, saw where the prize was, and then had the OPTION to offer a switch IF he wanted to, then the game is different. In theory, the host should only offer the option to switch if you have the prize - why offer to trade if you can only lose? We're into gambling territory here for what his strategies might be, and whether he's trying to trick you. Lots of options. But the upshot is - if the host has a free choice in whether to offer the switch, then the fact that the offer is made conveys information.
Second, the host does NOT open one of his doors randomly. The host NEVER opens a door with a prize behind it. He always knows, and he KNOWS the door he's opening is a loser. The game would be very different if the host's door was opened at random (i.e. if the host doesn't know what's behind the door he's opening). In that case, there's a 1/3 chance the prize would be revealed by opening the door (every door has a 1/3 odds for the prize before we open them), which makes the solution obvious - you'd ALWAYS trade if you could see the prize behind the host's door. If the host does NOT reveal the prize by opening his door, THEN the odds do indeed become 50/50 we gained new information when a randomly opened door was found not to contain the prize.
(Score: 2) by captain normal on Friday July 29 2016, @08:35PM
Nice WOT, but it totally misstates the actual problem. Please go back and read the Wikipedia link again.
Everyone is entitled to his own opinion, but not to his own facts"- --Daniel Patrick Moynihan--