OK I missed the fact that he opens doors to show you they are not winners. If Monty opens all the other doors except #73 then your odds of winning would be 50% whether you switch or stay.
That’s just if you have one choice. But the problem is about the chances of winning over a number of attempts.
The game gives you a chance to take advantage of your own low success rate at guessing the right door. When their are 100 doors, 99 times out of 100 your guess would be wrong. But 99 times out of 100 the door with the prize is one of the other doors. By opening all but one of the other doors, you are being shown which door the prize is behind 99 out of 100 times.
As stated above and below, this is incorrect. Suppose Monty didn’t open any of the doors - he just said you can keep your choice of door #1 or take the other 99 doors? Would you agree that you stand a much better chance to win if you switch?
The crux of this problem is understanding that opening all the doors except #73 is EXACTLY THE SAME as offering them all to you without opening any of them. It gives you no more information than you already had. The odds remain the same; 99 to 1 against you. You only had a 1 in 100 chance of guessing right at the beginning and nothing has changed since then. Monty is always able to open all the doors except 1; it’s built into the problem.
But if 98 doors are opened then they are shown not to be winners. At that point they can be discounted and your odds of winning go from 1 in 100 to 1 in 2.
We’ve done this before, probably in this thread, and there’s really no value in rehashing. Since you’re so sure that there’s no advantage to switching, go play the game and let us know how it works out.
I disagree. Having watched LMaD for years as a kid, it never occurred to me that Monty would be forced to always open a door. It’s not an ambiguity…it’s a contradiction to the actual underlying process she used to define the problem.
No. You are choosing between a 1 door set, and a 99 door set. Opening the doors has nothing to do with the odds. You either choose door #1, or all the other doors. So think about it this way. You get to keep everything behind door #1, or everything behind all the other doors. And you have to play the game over and over again.
I just played 16 times, rotating the door I pick, alternating my decision to switch, no significant difference.
You broke off the sentence half way through. I didn’t let anyone off the hook.
8 per strategy is a small sample. Try always switching for 30 or so rounds.
I didn’t disagree with the second half of your sentence.
KNEW that was coming! <Col. Jessep> “I have neither the time nor the inclination to do enough cases for a valid sample size.” <Col. Jessep>
Are you actually asking how the “Monty Hall Problem” is misleading, given that it bears no resemblance to Let’s Make a Deal or what Monty Hall did on the show?
If Monty opens a random door, there is no advantage to switching even when he opens a loser.
If (as in reality) there is no guarantee beforehand that Monty will show you a losing door, it doesn’t work. For example, he could simply show you a loser if and only if you chose the winner.
My favorite “solution” is as follows:
When your first pick is correct, there is a 2/3 chance Monty will show you a loser and let you switch. When your first pick is wrong, there is a 1/3 chance Monty will show you a loser and let you switch.
Therefore, if Monty shows you a losing door and allows you to switch, you have an exactly 1/2 chance of winning whether you switch or not.
As already said, this is wrong and understanding why it is wrong is the crux of the problem. The key is that not all doors are equal. Monty knows which door has the prize - he’s never going to open that door. The one remaining door that he is offering you is 99/100 likely to have the prize. The door you selected is 1/100 likely to have the prize. Open doors don’t do anything to change that.
Your initial choice of 1 door out of 100 had a 1% chance of being right, and it still does. The one remaining closed door (which is equivalent to all the doors you didn’t choose) has a 99% chance of being right. The fact that it’s a choice between two doors doesn’t mean those two doors are equal.
If Monty didn’t open those 98 empty doors you would choose the 99 vs the 1, right? Open doors doesn’t change the equation because you know Monty can do that. There’s no new information added to the situation, you know nothing more about your initial choice than before.
For 30 rounds I switched and won 18 times. For 30 rounds I stayed and won 11 times. I note for really no reason whatsoever that when I switched for ten rounds and stayed for ten rounds, my wins were the same: 4 for each.
BZZZZT! I’m sorry, that’s incorrect. But thanks for playing! ;0)
Okay, imagine Scumbag Steve here. He challenges you to a game with a standard 52 card deck, each card is spread out on the table. You bet 50 bucks, the bet is that if either of you get the Ace of Spades the winner gets 100 bucks, otherwise the bet is void.
So you pick a card, but don’t flip it yet. Steve picks a card and flips it, “nah, man, you gotta let me have do-overs I need this money,” he says. He flips the next card… and the next, and still the Ace of Spades hasn’t been revealed. There’s only one card left.
Would you say he “only” has a 50% chance of finding the Ace of Spades? Probably not. If he offered to swap you cards you’d be daft not to.
Monty does the same thing with all the doors, it’s just he’s:
- Guaranteed to flip all the cards except 1.
- Flips all the wrong cards at once, rather than one at a time.
Unless you believe that Steve knows which card has the AoS ahead of time, and he’s flipping the bad ones intentionally, then yes, I’d say he has a 50% chance of finding the Ace of Spades when we’re down to two. And, I’d be right.
But that’s not the game in question here. Monty Knows.
Again, the whole question is about a sequence of games. Any one decision like this is either right or wrong. That doesn’t tell you anything about your chances of winning.