Margery was the lady with the King of Clubs. She loved gardening, was a big Star Wars fan, worked in a school for the blind. She will be missed.
When I was choosing someone to stand up, really I was eliminating 50 choices.
IF the card I’d chosen had been the Ace of Spades, then I would have chosen someone at random. There was a ~2% chance of that happening, you agree?
IF the card I’d chosen had been anyone else, then of the 51 people who made a choice, there were exactly 50 people I could eliminate: I’d have no choice about who to leave standing. Do you agree? There was a ~98% chance of that happening, do you agree?
So when Margery found herself standing, either I’d chosen her at random (2% chance), or I’d chosen her because I had to (98% chance).
If we do this over and over and over, 98% of the time, the person I say to stand up is going to be someone I had to tell to stand up, because I couldn’t eliminate 50 people without keeping them in the game. 2% of the time, the person I say to stand up is going to be someone I chose randomly, because of the 51 standing, all were losers, so I eliminated 50 at random.
If you still don’t believe me, give your advice to Mr. 9 of Spades. He’s waiting.
I’ll play. We’ll say that I always choose door 1, and my friend chooses door 2.
#1: I chose the door with the car. Monty opens my friend’s door. If I switch, I lose. #2: I chose the door with the car. Monty opens my friend’s door. If I switch, I lose. #3: My friend chose the door with the car. Monty opens my door. I’m out of the game. #4: My friend chose the door with the car. Monty opens my door. I’m out of the game. #5: Neither of us chose the car. Monty opens my friend’s door. If I switch, I win. #6: Neither of us chose the car. Monty opens my door. I’m out of the game.
Of the possibilities where I haven’t been eliminated, I lose 2/3 of the time if I switch. I win 2/3 of the time if I don’t. Stay put.
FWIW in Let’s Make a Deal when you were picking a door you never played against another player. There were games against other players to see who got to go to the end game but the picking the door thing was always you against the house.
As far as why switching is best, the simplest way I have seen it explained that makes it clear why switching is better is by switching you are essentially getting the two remaining doors, you just happen to already know one is a loser. The math supports it as well but to me that explanation makes the answer feel intuitive too.
Yes it actually should be 1/3 to 1/6. Anyway - yes the contestant should stick. No switching. Monty opening the door not only eliminates one of the scenarios, it eliminates half of the other one.
This is the same as the classic Monty Hall problem. Just forget about a lot of the details, you have 3 doors, one has a prize, the other two have nothing. In the end Monty is offering you two choices:
Take what is behind the door you picked initially picked.
The answer as has already been said is that you should stick. Whether this increases your odds and by how much depends on whether Monty is fair or not when choosing (which as not stated in the OP I will assume is unknown) which contestant’s door to reveal, but you never decrease your chance by sticking and potentially decrease your chance of winning by changing.
Not trying to be snarky, but how much of the thread have you read? If you’re posting before reading the thread, I encourage you to go back and read it. If you’ve already read it, it’d be helpful for you to address one particular post–or just try the card game analogue.
Not to speak for TriPolar, but I think I get what he means:
The problem is the same as the original Monty Hall problem, in a sense, when you boil it all the way down: Getting two doors is better than getting one door.
It’s just that in the original problem you get two doors by switching, while in this version you do so by staying put.
Edit: Which TriPolar may be confused about, now that I read his post again. So maybe I didn’t get what he meant after all.
To me, the biggest difference is that in the Monty Hall problem, you’re guaranteed to get a chance to switch. In the new problem, half the time you get eliminated, and the other player gets a chance to switch. That changes the circumstances pretty significantly.
What I see in the end is that it doesn’t matter whether or not you switch. It is different because the 3rd door is never revealed. Obviously you lose if your door is opened first, but that’s irrelevant. Since the other contestant’s door is always revealed and it’s never the grand prize there’s an equal chance of it being the door you picked or the 3rd unopened door. The difference is subtle, but the cases where the grand prize is behind the other contestant’s door are taken away from you.
Care to try the card trick? It’s the exact same scenario, except instead of revealing one door that’s not right, I’ll reveal 50 cards that are not right, leaving you with the choice of directing the lucky schmuck to switch or stay.
If you’re right, there’s an equal chance of success whether you switch or stay. If you’re wrong, sticking with your original choice is the best bet.
Here, again, are the rules:
I’ll choose a card at random, the Prize Card.
You direct your 51 minions to choose 51 different cards, telling me the one unpicked card.
If the unpicked card is the Prize Card, then I’ll choose one of your minions at random, tell you that all the other minions are losers, and give that minion (also secretly a loser) the choice of switching to the unpicked card or staying.
If one of your minions picked the Prize Card, then I’ll tell you that all the other minions are losers, and I’ll give that minion (secretly a winner) the choice of switching to the unpicked card or staying.
Again, it’s identical to the OP, except that there are 52 choices instead of 3; in the end, there’ll be only two choices.
Would you like to play it? Rather than argue about it, a demonstration will be very helpful, I think.
I’m now convinced that the math works out to the same probability (except the advantage is to stay pat). However, the way you get there is definitely different.
Personally, while convinced, I’m still experiencing some cognitive dissonance because I haven’t yet talked myself out of all the arguments that I talked myself into… So to speak.
In any case, it’s pretty conclusive that switching is the wrong choice here. I ask you what I’ve asked Tripolar: have you read the thread? If so, can you respond to a particular line of argument and show what the flaw is?
I really like septimus’s dice setup: it gives Monty exactly the same behaviors, sets up the game exactly the same way, but you can clearly see by looking at each of the six possible outcomes how three of them are eliminated when you’re left in the game, and of the remaining three, two of them involve your sitting on a winning prize.