Nice try, Lefty. But there is one constant in the Monty Hall dilemma: The ones with the incorrect reasoning are precisely the ones who refuse to bet real money on it.
I don’t get to be 51 people, I get to be one person. Let’s say I’m the very first person. I say Ace of Spades.
Put a card to the side. The next card you deal is me. Let’s put that into the Folly circle of isolation.
You look at the remaining cards. By some miracle, contestant Folly has not been eliminated.
So, where’s the Ace of Spades. In the Folly circle of isolation? Or put to the side. What are the probabilities of both?
Someone else is welcome to try. We can do it many times. Septimus, you wanna make 51 guesses?
That’s fine. I’m asking you to tell me what all 51 people say.
MOST OF THE TIME YOU ARE GOING TO BE ELIMINATED. WE ARE NOT TALKING ABOUT THOSE TIMES.
But if you tell me what all 51 say, I’ll eliminate 50 of them, and give a choice to that one remaining person.
Try it my way, see what happens.
Try it my way. What are the probabilities between the Folly circle of Isolation and the card put off to the side.
The card I know you cannot check to see if its the Ace of spades to check if I’ve lost.
[quote=“Left_Hand_of_Dorkness, post:64, topic:719240”]
MOST OF THE TIME YOU ARE GOING TO BE ELIMINATED. WE ARE NOT TALKING ABOUT THOSE TIMES.
QUOTE]
That’s right we are choosing between a time I’m not eliminated because I got very lucky and a time I’m not eliminated because the “untouched card” is the winner.
Let’s.
… And ? Take your time. (Hint: what about scenario 2?)
Quod est demens ?
See post #57. So far I see it as the simplest explanation. Show me where it is wrong.
What about scenario 2? It is still viable. So is scenario 1. Either one can happen, and the chances of either are the same. That’s the point.
So I was wrong. It’s better to switch. It’s interesting how what seems insignificant turns out to make the difference.
Switching is better for the contestant give the offer to switch 2/3 of the time. I’m not understanding the 50% or the 80% figures some others have come up with.
It’s different because I get to be two people. Of course that would double my chances.
Then I take it you’ve never discussed probability before.
The greatest minds in the world have problems with probability.
But damn it, your post has swayed me. If I am the contestant left, 1/3 of all outcomes have been eliminated.
OK, now it’s you, **Folly **and me against everyone else.
I don’t understand your way, honestly–I don’t know what you mean when you say “Put a card to the side. The next card you deal is me. Let’s put that into the Folly circle of isolation.”
But I make a promise. If you’ll just goddamned give my way a go, and then explain your way, I’ll try your way. Just give a goddamned go already!
Are you talking about the original Monty Hall problem? Because that’s the answer to the original, not the OP.
Post 50 examines all the possibilities; take a look at that one to see where you’re wrong.
Or pick some cards, any 51 cards, and see how it plays out!
But that’s my point. In the scenario in the OP, you do effectively get to be two people.
Not quite. In addition to your incorrect answer of 50%, we have votes for 33% and 80%, but only a few for the correct answer: 67%.
@ Terr: If you’re sincere about wanting to learn, perform the step-by-step exercise in #47. That will lead you to your error.
Well…sort of. I think that confuses things.
SOMEBODY TAKE MY CARD CHALLENGE, DAMMIT! I really think the odds here will demonstrate, but if I make all the choices myself, it’ll be hard to follow. Choose 51 cards!
Post 50 is unnecessarily complex. Mine is simple. There are only 3 possibilities. No dice throwing two times. Monty, by selecting the other guy and not me eliminates one. 2 are left. Each one had equal chance of occurring before and still has equal chance of occurring. So the probability is 1/2.
Same with your card example. Instead of 3 possiblities, there are 52 each one with the initial probability of 1/52. By eliminating 50 players, Monty reduces the number of possibilities to 2. Each one is equally likely NOW. So the probability is still 1/2.
Put up or shut up. You’re giving orders to 51 people to choose different cards. Which one will they not choose?