But it did. It eliminated 33% of the options. Leaving win and lose and a choice to stay with your first choice or pick the other. One is win one is lose.
Ignore Bait: Highest IQ, Many Questions, Odds makers invited...
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The statement that it’s 50% is absolutely wrong.
It would only be 50% if the goat he opens is random.
And that’s only 50% of the 66% of the time he doesn’t randomly turn over the car.
But the odds are meaningless. Free car is a lot different than you bet 20K.
Monte starts with "twice as likely" but eliminates one of two losers leaving you each with one door, one winner and one loser, and you pick.The odds don’t change - Monty’s two doors are twice as likely to have the prize as your one door, both before and after he shows one of them.
Sorry, Jay, but I’ll take your bet too.
pj
chgo
That's an interesting rephrasing. How'd you get that figure?The odds don’t change - Monty’s two doors are twice as likely to have the prize as your one door, both before and after he shows one of them.
Sorry, Jay, but I’ll take your bet too.
pj
chgo
He's never going to open the car door, only a goat, every time, leaving one goat and one car, every time, and you get to chose one of the two remaining doors, 50/50.
Why?
I understand the risk/reward proposition, but are you really saying that you don’t care if you win a free car? That you wouldn’t do anything possible to improve the odds of doing so?
A non conditional scenario would be:
You pick a card at random.
I pick a card at random.
We turn over the last card.
33% of the time, that card is the car.
So, let’s say 100 times. So, you lose 33x right away.
Now, there’s 66 other times you get to pick 2 cards. You’ll win 33 times and lose 33 times if you either don’t change or you change 33x. Because it’s now a true 50/50. But, only 66x out of a 100 because you lose 33 times.
Now, here’s the conditional.
You pick a card randomly.
I look at the other two cards.
66% or 66/100, the car will be under one of them.
I’m *not* allowed to turn over the car.
So, 66 times out of 100, I will be *forced* to turn over a goat while the other card is the car. Because I have no choice. 66 times I will have the car and I’m not allowed to turn that one over.
So, I *literally* will tell you where the car is 66x out of 100.
You pick a card at random.
I pick a card at random.
We turn over the last card.
33% of the time, that card is the car.
So, let’s say 100 times. So, you lose 33x right away.
Now, there’s 66 other times you get to pick 2 cards. You’ll win 33 times and lose 33 times if you either don’t change or you change 33x. Because it’s now a true 50/50. But, only 66x out of a 100 because you lose 33 times.
Now, here’s the conditional.
You pick a card randomly.
I look at the other two cards.
66% or 66/100, the car will be under one of them.
I’m *not* allowed to turn over the car.
So, 66 times out of 100, I will be *forced* to turn over a goat while the other card is the car. Because I have no choice. 66 times I will have the car and I’m not allowed to turn that one over.
So, I *literally* will tell you where the car is 66x out of 100.
It’s all been explained, very clearly. You’ve been shown the trough, led right to it. Refusing to drink, just means you die thirsty
No. I'm considering this as an actual win lose situation not a game show gimmick. Even with LMAD analogy you guys (and the smart lady) bring up the 66% as a sure thing.
It’s a shame there isn’t something really on the line here
Seems simple to me, Monte knows where the goats are, he removes one of the doors you didn't choose to reveal a goat, leaving two doors, car and remaining goat. You get to pick between the two, goat or car.
Where did I go wrong?
Where did I go wrong?
Because you pick one first. It’s 1/3 or 33% that you pick the car.
Which means by default, you will lose 66%. Which means by default, he will have the car under one of the two cards 66% of the time.
He’s *NOT* allowed to flip over the car.
So, if he has the car 66% of the time, he has to show you where the car is 66% of the time via turning over the goat cause he has no choice.
You went wrong because you got to take on door away randomly.
He doesn’t get to make a random choice.
That’s where you went wrong.
His two cards are of random choice. But his choice is not allowed to be random.
Where’s Fatboy, he’s got Gambol
No, his choice is not random. He always has at least one goat and he always has to eliminate one goat. That always leaves one goat and the car. Always.
The numbers - even the 50/50 are meaningless except that if you switch, you played twice on the short end. Even on the game show itself, one time thrill.
Are you guaranteeing a win?
It’s a suckers bet the other way around too.
I tell you,
You take three cards. 2, 2, A.
You mix them up and I pull a card off to the side
You get to look at both cards. And you absolutely have to turn over a 2. You cannot turn over the A.
I will never look at the card I off to the side.
I then get to choose if I want to keep my card or swap it.
I turn over whichever of the two I want. If I turn the A over I win.
I’m *always* going to swap my card and turn over the one you didn’t turn over.
66% of the time, you’re gonna watch me turn over the A that you know for sure is there.
I tell you,
You take three cards. 2, 2, A.
You mix them up and I pull a card off to the side
You get to look at both cards. And you absolutely have to turn over a 2. You cannot turn over the A.
I will never look at the card I off to the side.
I then get to choose if I want to keep my card or swap it.
I turn over whichever of the two I want. If I turn the A over I win.
I’m *always* going to swap my card and turn over the one you didn’t turn over.
66% of the time, you’re gonna watch me turn over the A that you know for sure is there.
That is a great way of looking at it.
Let me try…
The answer is because the initial odds were 1 chance vs 2 chances of holding the car for the two sets.
The first set being your choice, and the second set being the other two doors combined.
You always knew going in that the second set might be comprised of one door that had the prize and one that did not.
So, when Monte shows you that one of the doors in the second set doesn’t have the car, it doesn’t change anything, because well, you already knew that. That set still has twice the likelihood of holding the car than the first set does, which means that you now know that the remaining door has twice the likelihood of holding the car…
Did that help?
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Nope. I’m guaranteeing I will win 66% of the time.