There is if you have to choose. On the other hand, the very act of switching implies by default, you will always pick wrong.
Ignore Bait: Highest IQ, Many Questions, Odds makers invited...
- Thread starter straightline
- Start date
It’s only 50/50 if Monte doesn’t know what the cards are and if somehow a goat was opened out of the last two cards.
Once he sees them and can’t open the car…..
The choice is no longer 50/50
It’s 66/33
DeeDeeCues
Well-known member
Wrong. But you can't be convinced, so I'll let you choose the other door...
Congrats, you win, you're correct!
It's going to be a tough life.
DeeDeeCues
Well-known member
No it doesn't. It means you will usually pick wrong. That is literally what statistics is.
I imagine that's how the change con works. Ok I get it. They actually spent $8.33 apiece for the pizza.
Yeah but you can't replace stats with actions and vice versa. Obviously you can act on stats as per this scenario but like I keep saying, you don't have the extra chances for the numbers to play out.
The numbers don’t have to “play out”. That’s not what statistics are. That’s not what odds mean.
That’s not how stats or life works.
Huh? Wait. 8.33 apiece, 25 bucks. Buck each back, 28 bucks, 2 buck for the delivery guy.
If you only played a mega millions lotto number once in your lifetime….
Are the odds 50/50 because you either win or you lose?
The stats don’t have time to play out……
Go back to my first post to you. You’re getting wrapped up in irrelevant math again.
The only math that matters is $25 for the pizza and $2 for the tip. Each guy spent nine dollars.
Yes I know. If you habitually gamble by the odds, then plunking down on any one shot with good numbers is a good bet. But here you are faced with a free car and 66% still guarantees nothing.
That doesn’t justify making a choice that gives you 33%.
The scenario doesn’t allow for 100%. So you have a choice.
33% or 66%
That leaves the missing dollar. $9.33 each plus the 2 bucks puts you back at 30.
By the numbers sure. But in an actual bet, No way Monty is trying to feed you the car. It's a battle of wits at as near as I can figure, 50/50.
Serious question. Why does it matter what Monty knows? Say he is clueless and just happened to open the goat?
The pizza was $25.
He kept $2 for a tip.
That’s $27. Leaves $3. Each one got a dollar back.
So, they each spent $8.33 on the pizza. And they each tipped him $0.66.
$8.33 + $0.66 = $8.99
Because he is not allowed to open the car.
So, 66% of the time, after you pick a door, the two doors left are a car and a goat.
33% of the time it’s a goat and a goat.
When we swap the door every time, we accept that we are going to lose 33% of the time. That is gone, lost, out of here.
But, 66% of the time he will have a car and a goat. And he’s not allowed to open the car door.
So, when he opens the door and shows you a goat, there is now a 66% chance the door he didn’t open is the car.
If he’s clueless and opens the goat…..
It’s now 50/50 if you have the car or he does.
His knowledge of what’s behind the doors is the “trick” part of the question. Along with not being allowed to open the door with the car.