Ignore Bait: Highest IQ, Many Questions, Odds makers invited...

Your original guess has 1/3rd chance of winning.
Just because Monty illustrated a door that does not contain the car, does not change the odds from your original Guess--it is still 1/3rd.

So now you are faced with the question of switching::
you have 1/3ed chance by sticking
you have 2/3rds change by changing

It really is that simple.

If you were faced with only 2 doors to start you would have a 50%-50% chance.
But you are not given this opportunity. You had to pick 1 or 3.
After reveling that door #k did not have the car your original guess is still at 1/3rd chance.
Your original guess percentage does not change by Monty reviling an irrelevant door.
If you switch you have 2/3rds chance of winning.
 
0274 still correct?

You always have a spot with me!
Yes sir, at least it forwards to my current phone. It's now a google voice number... What's funny is that those are the same last 4 for my AZ number too...

Jaden
 
Yes sir, at least it forwards to my current phone. It's now a google voice number... What's funny is that those are the same last 4 for my AZ number too...

Jaden
I’m good at keeping in touch with people. That’s my strong suit. Also I don’t disappear either. I might not be around 100% of the time, but I’d never just vanish. That’s not cool.
 
There is no missing dollar

The pizza cost $25. The tip was $2 dollars. Each person paid $10, but got $1 back, which means they each paid $9 total.

Combined they paid a total of $27. $25 for the pizza plus $2 dollars for a tip.
Yeah but the wording is they each put in 10 bucks therefore the math minus con is $9.33 each plus the 2 buck tip.
 
Like I said, “Monty’s two doors are twice as likely to have the prize as your one door”.

“The prize”, of course, is the car.

pj
chgo
Ok. The language is more compelling that way.
But, broken record time again, The math presumes or rather is interpreted as 'your first pick will always be wrong'.
What are the odds a person will get the car on one attempt both with and without the "required" switch?
 
He was trying to feed the car. They were making money based on how much they could charge for advertising. The advertising was based on viewership. The viewership rose because of big prizes.
That's what I said. This would be different in an actual proposition.
 
Ok. The language is more compelling that way.
But, broken record time again, The math presumes or rather is interpreted as 'your first pick will always be wrong'.
What are the odds a person will get the car on one attempt both with and without the "required" switch?
You’re completely missing the point and the math

The odds of picking correctly the first time is 1 in 3

It doesn’t change if you do not make the switch. It only changes if you do

That’s why the switch is mandatory to improve the odds
 
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