Because you don’t understand probability and gambling.
You either need to be the favorite to win or have the corrects odds.
You would need 2:1 odds to not swap doors
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Because you don’t understand probability and gambling.
You either need to be the favorite to win or have the corrects odds.
You would need 2:1 odds to not swap doors
MajorMiscue
Democat
And that doesn't matter at all. Once a goat is eliminated you are 50/50, regardless of whether you picked the car at first or not.
I have to admit, it‘s fun to see how hard some people will double down on the wrong answer, even when proven that it’s wrong, just out of hardheaded stubbornness.
I have no doubt if you bothered to play this to statistical relevance the probabilities would show true. That's a hypothetical situation that wouldn't make a good gamble at all.
MajorMiscue
Democat
You have a billion friends, just more smoke and mirrors.
No, because he doesn’t get to choose which card to eliminate 66% of the time.
66/100 times he will look at the two cards/doors and he will see one goat and one car.
So, those 66 times out of 100, he has *NO* choice which door or card to show you.
I’d suggest you never gamble then.
That’s literally how winners gamble. They pick the favorite over the long run. Unless they have the proper odds on money.
66% is 66%
The initial 33/66% can’t magically disappear unless the door he opens is random.
Someone is going to have a 66% chance in this scenario.
If you swap, you’ll be the one with 66%
If you don’t, you’ll be the one with 33%
Well you do know that LMAD is a 666 show. brrrmp
Like I said I'm not arguing the relevance of actual probabilities; just that it's win or lose one time.
IOW you and your 66% switch and lose. Now what?
MajorMiscue
Democat
Every time he will have a goat and will eliminate it, right?
Every time that will leave two cards or doors or whatever, one a winner the other the loser and you get to pick, right?
The 66% is just a distraction. It doesn't matter if 66% of the time he is looking at two goats, he is going to see at least one and eliminate it which will always leave a goat and a car, right?
MajorMiscue
Democat
He doesn't have the option to open a random door, he must reveal a goat and leave a goat and the car, every time, right?
This heavily presumes they have a long run. Slots for instance. They watched the machines and when the math says go, they'll empty out because they have ALL the remaining tries to hit it.
It’s effin hilariousI have to admit, it‘s fun to see how hard some people will double down on the wrong answer, even when proven that it’s wrong, just out of hardheaded stubbornness.
He doesn't have the option to open a random door, he must reveal a goat and leave a goat and the car, every time, right?
But 66% of the time he has to eliminate the goat when he has the car.
That’s the kicker. Because he has no choice in 66% of scenarios.
if I flip a coin 10 times and it comes up heads every one of those times, what are the odds of it coming up heads on the 11th flip?
Absolutely nothing to do with this scenario.
He doesn't have the option to open a random door, he must reveal a goat and leave a goat and the car, every time, right?
The only way this scenario could be 50/50 is if he turned over the goat, then mixed up the other two cards and you had to pick.
MajorMiscue
Democat
You're hung up on something that doesn't matter. Choice or no choice he will always eliminate a goat and leave a car and a goat. Every time.
He doesn't have the option to open a random door, he must reveal a goat and leave a goat and the car, every time, right?
66% of the time, the car will be in the card he didn’t turn over and you don’t have in your hand.