What if there were TWO contestants and each of these two contestants were ardent mathematicians, both with an innate understanding of the principle behind the Hall paradox. After each choose a door, and the third unchosen door was opened to reveal a zonk, would each of the two players, if given the chance, leap at the opportunity to accept the other fellow's door? (And in this scenario the 2 scholars are rivals who would not agree to sell the car and split the proceeds) Hmm?
Ok but this is assuming the host knows where the cars and goats are and will always pick a goat to show you. If he doesn't know then it's 33% no matter what you do.
Part2, it's not 50/50, and 66.6% represent the car b/c a goat was shown. E.g:
Event A) Picked car and stayed = 33.3% of picking car
Event B) Picked car and switched = 33.3% of picking car
Event C) Picked goat and stayed = still 33.3% of picking car
Event D) Picked goat and switched = still 33.3% of picking car plus the 33.3% of the shown goat resulting in 66.6% of picking car if you switched (if you stayed, you're back to 33.3% and NOT 50% because door1=33.3%, dr2=33.3%, dr3=33.3%).
Sorry for trolling your video (I am convinced of the switch will win with a 66 % chance btw) but what a terrible random generator (shuffle) you are using that the Ace is always the third card :P
Another way to view the problem is that your chances of picking correctly with your initial guess (1/3) are unchanged by Monty's revelation. Thus, after one goat is revealed, the chance that the "switch" is correct must be 2/3.
There are 3 doors, 2 incorrect (goats), 1 correct (car!). You choose one of them. One of the incorrect doors is revealed to be incorrect. You can then switch if you want to. Switching will only work if at first you chose one of the wrong doors. Seeing there are 2/3 wrong doors at the beginning, switching will work 2/3 of the time.
I made this video in response to those who *swear* up and down, on all that is holy, that the final probability of switching is 50/50 and not the correct 33/66. There is just nothing you can say to these people to convince them otherwise. I got so frustrated with one of these people that I made this video. If you ever get into this argument, send that person here and tell them to do this experiment.
@paddygilbert, I didn't make this for you, you get it. You need to hook up with one of these knuckleheads who swear the answer is 50/50. *Then* you'll see why I made it.
i dont understand why you went through such extremities to explain this problem. You could just keep the 3 cards face up like the later part and show what would happen if you switched or stayed starting from the first to the last card. To even do any kind of random generation I think kind of confuses than shows any kind of proof. I can flip a coin 100 times and show mixed results or just show that a coin can only fall 2 ways 100% of the time making the probability of landing on a side 50%
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What if there were TWO contestants and each of these two contestants were ardent mathematicians, both with an innate understanding of the principle behind the Hall paradox. After each choose a door, and the third unchosen door was opened to reveal a zonk, would each of the two players, if given the chance, leap at the opportunity to accept the other fellow's door? (And in this scenario the 2 scholars are rivals who would not agree to sell the car and split the proceeds) Hmm?
Johnny Radionic ™
Johnnyradionic 1 month ago
Ok but this is assuming the host knows where the cars and goats are and will always pick a goat to show you. If he doesn't know then it's 33% no matter what you do.
JohnnieNaked 1 month ago
thanks ;D
PeaceLemony 1 month ago
Thumbs up if you thought it was cool that this video was 3:33 before you even watched it;) (33.3333333. . . %)
danjor92 2 months ago 2
Part2, it's not 50/50, and 66.6% represent the car b/c a goat was shown. E.g:
Event A) Picked car and stayed = 33.3% of picking car
Event B) Picked car and switched = 33.3% of picking car
Event C) Picked goat and stayed = still 33.3% of picking car
Event D) Picked goat and switched = still 33.3% of picking car plus the 33.3% of the shown goat resulting in 66.6% of picking car if you switched (if you stayed, you're back to 33.3% and NOT 50% because door1=33.3%, dr2=33.3%, dr3=33.3%).
R2Mintus 2 months ago
Part1
It's easier to understand if looked from the car perspective (the prize u want):
1) picking a car = 33.3% (one door)
2) picking a goat = 66.6% (two doors)
If you stick to your 1st choice, your best is 33.3%,
but by switching you're trading your 33.3% for the remainder--the 66.6% (in effect, you are choosing TWO doors vs. one door).
So, staying = choosing one door = 33.3%.
switching = choosing two doors (opened door and switched door) = 66.6%.
Basically, opening two doors is better.
R2Mintus 2 months ago
Sorry for trolling your video (I am convinced of the switch will win with a 66 % chance btw) but what a terrible random generator (shuffle) you are using that the Ace is always the third card :P
pmelendezu 2 months ago
Not going to lie very well explained :)
Reckful222 3 months ago
U know U can even blame the weather for that.
ValmisFilm 3 months ago
Another way to view the problem is that your chances of picking correctly with your initial guess (1/3) are unchanged by Monty's revelation. Thus, after one goat is revealed, the chance that the "switch" is correct must be 2/3.
Solsane 3 months ago
@Solsane Nice. I like it.
kirkbocek 3 months ago
great proof...not.
SharinganTH14 3 months ago
It's quite simple to explain.
There are 3 doors, 2 incorrect (goats), 1 correct (car!). You choose one of them. One of the incorrect doors is revealed to be incorrect. You can then switch if you want to. Switching will only work if at first you chose one of the wrong doors. Seeing there are 2/3 wrong doors at the beginning, switching will work 2/3 of the time.
paddygilbert 4 months ago
@paddygilbert
I made this video in response to those who *swear* up and down, on all that is holy, that the final probability of switching is 50/50 and not the correct 33/66. There is just nothing you can say to these people to convince them otherwise. I got so frustrated with one of these people that I made this video. If you ever get into this argument, send that person here and tell them to do this experiment.
kirkbocek 4 months ago
@paddygilbert, I didn't make this for you, you get it. You need to hook up with one of these knuckleheads who swear the answer is 50/50. *Then* you'll see why I made it.
kirkbocek 3 months ago
i dont understand why you went through such extremities to explain this problem. You could just keep the 3 cards face up like the later part and show what would happen if you switched or stayed starting from the first to the last card. To even do any kind of random generation I think kind of confuses than shows any kind of proof. I can flip a coin 100 times and show mixed results or just show that a coin can only fall 2 ways 100% of the time making the probability of landing on a side 50%
Sonarghosts 4 months ago
@Sonarghosts take a look at my reply to paddygilbert
kirkbocek 4 months ago
lol @ the video being 3.33 sec long
TheWatchOrgie 4 months ago
@TheWatchOrgie -- Funny, but a coincidence. Wish I'd planned it that way.
kirkbocek 4 months ago