I’ve read a few more pages since I first posted.
I keep hoping to god someone takes you up on betting money.
I’d like to be in but it’s your gig.
Unbelievable how stubborn people are even with the answer.
Ignore Bait: Highest IQ, Many Questions, Odds makers invited...
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I’ve read a few more pages since I first posted.
I keep hoping to god someone takes you up on betting money.
I’d like to be in but it’s your gig.
Unbelievable how stubborn people are even with the answer.
Lordy.
I don’t know if I can make it to the end of this thread.
Are people really this ignorant and stubborn?
Yes, yes they are.
I'd like to kick the person that started this thread right in the nuts. I'm convinced the right thing to do is switch doors, but I still cannot wrap my head around the reason why. It's driving me nutser.Lordy.
I don’t know if I can make it to the end of this thread.
Are people really this ignorant and stubborn?
Yes, yes they are.
Read the whole thread.
Kudos to MajorMiscue on his admission.
It was still a tough read.
Kudos to MajorMiscue on his admission.
It was still a tough read.
You have 3 doors to randomly choose one.
There’s only one car. That’s a 1/3 chance to choose right. Most everyone immediately know this.
So, when you pick a door and it’s off to the side, that by default means there’s two other doors that don’t belong to you.
Which means if you take your 1/3 out, there’s a 2/3 chance the care is behind one of those.
Monty *knows* what’s behind all the doors. And if the car is one of “his” two doors, he will always open and show you a goat.
So, if 2/3 of the time the car is behind the 2 doors you didn’t pick, Monty is basically allowing you to swap one door for his two. As he always eliminates a goat.
So, if you switch, the *only only only* way you can lose is when both his doors have goats.
Another way to look at it, since he will always show a goat…..
If you never swap doors, you’re stuck with the 1/3 odds you started with.
Another way to look at it……
If you never switch, the only way you win is when your first pick has the car. 1/3
And since he always shows you a goat and gets ride of a goat by showing you, that gets rid of one of his two doors and you get to swap for his other door.
It’s effectively letting you trade your one door for two doors.
Let’s change the scenario and it would still be the same.
The rules say:
-You pick a random door and it’s eliminated. It doesn’t exist anymore.
-You now get the other two doors. They are yours.
-Monty will look behind both and will then open a goat.
-The last door is yours and you have to open it and keep what’s inside.
Well, the only time you lose is when the first door you picked was the car and it was eliminated immediately. So you never had a chance.
That random pick was obviously 1/3 to be a bad pick.
So, by default, you get the leftover 2/3 odds.
MajorMiscue
Democat
Let me try: Every time, Monte gets to see 2 of the three choices.
2 out of 3 times he will have the car but he must remove a goat.
You will have correctly selected the car door 1 in 3 times.
Stick with your door and you win 1 in 3 times.
If you always switch to Monte's door you win 2 out of three times.
No skill or luck involved. Just statistics.
MajorMiscue
Democat
Regardless, he gets twice as many cars as you because he gets twice as many doors. When you switch you take his two door advantage in exchange for your one door.
Yes, I know the answer. I went to lots of college and stuff.
It's an old problem and the only fun left in it is watching people who don't understand it try to explain it. Not directing that at you....
Here's a chart that usually turns the doubters around,
Sent from my SM-T830 using Tapatalk
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The original scenario then swapping scenarios in midstream makes this deliberately confusing. If you focus on the 66% instead of the 33% it is easier to understand or so it seems to me. Also, the important thing is what Monte knows when he bends the scenario, not what the player knows.
I was long in the 50-50 camp myself years ago. Not convinced I fully understand things now. I am convinced that if you let somebody swap you are on the short end of the deal! I don't know that I fully understand the "why" but I understand the "how"!
Hu
That would be me. I got the premise off Pocket in Firefox IIRC. Only choosing from 3 and THEN when offered, switching, changes the statistic from 33% to 66% your favor. It doesn't mean you win. Even if you are able to play a thousand times times you may still come up 100% loser. If you play infinitely, your numbers would eventually hover around 66% but notice, you lost your soul!
My point is you get one shot. This isn't LMAD, and the dealer is now guaranteed trying to beat you. You now are in a battle of wits at 50/50.
Dealer want's you to switch because you have the winner,
Or dealer wants you stay because you have the loser.
Alternatively, If the 66% method caught on, the cardboard box guys could clean up on the mathmeticians.
That's all the gambol I gots.
If you change your underwear once every 3 days what are the odds you will have a car accident?
It's really a simple concept and the best way to look at it has already been stated. 3 choices. You make 1, 33.333%, the remaining choices have a 66.666% chance of having the car. Once one of those choices is revealed to NOT be the car, the sole remaining choice STILL has a 66.666% chance of being correct, and your original choice still has a 33.333% chance of being correct.
It becomes more apparent (also already pointed out) when you look at more original choices.
There are 100 doors. You choose one of those, there was a 1/100 chance that you chose the correct door, and the other 99 doors have a 99/100 chance of being correct. If 98 of those doors are revealed to NOT be the correct door, then the remaining door out of those 99 STILL has a 99/100 chance of having been the correct one.
Jaden
It becomes more apparent (also already pointed out) when you look at more original choices.
There are 100 doors. You choose one of those, there was a 1/100 chance that you chose the correct door, and the other 99 doors have a 99/100 chance of being correct. If 98 of those doors are revealed to NOT be the correct door, then the remaining door out of those 99 STILL has a 99/100 chance of having been the correct one.
Jaden
Yes, I know the answer. I went to lots of college and stuff.
It's an old problem and the only fun left in it is watching people who don't understand it try to explain it. Not directing that at you....
Here's a chart that usually turns the doubters around,
Sent from my SM-T830 using Tapatalk
Exactly. No need for 1000 simulations or other nonsense that people post to show off that they know some statistics. This has nothing to do with statistics, it is a pure probability problem. A simple enumeration, such as yours, is an analytically comprehensive proof, one that would be acknowledged as a proof by any mathematically literate person.
It's not Math 55.
️That would be me. I got the premise off Pocket in Firefox IIRC. Only choosing from 3 and THEN when offered, switching, changes the statistic from 33% to 66% your favor. It doesn't mean you win. Even if you are able to play a thousand times times you may still come up 100% loser. If you play infinitely, your numbers would eventually hover around 66% but notice, you lost your soul!
My point is you get one shot. This isn't LMAD, and the dealer is now guaranteed trying to beat you. You now are in a battle of wits at 50/50.
Dealer want's you to switch because you have the winner,
Or dealer wants you stay because you have the loser.
Alternatively, If the 66% method caught on, the cardboard box guys could clean up on the mathmeticians.
That's all the gambol I gots.
This is again full of wrong.
The odds of never winning 1/1000 tries is beyond astronomical.
This also has nothing to do with the cardboard box guys which is literally a scam. It’s not based on any math.
You wouldn’t remotely ever be close to 50/50 with the cardboard box guys. You’d win when the bet was low long enough for them to hook you for a big bet.
You’d only win exactly when they want you to win on the box.
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