Posted on 10/12/2017 6:23:59 AM PDT by ConservativeStatement
Forget Fermat’s last theorem. The most vexing challenge in mathematics just might be the Monty Hall problem. Monty Hall — born Monte Halparin — presented nearly 5,000 episodes of Let’s Make a Deal, the US game show that inspired the puzzle. It is an onion of a conundrum; layer after layer, and guaranteed to make you cry. The puzzle is this: a contestant faces three doors. Behind one of them is a big prize such as a Cadillac. Each of the other two doors conceals a booby prize such as a goat.
(Excerpt) Read more at ft.com ...
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Really? It worked for me and I am not a subscriber. Interesting debate on the issues of probabilities. Do you switch your choice you made when it was 1 out of 3 and is narrowed to 1 out of 2? That was the topic at hand.
If you stick with your original choice, you have a 33% chance of being correct.
If you decide to switch to one of the other two, your chances of winning are 66%......................
Just search it on youtube.
I discovered this a couple of years ago. The key is that Monty KNOWS which door the prize is behind, otherwise the odds don’t change. It’s complicated, but does actually make sense once you really look at it.
What’s weird is that if Monty were to give this choice every week, no matter what, then the odds would be as expected. But because he didn’t, well...
Only if you assume your original pick was wrong. It’s still 1 out of 3, it doesn’t matter how many times you switch. But i could be wrong, new math and all.
They did an entire episode on this problem on the TV show "Mythbusters" (Episode 177), where they explain the math and even demonstrate how it works by staging a fake game show where contestants were directed to either stick with their original answer or switch. I was shocked by the result, not only was it advantageous to switch but it was actually by a pretty big margin. They also had a segment where they demonstrate that the vast majority of people stick with their first answer even when they know the mathematical advantage of switching.
Doesn’t matter. The results are rigged when the show switches the prize behind the scene during the build-up to the reveal.
it is always a good idea to switch. it is more obvious if it starts with, say 100 doors.
Yep. Covered right here: https://en.wikipedia.org/wiki/Monty_Hall_problem
Monty, the consummate showman, passed away just a few days ago...................
Rest in piece, Monty. Do you pick casket #1, #2 or #3?..............
https://en.wikipedia.org/wiki/Monty_Hall
I remember long ago when one of the booby prizes was an oil field pump jack. The contestant elected to give it up. Monte about crapped himself when he found out it was a $40,000 prize. That was way more than the cars of that era.
Just watch the movie scene in 21 on this:
https://www.youtube.com/watch?v=Zr_xWfThjJ0
Explains it in full
I’m the exception that proves the rule, because I’d win either way. I love animals and goats are among my favorites. :)
Assuming that the host will always open a door with a goat, and will always allow you to change your door, you should always change your door.
Think of it this way: when you choose one of three doors, your odds are 1/3 and the odds the prize is behind one of the other two doors is 2/3. That is pretty simple to understand. By changing doors, you in effect picked two doors, the mystery door, and the one the host exposed.
Cars come and go. I have a bunch of cars.
I have some acreage now and would probably do better to have a goat. Or 2.
I think you and I solved the puzzle. Cadillacs are dumb. A goat can feed a family, or at least a baby with digestive issues which lemme tell you IS THE WORLD WHEN IT HAPPENS TO YOU.
One canonical answer is: always switch.
The assumption is that there is one valuable prize and two gag prizes. When you make your choice, you have one chance in three of being correct, two out of three times you will select a gag prize. If you pick the valuable door, Monty will simply select a gag prize door at random. If you pick a gag prize door Monty will reveal the other gag prize every time.
A priori, you know that 2 out of 3 times the valuable prize is behind one of the two doors you did not select. By switching you will win 2 out of three times.
I read about this (the Wikipedia article is pretty good) and completely failed to understand the logic behind it. But the results speak for themselves and are not subtle, they are massively in favor of switching.
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