@Fast Girl: I attended a speech by a lady who worked with people who had various mental disabilities. Her films and accounts of them in action were astonishing. She felt that they know things we that we don't. Savants at math and music are easier to understand because the operate at a level we can appreciate, but some may have insights that are beyond our understanding.

Most useless organization I've ever been a member of. Went to one meeting. The only thing that the average member is better at compared to the rest of the population is solving IQ tests.

Maybe some of you Mensa geniuses can give a succinct reason for the "Monty Hall" scenario:
You're a contestant on a game show (like "Let's Make a Deal). You are given the opportunity to win a car if you pick the correct one of three. The other two doors have a goat behind them. You pick Door 1.

The host knows which door the car is behind, and can only spontaneously reveal the doors with a goat behind them. He raises the curtain on Door 3, revealing a goat.

He asks you, "do you want to stay with Door 1, or change to Door 2?"

@rxlawdude:
You always switch doors.
With your original guess, you have a 33% chance of guessing right - pure probability
However, since they always eliminate one of the loser doors, they have given you more information and have given you a great chance to improve your odds - and the new door has a 67% chance of having the grand prize.

This is a happy tour for me to read on this post. After reading the whole post, we can get solutions to deal with the trouble quickly and safely. Thanks for your nice sharing.

If there is an equal chance of the grand prize bein behind any of the three doors, there is a 33% chance of it being behind A, 33% chance of it being behind B and 33% chance of being behind door C. There is a 67% chance of it NOT being behind Door A.

If you choose A, you have a 33% chance of being correct and a 67% chance of being wrong. The act of them purposefully (and not randomly) displaying a goat does not change the odds that you have a 67% chance of being right by switching.

Change the scenario to 10 doors, and your choice of Door A had a 10% chance of being right. Then Monty Hall starting opening 8 of the remaining doors one by one to display goats (knowing that he will not open the door with the grand prize). When there is only Door A and Door J left - there is a 90% chance that the prize is behind J (and not only a 50% chance).

How about if there were 10,000 doors .... you should always switch from your first choice to the last door available.

I posted that because it struck me as analogous to your explanation of the 67%-- which I think is wrong. What happened before there were 2 doors and one goat does not matter, the way I read it.

On the horses, a lot of people say the man made $10 because he "lost $10" when he bought the horse back. But read the problem as this, and that disappears: "A man buys a horse for $60 and sells it for $70. Then he buys A DIFFERENT horse for $80 and sells it for $90. How much money does he make in the horse-trading business?"

With that language, people answer "$20." Yet it is the same problem.

I agree with the $20. Another way to look at it is if he started with $100, then after spending $60 and $80, receiving $70 and $90 - he ends up with $120 - a net gain of $20 (though I do not see the parallel between the two).

So, with probability - When in doubt, play it out.

You originally pick Door A and choose NOT to switch at the last minute.
Scenario 1 (33% chance) - It was behind Door A - You WIN
Scenario 2 (33% chance) - It was behind Door B - You LOSE
Scenario 3 (33% chance) - It was behind Door C - You LOSE
Therefore by not switching, you have a 33% chance of winning and 67% chance of losing

You originally pick Door A and CHOOSE TO SWITCH at the last minute.
Scenario 1 (33% chance) - It was behind Door A - They show either the goat behind B or C, You switch - You LOSE
Scenario 2 (33% chance) - It was behind Door B - They show you the goat behind Door C - You switch to B - You WIN
Scenario 3 (33% chance) - It was behind Door C - They show you the goat behind Door B - You switch to C - You WIN
Therefore by switching, you have a 33% chance of losing and 67% chance of winning.

mntlvr23 -- Bighorn "I would have said 50 as well." I have to agree with Bighorn - A. He is always right :-).
B. if you change the problem to only have 2 options, it becomes a 50/50 chance. Really don't have time to explain probability right now.

I give up.
I think there will have to be a Tesla behind one of the doors for people to really spend the time to work out the math on this one .... or you can read here for more explanation ....
en.wikipedia DOT org/wiki/Monty_Hall_problem

I have been a member of Mensa on and off over the years. My experience was that the character of the group tended to be regional.

During the '80s, my experience with San Francisco Regional Mensa (SFRM) was quite good. I suspect it still is, but I haven't lived in the area since then. The group tended to be a nice social environment with folks who just happened to be good at taking tests that are broad measures of intelligence (and there are many tests which can be used for admittance: us.mensa.org/join/testscores/qualifying-test-scores/). Not many that I ran into considered themselves "better" than others.

I had a very different experience with an east coast region. The events I went to there tended towards mutual admiration societies.

Another west coast region just didn't seem to have as much going on as SFRM.

One common thread is that Mensans tend to come from all walks of life. The only overt selection criteria is scoring in the 98th percentile of some test. Any other criteria are entirely self-selection as to whether Mensa resonates with you.

@Bighorn I have more faith in you than you do! Read and see where you were correct to begin with.

http://www.realclearscience com /articles/2015/02/25/the_monty_hall_problem_everybody_is_wrong_109101.html
or https://ablestmage.wordpress DOT com/2013/04/09/5050-is-king-classic-monty-hall-problem-re-addressed-and-re-debunked/

My old girlfrield said she was. Ran her own company with 30 employees in her. mid 20s and then dated a guy in his 50s who owned a bunch of companies and said she could probably run a Fortune 500 company by the time she was 40. He died, she got married to one of her former employees that she fired, and had kids and then got divorced. But then she got sick about 6 months affter we started dating and had brain damage and I couldnt tell if she was extremely smart or just crazy.
I decided to err on crazy and dumped her.

## Comments

You're a contestant on a game show (like "Let's Make a Deal). You are given the opportunity to win a car if you pick the correct one of three. The other two doors have a goat behind them. You pick Door 1.

The host knows which door the car is behind, and can only spontaneously reveal the doors with a goat behind them. He raises the curtain on Door 3, revealing a goat.

He asks you, "do you want to stay with Door 1, or change to Door 2?"

What should you do?

You always switch doors.

With your original guess, you have a 33% chance of guessing right - pure probability

However, since they always eliminate one of the loser doors, they have given you more information and have given you a great chance to improve your odds - and the new door has a 67% chance of having the grand prize.

RX this isn't Bayesian probability is it? P of A given B = P (A)?

I would have said 50 as well. I vaguely remember this riddle though from long ago.

If you choose A, you have a 33% chance of being correct and a 67% chance of being wrong. The act of them purposefully (and not randomly) displaying a goat does not change the odds that you have a 67% chance of being right by switching.

Change the scenario to 10 doors, and your choice of Door A had a 10% chance of being right. Then Monty Hall starting opening 8 of the remaining doors one by one to display goats (knowing that he will not open the door with the grand prize). When there is only Door A and Door J left - there is a 90% chance that the prize is behind J (and not only a 50% chance).

How about if there were 10,000 doors .... you should always switch from your first choice to the last door available.

A man buys a horse for $60 and sells it for $70. Then he buys the same horse back for $80 and sells it for $90.

How much money does he make in the horse-trading business?

I posted that because it struck me as analogous to your explanation of the 67%-- which I think is wrong. What happened before there were 2 doors and one goat does not matter, the way I read it.

On the horses, a lot of people say the man made $10 because he "lost $10" when he bought the horse back. But read the problem as this, and that disappears: "A man buys a horse for $60 and sells it for $70. Then he buys A DIFFERENT horse for $80 and sells it for $90. How much money does he make in the horse-trading business?"

With that language, people answer "$20." Yet it is the same problem.

So, with probability - When in doubt, play it out.

You originally pick Door A and choose NOT to switch at the last minute.

Scenario 1 (33% chance) - It was behind Door A - You WIN

Scenario 2 (33% chance) - It was behind Door B - You LOSE

Scenario 3 (33% chance) - It was behind Door C - You LOSE

Therefore by not switching, you have a 33% chance of winning and 67% chance of losing

You originally pick Door A and CHOOSE TO SWITCH at the last minute.

Scenario 1 (33% chance) - It was behind Door A - They show either the goat behind B or C, You switch - You LOSE

Scenario 2 (33% chance) - It was behind Door B - They show you the goat behind Door C - You switch to B - You WIN

Scenario 3 (33% chance) - It was behind Door C - They show you the goat behind Door B - You switch to C - You WIN

Therefore by switching, you have a 33% chance of losing and 67% chance of winning.

Always switch doors.

B. if you change the problem to only have 2 options, it becomes a 50/50 chance. Really don't have time to explain probability right now.

I think there will have to be a Tesla behind one of the doors for people to really spend the time to work out the math on this one .... or you can read here for more explanation ....

en.wikipedia DOT org/wiki/Monty_Hall_problem

During the '80s, my experience with San Francisco Regional Mensa (SFRM) was quite good. I suspect it still is, but I haven't lived in the area since then. The group tended to be a nice social environment with folks who just happened to be good at taking tests that are broad measures of intelligence (and there are many tests which can be used for admittance: us.mensa.org/join/testscores/qualifying-test-scores/). Not many that I ran into considered themselves "better" than others.

I had a very different experience with an east coast region. The events I went to there tended towards mutual admiration societies.

Another west coast region just didn't seem to have as much going on as SFRM.

One common thread is that Mensans tend to come from all walks of life. The only overt selection criteria is scoring in the 98th percentile of some test. Any other criteria are entirely self-selection as to whether Mensa resonates with you.

A switch does double your odds of being right. Here's a pretty good explanation:

https://betterexplained.com/articles/understanding-the-monty-hall-problem/

http://www.realclearscience com /articles/2015/02/25/the_monty_hall_problem_everybody_is_wrong_109101.html

or

https://ablestmage.wordpress DOT com/2013/04/09/5050-is-king-classic-monty-hall-problem-re-addressed-and-re-debunked/

I decided to err on crazy and dumped her.