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Found the Monty Hall thread very interesting after my flatmate referred me to the thread on this site about it.
Never really got it myself (lots of arguments) until I read an explanation that apparently has had great success in convincing people, even some maths academics, that it is actually a 33/67 proposition:
Source: Understanding the Monty Hall Problem | BetterExplained
One explanation was as follows:
Perhaps an easier way to get people to understand the problem is to disprove the 50/50 theory rather than attempt to prove the 33/67 theory.
A simple way might be to ask the proponent of 50/50:
"If you ran a game, such as on "Let's Make a Deal", where contestants are given a one-in-three chance of picking a prize but in this case they are not allowed to switch, on average, how many contestants would you expect to win the car based on their first random pick from three doors?"
Hard to imagine that anyone, even a 50/50 proponent, could answer anything other than "33" or "About 33".
You would agree with this and perhaps say:
"Then how can each of the two remaining doors in the Monty Hall problem offer a 50% chance of winning?
If this were so, 50 contestants would each win a car if no contestant switched and, ipso facto, 50 contestants would each win a car if all contestants switched."
"However, as we've agreed, only 33 contestants on average can win a car for every 100 contestants when no one switches. So how can the number of contestants who pick the correct door in a one in three guess suddenly increase from 33 to 50?"
Presuming the dropping of the naysayer's jaw will prevent them from attempting to explain this conundrum (and we hope they won't persist in claiming that the order in which the unopened doors are opened somehow increases the odds of winning...) one might say:
"If 33 win when no contestant switches, 67% must win when everyone switches (because the 33 who won when no one switched will now lose and the 67 who lost when no one switched will now win)."
So not only can it not be a 50/50 proposition, it must be a 33/67 proposition as the number of contestants who win can be expected to double when all contestants switch doors.
Even the proponents of 50/50 would seem to have to agree with the proposition.
I have read many explanations of the Monty Hall problem but never one that focuses on why "50/50" cannot be right rather than why "33/67" is right.
It would seem that if you can get the 50/50 folks to agree that you do not have a 50% chance of winning if you do not switch, then that would seem to leave 33/67 as the only plausible theory.
If the chances of winning a 1-in-3 guessing competition cannot be 50%, they must be 33% for the first door picked (i.e. 1/n, where "n" is the number of options from which it was chosen), and 67% for the other (n-1/n), as the sum of all probability must add up to 1 or 100%.
The larger the value of "n", the more advantageous it is to switch. Many people have shown this when they point out the analogous scenario of a game comprising 100 or one million then, usually in desperation, one billion or perhaps, a Googol doors to show that the probability of the first door being a winner very much depends upon the number of doors from which it is chosen, as with any random guess, not the number of options remaining after all of but one of the duds have been eliminated.
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I thought that explanation was pretty hard to argue against.
Never really got it myself (lots of arguments) until I read an explanation that apparently has had great success in convincing people, even some maths academics, that it is actually a 33/67 proposition:
Source: Understanding the Monty Hall Problem | BetterExplained
One explanation was as follows:
Perhaps an easier way to get people to understand the problem is to disprove the 50/50 theory rather than attempt to prove the 33/67 theory.
A simple way might be to ask the proponent of 50/50:
"If you ran a game, such as on "Let's Make a Deal", where contestants are given a one-in-three chance of picking a prize but in this case they are not allowed to switch, on average, how many contestants would you expect to win the car based on their first random pick from three doors?"
Hard to imagine that anyone, even a 50/50 proponent, could answer anything other than "33" or "About 33".
You would agree with this and perhaps say:
"Then how can each of the two remaining doors in the Monty Hall problem offer a 50% chance of winning?
If this were so, 50 contestants would each win a car if no contestant switched and, ipso facto, 50 contestants would each win a car if all contestants switched."
"However, as we've agreed, only 33 contestants on average can win a car for every 100 contestants when no one switches. So how can the number of contestants who pick the correct door in a one in three guess suddenly increase from 33 to 50?"
Presuming the dropping of the naysayer's jaw will prevent them from attempting to explain this conundrum (and we hope they won't persist in claiming that the order in which the unopened doors are opened somehow increases the odds of winning...) one might say:
"If 33 win when no contestant switches, 67% must win when everyone switches (because the 33 who won when no one switched will now lose and the 67 who lost when no one switched will now win)."
So not only can it not be a 50/50 proposition, it must be a 33/67 proposition as the number of contestants who win can be expected to double when all contestants switch doors.
Even the proponents of 50/50 would seem to have to agree with the proposition.
I have read many explanations of the Monty Hall problem but never one that focuses on why "50/50" cannot be right rather than why "33/67" is right.
It would seem that if you can get the 50/50 folks to agree that you do not have a 50% chance of winning if you do not switch, then that would seem to leave 33/67 as the only plausible theory.
If the chances of winning a 1-in-3 guessing competition cannot be 50%, they must be 33% for the first door picked (i.e. 1/n, where "n" is the number of options from which it was chosen), and 67% for the other (n-1/n), as the sum of all probability must add up to 1 or 100%.
The larger the value of "n", the more advantageous it is to switch. Many people have shown this when they point out the analogous scenario of a game comprising 100 or one million then, usually in desperation, one billion or perhaps, a Googol doors to show that the probability of the first door being a winner very much depends upon the number of doors from which it is chosen, as with any random guess, not the number of options remaining after all of but one of the duds have been eliminated.
: : :
I thought that explanation was pretty hard to argue against.
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