by stevenyc 26 Replies latest jw friends
You're a contestant on a TV game show.
The host has three boxes, A,B, and C.
One of the boxes contains keys to the star prize, a car and the other two are empty
The host asks you to select one box out of the three
You select box C
He then removes one box he's holding, box A, which he knows is empty and then asks you from the two boxes remaining
do you want to keep your box C or swap for the box B.
What do you do?
- based on a real story -
steve
Keep C.
Swap for box B.
The odds are better for changing.
3rd
Yes, the odds are slightly in your favor to change. They covered this on NUMB3RS just a few weeks ago.
Yep, its called the monty hall problem.
By swapping you increase your odd of winning.
heres how:
Box A = 1/3 or a chance
Box B = 1/3 of a chance
Box C = 1/3 of a chance
By selecting box C you hold 1/3 of a chance of winning and the host holds 2/3 chances of winning. When he removes box A that he knows is empty, his remining box B still has a 2/3 chance of winning.
Box B = 2/3 of a chance
Box A = 1/3 of a chance
steve
Actually, once A is removed your odds of winning become 50/50. It is a showman trick.
You have already selected one third of the original options.
When one third is removed, it would seem your odds have increased to 2/3 if you switch. But what has actually happened? One box that was empty (could not ever have been a winner) has been removed. So, it is revealed that your odds of winning are now 50/50.
Of the two remaining boxes (which are the only two possible winners there ever were anyway), either one could be the winner. 50/50
If someone disagrees, please show mathematically how your odds of winning increase from switching. Keep in mind that A is as if it never was as soon as it is removed from the possibilities.
Respectfully,
OldSoul
Lets suppose the prize happens to be in box C. No matter what happens with the maths or the odds, if you stick with C you will win, and if you change to B you lose. I agree with oldsoul that if you have no information about which one of 2 boxes holds the prize, the odds are 50/50.
If A is removed, and A could never have been the winner, what is your chance of selecting the winner?
I look at it this way.
Chances are against me picking the right one with A B C.
So, when I do pick, odds are I picked wrong.
Faced with another choice (keeping in mind I probably picked wrong) it is to my advantage to get ANOTHER chance to pick right.
So, I'll ALWAYS swap what I've got.
My odds go up when I do.
T.
If one option was not removed, would you pick again when given the option? If not, why not?
The odds for each option would be equal to the odds of each other option. A=1/3, B=1/3, C=1/3, that you picked right. A=2/3, B=2/3, C=2/3, that you picked wrong.
If one is removed some say they would pick again. Why? With "A" removed, what changed about the odds that would indicate picking again?
The odds for each option would be equal to the odds of each other option. B=1/2, C=1/2, that you picked right. B=1/2, C=1/2, that you picked wrong.
As soon as A is eliminated as a possibility, it ceases to exist in the equation. You are no longer dealing with three parts of a whole. You are now dealing with two parts of a whole. By any mathematical standard I know of that becomes a 50/50 chance.
At every stage each available option has odds of being correct or incorrect that are equal among the available options. At every stage in the logic problem, the odds of each option being the winner or loser is equivalent to the odds of every other available option being the winner or loser.
Respectfully,
OldSoul
