In a game show, there is a prize in one of the 3 boxes, A, B and C. After the contestant chooses one, the host opens one of the 2 other boxes that has nothing. The contestant is given a choice of changing their mind to the other box left or stick to their original choice. Which option will give a higher probability of winning the prize?
The probability of winning a prize out of 3 options with the option of changing your selection?
This called the Monty Hall problem. It has been the source of much debate and issues. The problem becomes very complicated with you consider all the different possible assumptions that have to go into solving the problem.
Check this site for some history on the question and links to the solutions.
http://math.rice.edu/~pcmi/mathlinks/mon鈥?/a>
The probability of winning a prize out of 3 options with the option of changing your selection?
This involves variable chance. When he chose a box say Box A, his chances of winning is 33.33% against 66.66% but when one of the other boxes was opened, his chances for box A became 50%. I would change my mind and select the other box because it's one of the box that has 66.66% winning against the 50% of box A. Smile..
Switch... for pete's sake, switch!
The probability of picking the prize originally is 1/3.
So 1/3 of the time, you pick the prize and switch, you lose.
(1/3 of the time, you pick the prize and stay and win)
The probability of NOT picking the prize is 2/3.
Since the host shows the other empty box, if you switch, you win.
So 2/3 of the time, you pick nothing, switch and win.
(2/3 of the time, you pick nothing, stay and lose)
Combining this: 2/3 of the time, if you switch you win.
1/3 of the time, you switch and lose.
2/3 of the time, you stay and lose
1/3 of the time you stay and win.
The odds favor switching (because the host will always show a box with nothing in it)
This is a great question for causing arguments!
