Monty Hall problem - visualize it as a Minesweeper game
A few months ago, I came across an article about the Monty Hall problem.
In the problem, a prize is hidden behind one of three doors (as in the Let's Make a Deal game show).
The contestant picks a door, and then the game show host uncovers one of the other doors - to reveal that the prize is NOT hidden behind it. The contestant can choose to stick with their first choice or pick the other door.
If they pick the other door, it is claimed, their chance of winning is higher - seemingly against reason, which suggests that the odds of the prize being behind each remaining door are equal.
I had trouble imagining why this was the case - until I visualized the problem using the Minesweeper game in Windows, and with 100 doors (or squares) instead of just three:
Imagine that you have a 10 by 10 minefield with 100 squares.
You're told that there's only one mine under all the squares.
You mark one square as the one you guess it's under. Your chances of getting it right are 1 in 100.
Next, you click an imaginary button that automatically uncovers all squares but two - the one you marked, and the one the mine is under.
Pretend, for instance, that the square you picked is near the top left corner, and the only other uncovered square is near the bottom right corner.
If you were given a chance to guess again, almost everyone I know would pick the other uncovered square near the bottom right corner - and they'd have a very good chance of being right.
Calculating the odds - why you have a 2/3 chance if you switch doors
Using Minesweeper as an analogy helps to make the problem more intuitive.
However, it doesn't explain the odds. To explain why the odds go from 1/3 at the start of the game to 2/3 (after Monty removes a door AND the contestant switches doors), it's easiest to show by outlining all the scenarios - and their possible outcomes:
Digg this article
|
Add to del.icio.us
In the problem, a prize is hidden behind one of three doors (as in the Let's Make a Deal game show).
The contestant picks a door, and then the game show host uncovers one of the other doors - to reveal that the prize is NOT hidden behind it. The contestant can choose to stick with their first choice or pick the other door.
If they pick the other door, it is claimed, their chance of winning is higher - seemingly against reason, which suggests that the odds of the prize being behind each remaining door are equal.
I had trouble imagining why this was the case - until I visualized the problem using the Minesweeper game in Windows, and with 100 doors (or squares) instead of just three:
Imagine that you have a 10 by 10 minefield with 100 squares.
You're told that there's only one mine under all the squares.
You mark one square as the one you guess it's under. Your chances of getting it right are 1 in 100.
Next, you click an imaginary button that automatically uncovers all squares but two - the one you marked, and the one the mine is under.
Pretend, for instance, that the square you picked is near the top left corner, and the only other uncovered square is near the bottom right corner.
If you were given a chance to guess again, almost everyone I know would pick the other uncovered square near the bottom right corner - and they'd have a very good chance of being right.
Calculating the odds - why you have a 2/3 chance if you switch doors
Using Minesweeper as an analogy helps to make the problem more intuitive.
However, it doesn't explain the odds. To explain why the odds go from 1/3 at the start of the game to 2/3 (after Monty removes a door AND the contestant switches doors), it's easiest to show by outlining all the scenarios - and their possible outcomes:
Digg this article
|
Add to del.icio.us
4 Comments:
You might want to remove the second circle on the seventh scenario. ;)
Very good visuals!
Doh - thanks...I've fixed the image.
why there are only 9 scenarios in the third column ?
In the first line of column 2 (first case) Monty can remove the third door too...and the contestant loose.
In the second case Monty can remove the third door...and the contestant loose.Same apply for the last case.
So 12 scenarios total, 6 win.
odds:1/2
Am I wrong?
Yes, you are wrong. Monty hall is a good man and never takes away the grand prize. he only tells you which was a bad door.
Post a Comment
Links to this post:
Create a Link
<< Home