can someone solve this math riddle for me - Page 5

On July 23 2007 21:27 geometryb wrote:


if you want more experimental results then i guess.

On July 23 2007 17:44 JeeJee wrote:
i hated explaining monty hall problem to others, but i've foudn a really simple solution that everyone can clearly understand

you only lose by switching if you picked the right box to begin with. what are the odds of picking the right box to begin with?

that's usually enough =p

On July 23 2007 21:47 evanthebouncy! wrote:


Hun no lol It's merely there to help ppl visualize things better by exaggerating the situation, who's gonna experiment? We're talking theory here!

On July 23 2007 20:56 [r]h_probe wrote:


Wrong!

Look at it this way:
10 doors
You choose one door ----> 1/10 chance of getting it right no matter what!
Out of the 9 other doors the host reveals 8


So essentially you are either choosing 1 of two groups:
The group of one or the group of 9 (of which 8 have been revealed)

Therefore it is a 90% prob of getting it right if you switch!
There is no "dividing the chance by that added mount"!

To further illustrate my point, lets make the problem even bigger: say you to choose a number between 1 in a million. There is 1 random correct number.

Then someone (who obviously knows what the number is) says it is either your number or it is a different one they tell you. You would have to be absolutely insane to stick with your number because it is a 1/1000000 shot no matter what where as the other number is 999999/1000000