There are three doors. one doors have money and two has nothing. I pick door two. Door one opens and it empty. If i switch to door three what % chance do I have getting the money door.
Anyone know?
if you switch you have a 1/2 chance of prize if you do not switch you have a 1/3 chance of prize
Incorrect
why incorrect?
They were not the right percents. P.s. its not because they were in fraction form.
would you accept 50% and 33.3%
They are incorrect.
ok well i still dont see why , do you have evidence to suggest otherwise?
2/3?
lol Monty Hall principle...just saying
Jazy is correct.
but that is only a single fraction
Three scenarios: - You choose one of the two 'nothing.' The other door reveals nothing and you switch to the money. - You choose the other 'nothing' and the first 'nothing' is revealed. You switch to the money. - You choose the money and nothing is revealed, and switch to nothing, losing. Of course, you win 2/3 of the time in this set up.
If door 1 was opened intentionally BECAUSE it had nothing behind it, you increase your odds to 2/3 if you switch. If it was just opened, you actually still have 1/2 chance with both doors available to you.
Its the Monty Hall principle :)
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