Quote:
Originally posted by Equinox
flipping a coin... of course it's .5 for either side, because you do have two sides which equates to 1/2. In the dice case, P(1)=P(2)=1P(3)=P(4)=P(5)=
P(6) = 1/6.
I believe the empirical evidence that P(win|switch)= 2/3 - however let me try to work this out by myself.
P(win at first try) = 1/3
P(miss at first try) = 2/3 - evenly distributed.
You do not switch - P(win) remains at 1/3.
If you missed with P(miss)=2/3, the prize is behind one of the remaining doors. If you miss and switch, your probability to win equates
1 - P(picked winner at first try) = 1 - 1/3 = 2/3.
So if you don't switch, your P(win) equates P(win picked at first try) which is constant at 1/3.
That sounds reasonable to me. 
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does your 66% chance win 100% of the times ? if so you cant lose if not you have 50% chance of the 66% winning