Originally posted by BlauK
dedalos,
ok... show me the formula all the way from the beginning with an accidentally lost training bomb. If you lose a real bomb, you can disregard those cases, because they do not fulfill the stated case.
How will you end up with 50:50 for switching case?
What is your bases for the statement that this does not apply for an accidentally lost bomb (which WAS already lost and therefore happens in 100% of the possible relevant cases!!!)
I don't think I can because of this line:
The probability that Monty Hall opens door B if the prize were behind C,
P(Monty opens B|C) = 1 If the location of the real bomb/prize is not known, then P(Monty opens B|C) = .5 and therefore the result of the the formula will not be 2/3.
The base of this prob problem is the fact the who ever opens the door or drops the bomb, knows where the real one/prize is. If not, then it is just random acts and at the end you have 2 choices to make a random decison.
Lets try it anyway, without the knowlege of where the prize is:
The probability that Monty Hall opens door B if the prize were behind A,
P(Monty opens B|A) = 1/3
The probability that Monty Hall opens door B if the prize were behind B,
P(Monty opens B|B) = 1/3
The probability that Monty Hall opens door B if the prize were behind C,
P(Monty opens B|C) = 1/3
The probability that Month Hall opens door B is then
p(Monty opens B) = p(A)*p(M.o. B|A) + p(B)*p(M.o. B|B) + p(C)*p(M.o. B|C)
= 1/9 + 1/9 + 1/9 = 3/9 = 1/3
Then, by Bayes' Theorem,
P(A|Monty opens B) = p(A)*p(Monty opens B|A)/p(Monty opens B)
= (1/3)/(1/2)
= 2/3
and
P(C|Monty opens B) = p(C)*p(Monty opens B|C)/p(Monty opens B)
= (1/3)/(1/2)
= 2/3
All choices are equal and therefore the selction will be random.