Answers

1. Let p(C1) be the probability that the car is behind door 1, p(O3) the probability that the host opens door 3, and p(O3|C1) the probability that the host opens door 3 given that the car is behind door 1. We have

p(O3) = p(O3|C1) p(C1) + p(O3|C2) p(C2) + p(O3|C3) p(C3)

as well as

p(O3|C2) p(C2) = p(C2|O3) p(O3).

If the first choice is door 1, then p(O3|C1)=1/2, p(O3|C2)=1, and p(O3|C3)=0. Hence

p(O3) = (1/2) x (1/3) + 1 x (1/3) + 0 x (1/3) = 1/2

and thus

p(C2|O3) = [p(O3|C2) p(C2)]/p(O3) = [1 x (1/3)]/(1/2) = 2/3.

In words: If the player's first choice is door 1 and the host opens door 3, then the probability that the car is behind door 2 is 2/3, whereas the probability that it is behind door 1 is 1−2/3=1/3.

A quicker way to see that switching doors doubles the chances of winning is to compare this game with another one, in which the show host offers the choice of either opening the originally chosen door or opening both other doors (and winning regardless of which, if any, has the car).

2. The average number of tosses until HTT occurs, A(HTT), equals 8, whereas A(HTH)=10. To see why the latter is greater, imagine you have tossed HT. If you are looking for HTH and the next toss gives you HTT, then your next chance to see HTH is after 6 tosses, whereas if you are looking for HTT and the next toss gives you HTH, then your next chance to see HTT is after 5 tosses.

3. The answer depends on how rare the disease is. Suppose that one in 10,000 has it. This means 100 in a million. If a million are tested, there will be 99 true positives and one false negative. 99% of the remaining 999,900 — that is, 989,901 — will yield true negatives and 1% — that is, 9,999 — will yield false positives. The probability that a randomly picked person testing positive actually has the disease is the number of true positives divided by the number of positives, which in this particular example is 99/(9999+99) = 0.0098 — less than 1%!