This Quantum World/Appendix/Probability/Problems

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Some Problems

Problem 1 (Monty Hall). A player in a game show is given the choice of three doors. Behind one door is the Grand Prize (say, a car); behind the other two doors are booby prizes (say, goats). The player picks a door, and the show host peeks behind the doors and opens one of the remaining doors. There is a booby prize behind the door he opened. The host then offers the player either to stay with the door that was chosen at the beginning, or to switch to the other closed door. What gives the player the better chance of winning: to switch doors or to stay with the original choice? Or are the chances equal?

Problem 2. Imagine you toss a coin successively and wait till the first time the pattern HTT appears. For example, if the sequence of tosses was

H H T H H T H H T T H H T T T H T H

then the pattern HTT would appear after the 10th toss. Let A(HTT) be the average number of tosses until HTT occurs, and let A(HTH) be the average number of tosses until HTH occurs. Which of the following is true?

(a) A( HTH) < ( HTT), (b) A(HTH) = A(HTT), or (c) A(HTH) > A(HTT).

Problem 3. Imagine a test for a certain disease (say, HIV) that is 99% accurate. And suppose a person picked at random tests positive. What is the probability that the person actually has the disease?

Solutions

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

${\displaystyle p(O3)=p(O3|C1)\,p(C1)+p(O3|C2)\,p(C2)+p(O3|C3)\,p(C3)}$

as well as

${\displaystyle p(O3|C2)\,p(C2)=p(C2|O3)\,p(O3).}$

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

${\displaystyle p(O3)={\frac {1}{2}}\times {\frac {1}{3}}+1\times {\frac {1}{3}}+0\times {\frac {1}{3}}={\frac {1}{2}}}$

and thus

${\displaystyle p(C2|O3)={\frac {p(O3|C2)\,p(C2)}{p(O3)}}={\frac {1\times {\frac {1}{3}}}{\frac {1}{2}}}={\frac {2}{3}}.}$

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 ${\displaystyle 2/3,}$ whereas the probability that it is behind door 1 is 1 – 2/3 = 1/3. A quicker way to see that switching 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).

Note: This result depends on the show host *deliberately* opening only a door with a goat behind it. If she doesn't know - or doesn't care (!) - which door the car is behind, and opens a remaining door at random, then 1/3 of the outcomes that were initially possible have been removed by her having opened a door with a goat. In this case the player gains no advantage (or disadvantage) by switching. So the answer depends on the rules of the game, not just the sequence of events. Of course the player may not know what the 'rules' are in this respect, in which case he should still switch doors because there can be no disadvantage in doing so.

Problem 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 a total of 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 a total of 5 tosses.

Problem 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%!

Moral

Be it scientific data or evidence in court — there are usually competing explanations, and usually each explanation has a likely bit and an unlikely bit. For example, having the disease is unlikely, but the test is likely to be correct; not having the disease is likely, but a false test result is unlikely. You can see the importance of accurate assessments of the likelihood of competing explanations, and if you have tried the problems, you have seen that we aren't very good at such assessments.