Kolby
secondary (1012)
Other is asking the question
Maths Problem Question:
Through an informer from the underworld, the police know the meeting place of a gang. The identity of the different gang members, however, is unknown. it is the duty of a policeman to shadow the leader of the gang. The policeman knows that the leader is the tallest of the five persons, all of whom have different heights. after the meeting, the gangsters, as a safety measure, leave the building separately at intervals of 5 minutes. As the policeman cannot see who is the tallest, he decides to let the first two gangsters go and shadow the one after that who is taller than all those who left before. What is the probability that the policeman will shadow the correct person?

Hi Kolby. That's an interesting puzzle.
I think there are only a couple of scenarios in which the police officer will get it right. I'm going to number the suspects by height: 1, 2, 3, 4, 5 with 5 being the leader. Each of these following five scenarios is equally likely (1/5).
 5wxyz: In this situation, the police officer will certainly not trail the right person because she lets the first two people go. So add 0 to our total chances.
 w5xyz: Same as (a) for the same reason. Add 0 to our total chances.
 wx5yz: The police officer gets the leader for sure. Add (1/1) * (1/5) = 1/5 to our total chances.
 wxy5z: The police officer gets the leader only if y is less than w or y is less than x. The y has to be 1, 2, 3, or 4 of course. So there are four choices If y is 1 or 2, the officer gets the leader. Add (2/4)*(1/5) = 1/10 to our total chances. If y is 4, the officer gets the wrong person. Add 0 to our total chances. If y is 3, the officer only gets the right person if 4 was one of the first 2. Since there are three spots left for the 4 to be in, then we add (2/3)*(1/4)*(1/5) = 1/30 to our total chances.
 wxyz5: The police officer only gets the leader if w=4 or x=4. Since there are four spots for the 4 to be in, we add (2/4)*(1/5) = 1/10 to our total chances.
The total probability the police officer chooses the right person is therefore 1/5 + 1/10 + 1/30 + 1/10 = 13/30.
Stephen La Rocque.>
This is a nice problem, and a nice setup for it. In "Innumeracy" by John Allen Paulos, it is a woman who want to find the best possible husband, but she has to accept a proposal without knowing who else would propose afterwards, and she can't ever go back to former lovers who have had their proposal declined. Alternatively, it could be a hunter who wants to kill the largest possible duck, but he is allowed only one shot, and the ducks that fly by never come back. In all cases, it is the same problem: The candidates go by one by one, and you want to pick the biggest, but you have to make a snap decision, and you can never go back to the candidates that have gone by once before.
In your case, there are many possible strategies for the policeman: He could decide to follow the first person that comes out of the building; that way he would have one chance out of five of following the leader. Alternatively, he could let go the first one and follow the first one after that that is taller than all the ones that went before. What are his chances then?
 With probability 1/5, the tallest walks out first, so then the policeman will watch all the others go by and never follow any of them.
 With probability 1/5, the second tallest walks out first, so then when the tallest walks out some time later, the policeman will follow him.
 With probability 1/5, the third tallest walks out first, the policeman will then later follow the second tallest or the tallest, and the latter happens only if the tallest walks out at some time before the second tallest; the probability of that happening is 1/2.
 With probability 1/5, the fourth tallest walks out first, and then the policeman follows the right guy only if he walks out before the second tallest and the third tallest; the probability of that happening is 1/3.
 With probability 1/5, the smallest walks out first, and then the policeman's only chance is that the tallest is the next one to come out, and the probability of that happening is 1/4.
So all in all, the probability that the policeman follows the leader is
(1/5) 0 + (1/5) 1 + (1/5) (1/2) + (1/5) (1/3) + (1/5) (1/4) = 5/12.
This is much better than the 1/5 probability he gets with the first strategy. The probability with the strategy described in your problem is higher still. You can compute it in much the same way I have done above.
This being said, there is more to say about the non computational aspects of the problem: I note that even though the problem is very well stated, original and entertaining, the setup is a bit unrealistic: If we are supposed to admit that the policeman knows the hiding place of the gang, the number of members, the exit procedure and the fact that the leader is the tallest, without having any physical description of the leader or even his height, can't we also decide that the gangsters will be able to solve this math problem and then send the leader out first instead of exiting in random order to outsmart the policeman? Can't the policeman guess that and decide to outsmart the gangsters by following the first one who comes out after all? In some aspects, the duckhunting setup is more realistic, because then the order of the weights of the ducks that fly by is truly random. But then, we cannot be sure that exactly five (or ten?) ducks will fly by; we can only make a guess as to the expected number of ducks that will fly by. Also, if the hunter ends up shooting down a duck that is only two ounces lighter than the heaviest duck, is it really that bad?
In real life, there are lotteries, casinos, lowprobability catastrophes that make the headlines, random processes at work in nature... So besides the computational aspects of probabilistic models, I think it is important that probability courses also show how to recognize the manifestations of randomness in real life.
Claude
