Taxi driver problem

There’s a guy who travels by taxi every day from his work to his home (20km). One day he finds out that a colleague of his also takes taxi every day from work, the same direction, but he lives only half a way of the first guy (10km). So, they decide to share a taxi and also to share the costs. How should they share the costs?

First answer could go like this: First guy travels twice longer so he should pay twice as much as the second guy. So, the first guy should pay 2/3 of the total cost and the second guy only 1/3.

Second answer could go like this: Both of them travel the first half together, so they should split this part in half. The first guy travels alone the second half of the way, so he should pay the full amount of the second half. In total, first guy should pay 3/4 and the second one only 1/4.

Which one is fairer?

Let’s assume that first guy pays double for his trip as compared to the second one, when they travel separately – in two taxis. This is usually not correct, because of the initial charges, but let’s assume that the charge is proportional to the distance. Let’s assume that it’s fair if both of them get the same “discount” (percentage wise) when joining together. The first one would have to pay 2p(1-k) and the second one p(1-k), where k is the discount, p is what the second guy usually pays and 2p is what the first guy usually pays. So the first one should still pay twice as much as the second one, which leads to the first answer.

Let’s assume now that it’s fair if they all get the same discount (absolute cost reduction). So, the first one would pay 2p-q and the second one p-q, where q is the discount amount. Then (2p-q)+(p-q) = 2p => p = 2q => q = p/2, which leads to the second answer.

Let’s say that one day the first guy finds out that his neighbour also takes taxi every day from work home and accidentally, he works very close to where the second guy lives, just one block away, so he comes to a cunning idea. He will share the second part of his trip with his neighbour and he will reduce his costs even more. Obviously, he would prefer for the first formula to be applied, because in that case his colleague will pay 1/3, his neighbour also 1/3, which lives him with 1/3 as well. The problem now is that his colleague must not know about his neighbour and vice versa, because in that case they would ask for splitting the costs proportional to the distance, which would be 1/4 for the colleague, 1/4 for the neighbour and 1/2 for the first guy. Obviously, the first formula is not transparent and it allows for speculation. Second formula would give the totally transparent solution in which case the first guy wouldn’t care if his colleague and his neighbour know about each other.

Moral question: In case of applying the first formula in case of 3 guys sharing the taxi, when everybody pays 1/3 of the total price, do you think the first guy would be a cheater or a good businessman? How do you think the other two guys would react when they find out about it? Would they be angry? Would they be like: “Wow, you really tricked us professionally. Well done!”. Or would they just be glad they found it out, so from now they would pay less.

From where I come, he would definitely be a cheater, because the base of his “success” is in hiding the truth from the other two guys. On the other hand, we’re all aware that much of today’s businesses are based on hiding the truth from the others. Sometimes it’s some intellectual property information, but very often it’s just a cheap fact. If a famous chef hides the details of his recipe, that’s kind of fine. But if a job agent hides the details about how much of your contract he gets from your employer, that’s kind
of completely different story, isn’t it? This brings up more serious question: Is hiding the truth in today’s businesses unethical? Is hiding the truth in politics and diplomacy unethical?

I could write a lot about this and I already went quite far away from initial “logic” problem to an ethical one, that’s probably the reason why I post so rarely, so I’ll stop here

Like Father Like Daughter

We worked hard on the extra second kick, now we need to remove that extra arm stroke

Recursive expectation (part 2)

This post follows the previous recursive expectation post.

Problem:

There is an equilateral triangle ABC and the bug is in the vertex A. The bug can move only along edges of the triangle. Once it starts going along an edge, it doesn’t change its moving direction until it reaches the next vertex. It starts walking from the vertex A. It takes bug 1 minute to walk between any two vertices. Every time it comes to a new vertex, it makes a random choice about which of other two vertices to go to next. What is the expected time for the bug to return to the starting vertex A?

Solution 1:

First, completely intuitive solution, without (much) mathematics. It takes one minute for the bug to come to the edge BC. From that edge, each minute there is 1/2 chance that the bug will return to A. Basically we start the experiment that has 1/2 chance of success each time. So, according to the previous post, expected time to get back to A from BC will be 2 minutes, so all together with the first minute, we get the final result – 3 minutes.

Solution 2:

And now, let’s try to do it pure mathematical way, without much thinking.

Possible solutions for number of minutes (length of trip) to return back to A are 2, 3, 4, … until infinity. If we take one of the possible solutions, let’s say n minutes, how many different paths do we have for this to happen? Basically, we can go from A to B and then go n-2 times between B and C and then the last time, we go back to A, either from B or from C. Another path would be from A to C and then n-2 times between B and C and then the last time, we go back to A. So, for each n, number of possible paths is always 2. Probability for each path of length n is:

So, probability for n-minutes solution to happen is:

Now, the solution for the expectation becomes:

which requires some mathematical knowledge, but doesn’t require much of the intuition.

Solution 3:

This one will actually use recursive expectation (please see the previous post). Expected time from A back to A is equal to expected time from A to edge BC + expected time from edge BC back to A. Expected time from A to BC is 1. Expected time from BC to A (following the logic from the previous post) is:

So, the overall expectation is:

Recursive expectation once again, provides much easier solution than the mathematical one. Additionally, it’s much more intuitive.

Monster fly

Now I understand my daughter when she asks me: “Daddy, why is your face looking like a monster while swimming?”

Rolling dice expectation

The day after my previous expectation related post, I kind of realised what lies behind intuitive answer to the question:

“What is the expected number of times we need to roll a dice to get ’6′ for the first time?”

Answer 1:

The question above is equivalent to the following. Let’s roll the dice large number of times. We will get long sequence, like:

3, 4, 6, 5, 6, 1, 2, 4, 3, 5, 1, 4, 5, 6, 6, …

Next, let’s cut the tail off after the last ’6′. Let’s mark the length of the remaining sequence with N. Now we need to count “distances” between every two consecutive ’6′s. In our example above, we will get 3, 2, 9, 1, … Distance is defined as the number of rolls between two consecutive ’6′s. Since ’6′s are at positions: 3, 5, 14, 15, … in the sequence above, distances are 3, 5-3=2, 14-5=9, 15-14=1, … The answer to the question, about the expected number of times, is the same as the average distance between every two consecutive ’6′s. Let’s mark distances with the sequence: . is the total number of distances. Basically, will correspond to the number of experiments performed. The cut off tail is unfinished last experiment, which we decided to ignore. If X is the number of tries to get ’6′ for the first time, we can write:

Furthermore, sum of all distances must be equal to the total number of tries – N.

Since is the number of distances, it is also the number of ’6′s in the whole sequence. Since there is no reason for any number on the dice to appear (by expectation) different number of times than any other number (on the dice), we expect for to be:

 If we replace last two equations in the formula for E(X), we get:

and no mathematics knowledge was needed for this proof. So, maybe this is what lies behind the instinctive answer.

Answer 2:

If we write down the results of each experiment (experiment is said to be finished once we get ’6′ for the first time):

Since there are experiments, we have exactly 6′s, but we also expect to have each other dice number also times, so the sum of all lengths of all experiments is expected to be , which makes average length to be .