Car Sharing Exercise - Probability

[tmtc1989]

Let us calculate liquidity on a simplified analytical model. In this model, a day is divided into 24 slots, each of one hour in length. Then, if a driver posts on the car share platform that they are driving from point A to point B in a given slot, and there is a passenger that wants to go from point A to point B in that same slot, a match occurs, and the ride happens. If a driver proposes a given slot and no passengers appear or, the other way around, if a passenger does not find a driver in the slot he/she is interested in, then they leave the platform, and a match does not occur. To simplify things, in this “first cut” model, each driver can only take one passenger. For example, if there are two drivers in one slot and only one passenger, or two passengers in a slot and only one driver, only one ride occurs. For the questions below, assume that drivers and passengers choose slots with the same probability - all slots are equally likely to be selected (they are selected uniformly at random). Please answer the following questions:

1. If a single passenger arrives on the platform and is looking for a ride on a specific slot, what is the minimum number of drivers that should be in the system so that he/she finds a ride with probability at least 80%.

2. Note that the scenario above is great for the passenger, but terrible for the drivers. A well-managed platform that has a large number of matches depends not only on the number of drivers and passengers, but in the balance between them. Thus, with the same assumptions as before, what is the minimum number of drivers and passengers needs such that at least 80% of passengers and drivers find a match?

 

[JeffM]

Great question. Here is one strategy to use.

Suppose there are only two slots. what is the probability that a specific slot is not chosen when there are n drivers?

Suppose there are three slots. What is the probability that a specific slot is not chosen when there are n drivers.

Now can you generalize to m + 1 slots and n drivers?

 

[tmtc1989]

I didn't get your explanation

I know that for the first case, to reach 80% probability for 1 passenger, I'd need 20 drivers given 24 hour slots

=20/24 = 0,8333...

 

[nasi112]

I could solve this problem easily. My problem is that I am so lazy to read lengthy questions. I got sleepy when I completed the second line, so I stopped.

 

[JeffM]

I didn't get your explanation

I know that for the first case, to reach 80% probability for 1 passenger, I'd need 20 drivers given 24 hour slots

=20/24 = 0,8333...

How do you know that?

It is not the answer I get.

 

[tmtc1989]

How did you solve the first one then?

I was thinking that once a passenger arrives, whenever is his first time slot, he would need at least 1 driver per slot time
Given that, I'd say that to reach a probability of 80% he would need 20 drivers spread into those 24 hour slots.

 

[jonah2.0]

Beer soaked ramblings follow.

I could solve this problem easily. My problem is that I am so lazy to read lengthy questions. I got sleepy when I completed the second line, so I stopped.

That is beautiful.
It's not supposed to be funny but it made me laugh so hard that I lost control of my fingertips for a few minutes. It reminded me a lot of these two guys I saw on tv years ago who were having fun with their unicycles for a few hours and then suddenly stopped; saying afterwards: What are we doing? Definitely a life changing epiphany.

 

[JeffM]

How did you solve the first one then?

I was thinking that once a passenger arrives, whenever is his first time slot, he would need at least 1 driver per slot time
Given that, I'd say that to reach a probability of 80% he would need 20 drivers spread into those 24 hour slots.

My reading of the first question is different.The single passenger is looking for a specific slot, not a sequence of slots. The passenger’s “first time slot” is a misnomer for “only time slot. Moreover, you seem to be assuming that no driver selects a slot selected by another driver until all twenty-four slots have been chosen by “at least one” driver. But the drivers are selecting at random according to a uniform distribution. They are not selecting based on “filled” or ”not filled.” There is no reason to believe that having 20 drivers means that 20 slots are filled.

I can think of at least two ways to make the answer intuitive. One is to think about smaller cases. Let’s take three slots, A, B, and C and assume that the passenger chooses A.

What is the probability that a driver #1 picks A Obviously 1/3. Way less than 80%. If driver # 1 picks B or C, that is a probability of 2/3. What is the probability that driver #2 picks A. Obviously, 1/3. So we have four possibilities. Both pick A. Probability 1/9. Driver 1 chooses A while driver B selects B or C. Probability 2/9. Driver 1 chooses B or C and driver 2 selects A. Probability 2/9. Neither pick A. Probability 4/9. So the probability that at least one of two drivers will pick A is 5/9, which is far less than 80%.

Notice 5/9 = 1 - 4/9

Now if we do this assuming 3 drivers, we get

all three pick A 1/27
two pick A 6/27
one picks A 12/27
none picks A = 8/27

So the probability that at least one picks A is 19/27, 1 - 8/27. That is approximately 70%.

How about 4 drivers.

Work it out, but the probability of at least one selecting A is 1 - 16/81, which is a tad above 80%.

Now how might you attack the actual question?