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Problem: A high school locker room owner has a locker room with 1000 spaces. Each space costs Rs. 100 a day. He has already sold monthly permits to 1001 high school members (knowing that it is likely that not all the high school members would want to keep their bags there at the same time).

If someone with a permit arrives to keep their bag and there are no spaces, the owner will refund Rs. 200 for that day (penalty of Rs.100).

Parking on any given day of the month is independent of every other day.

Has he made the smart decision by selling an extra space?

What I am stuck at is the issue that there are no probabilities given in this question. As we are talking about high school members that means everyone will have lectures on different time of the day (6 working days-from monday to saturday). I think this question will not be requiring any numerical values to be solved. As expected profit is the part of this problem so E[random variable (RV] will be useful in solving this problem. Now If I am not wrong I think binomial RV will also play a major role in getting a general solution for this problem (because if we are interested in that 1 extra member then we will be interested in only two outcomes. 1. if that member will come 2. if that member will not come). That's all I know. I have no idea from where to start. I have no idea how to use CDF(Cumulative distribution function) to my advantage or how to even start this problem and get to the desired results. I am trying to find solution in terms of concepts related to RV not simply probability problems. I would appreciate if anyone can share their thoughts on this problem.

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  • $\begingroup$ Agreed, there's clearly not enough information to solve. However, maybe you can assume a binomial distribution with $p = 0.99$ and show that there will very likely be open lockers? $\endgroup$
    – Kaynex
    May 14, 2018 at 15:40
  • $\begingroup$ well the only thing that is stopping me from assuming anything is that its somewhat a realistic model and in real model assuming probabilities is not a good choice to make. I am trying to take it as real as possible and trying to get a general solution that can meet my needs related to this problem. -thanks for your time that you put on writing this comment. $\endgroup$
    – probablisticmodel
    May 14, 2018 at 15:46
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    $\begingroup$ Too bad, you're not going to get that. There's nowhere near enough info to make a suitable model. For example, do people show up 100% of the time? Then he's made a very bad decision. $\endgroup$
    – Kaynex
    May 14, 2018 at 15:51
  • $\begingroup$ that's where I am stuck at. if the owner already knew that 100% of the people he gave permits to will show up than he made a bad decision but why would any owner do that. $\endgroup$
    – probablisticmodel
    May 14, 2018 at 15:53
  • $\begingroup$ Airlines routinely do this. Since they have, at times, incredibly low profit margins, I would expect them to have done quite a lot of research into how to predict the number of people who will actually turn up. $\endgroup$
    – Arthur
    May 14, 2018 at 16:32

4 Answers 4

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As true blue anil says you need a lot more information to make a model, then you can use the model to decide how many lockers to rent. It sounds to me like he rents the lockers on a monthly basis, or on some term longer than a day. Each day, some of the renters show up and use lockers. The simplest assumption is that each person shows up independently with probability $p$. If he has rented $1001$ so far and recorded how many people come each day he has a measure of $p$. We have enough people that the normal approximation is reasonable, so if he rents to $n$ people he would expect a mean of $np$ to show up with a standard deviation of $\sqrt{np(1-p)}$. If $k$ people show up his revenue for the day is $$r(n,k)=\begin {cases} 100n&k \le 1000\\100n-200(k-1000)& k\gt 1000 \end {cases}$$ We want to maximize the expected value of the revenue $E(r(n,k))$ by choosing $n$ given the probability distribution of $k$. We can convert the sum over $k$ to an integral in the continuous approximation and we get $$E(r(n,k))=100n-200\int_{1000}^\infty(k-1000)\frac 1{\sqrt {2\pi np(1-p)}}e^{-\frac{(k-np)^2}{2np(1-p)}}dk$$ Evaluate the integral, take the derivative with respect to $n$, set to $0 \ldots$

As a mathematical approach this is defensible if the model is right. Two problems are that there could be special events where $p$ gets much higher. Say there is a big soccer match in town. Maybe all you customers come in for the match and your refund loss is much higher. It could be that somebody watches the clients arrive and depart and discovers that most people show up by $9$ AM and nobody leaves before $11$. He signs up $50$ friends who rent lockers, then all show up between $9$ and $11$ hoping to collect refunds.

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  • $\begingroup$ how you computed E(r(n,k)). everything else is explained pretty good. just this E(r(n,k)) part is hard for me to understand. because I haven't worked with expected value calculation in such detail but from the structure of E(r(n,k)) I can see that it will be pretty good for the stated problem. -thanks for your time $\endgroup$
    – probablisticmodel
    May 15, 2018 at 13:08
  • $\begingroup$ It is just the usual expected value calculation, integrating the possible values times the probability of that value. The probability factor comes from the normal distribution. $\endgroup$
    – Ross Millikan
    May 15, 2018 at 13:41
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You want a model which means that I need to make a bunch of assumptions, given below.

  • The owner can charge for the space to a person only if the person uses it on a particular day, so nothing can be done about "no shows".

  • Each person has a uniform probability $p$ of needing a space on any day.

With the above assumptions, the "break even point" will occur if
P(all 1001 spaces are needed on a day) $ = 0.5$

For this, $p^{1001} = 0.5$ which yields $p = 0.999308....$

This is an extremely high value for $p$ which is most unlikely,
hence the owner has indeed made a smart decision !

Added

If we assume $p=\frac{1000}{1001}$, which is still highly iunrealistic,

P(all 1001 spaces are needed) $\approx= 0.3677$, so the owner will lose , say $37$% of the time, and gain equivalent amounts $63$% of the time.

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  • $\begingroup$ everything you mentioned is clear but I am trying to approach this problem using binomial/Poisson RV. p.s. for simple probabilistic techniques your answer is very decent and easily understandable. -thanks for your time that you put on writing this. $\endgroup$
    – probablisticmodel
    May 14, 2018 at 19:22
  • $\begingroup$ Indeed, I have used the binomial distribution. The owner loses if $X=1001$ for which $Pr = \binom{1001}{1001}p^{1001}(1-p)^0 = p^{1001}$ $\endgroup$
    – true blue anil
    May 14, 2018 at 19:30
  • $\begingroup$ this added part, don;t you think its too much unrealistic because p=0.9990. that means like probability of needing a space on any day is 0.9990. its shows a pure loss to the owner. isn't it ? $\endgroup$
    – probablisticmodel
    May 14, 2018 at 19:30
  • $\begingroup$ oh the more I am talking about this problem the more things are getting clearer and clearer. everything else is clear. the only thing that is confusing me is that break even point. if you don't mind can you explain it. I will really appreciate that. I am sorry if I am being dense. $\endgroup$
    – probablisticmodel
    May 14, 2018 at 19:33
  • $\begingroup$ If the individual probability $p= 0.999,$ all spaces will be needed only 37% of the time; and if $p = 0.999308$, all spaces will be needed 50% of the time, so lose 100 dollars 50% of the time and gain 100 dollars 50% of the time $\endgroup$
    – true blue anil
    May 14, 2018 at 19:40
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There is insufficient information here, but one can make straightforward assumptions and answer the question.

One assumes the distribution of users follows a Poisson distribution with a given mean $\mu$. Suppose you assume that $\mu = 1000$. That means some days there will be fewer than $1000$, some days more.

To get the expected return, you integrate (actually, it is a discrete distribution, so you merely sum) the values in your distribution (times 100 Rs) up to $x = 1000$ to get the expected income, and then sum your distribution from $1001$ (times 200 Rs) to infinity to find the expected loss. Subtract these two numbers to get the expected overall profit.

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  • $\begingroup$ still scratching my head just to get a hold of this problem but I still cannot make/get a possible and convincing solution. $\endgroup$
    – probablisticmodel
    May 14, 2018 at 18:36
  • $\begingroup$ So what did you find confusing or unconvincing about my solution? $\endgroup$
    – David G. Stork
    May 14, 2018 at 18:43
  • $\begingroup$ the confusing part is Poisson distribution itself. don't we use Poisson RV when we have to deal with some kind of rate like receiving bits/min or calls/minute. I don't understand how are we trying to use Poisson RV here when there is no rate involved. I am sorry if I am asking very dense questions. its just because this questions is taking too much time to be solved. $\endgroup$
    – probablisticmodel
    May 14, 2018 at 18:47
  • $\begingroup$ The Poisson distribution is appropriate for this case. On the first day we might get 873 people. On the second day 952 people. On the third day 1013 people. There is an assumed average rate, and variations in numbers. We never get negative numbers of people. And the number on each day is independent. Exactly what we mean by a Poisson distribution. $\endgroup$
    – David G. Stork
    May 14, 2018 at 18:52
  • $\begingroup$ If we are using poisson RV then the expected value or E[X] of poisson RV is simply λ. how can we compute λ. In normal cases, λ is usually (rate)(time) $\endgroup$
    – probablisticmodel
    May 14, 2018 at 18:59
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To begin with, a month has thirty days. You have called binomial distribution correctly - this is $X=B(30,p)$ that you're studying, wherein $p$ is the probability that all 1001 spaces are used at the same day. Calculate the latter, then the expected value of your variable times 100 is a good measure of whether the owner made a smart decision (I believe he did).

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  • $\begingroup$ The probability that all lockers are used is not at all obvious. $\endgroup$
    – Kaynex
    May 14, 2018 at 15:44
  • $\begingroup$ If you don't mind can you explain a bit more? as far as I know, by using binomial RV I can get the PMF which will be Px(k=x) for different values of k. which I can use in E[X] formula to get my expected profit. Is it correct? -thanks for your time that you put on writing this comment. $\endgroup$
    – probablisticmodel
    May 14, 2018 at 15:50
  • $\begingroup$ yes @kaynex is correct. We are not even sure that all the locker are full. they may be like half full or quarter. Most of the important stuff is not known $\endgroup$
    – probablisticmodel
    May 14, 2018 at 15:52
  • $\begingroup$ Well, since it is not stated otherwise, I assume that each student has a probability of $1/2$ of appearing each day - that means that the probability of all of them appearing at the same day is $\frac{1}{2^{1001}}$, Recall that $E[x]=np$ when the distribution is binomial. And I think each person present gets exactly one locker $\endgroup$
    – George K
    May 14, 2018 at 16:03

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