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I was reading about pigeon-hole principle from this paper.

If n items are put into m pigeonholes with n > m ( m,n ∈ N ∗ ), then at least one pigeonhole must contain more than one item.

One of the applications of the principle is stated as:

Pitter is the boss of a lotto games company, the lottery is a number which contains 5 digits. Every month, the machine picks up 1 number randomly and the owner of the lottery ticket with the same number will win one million dollars. In order to demonstrate the justice of the lottery, there must be at least one winner every month.

Based on the former condition, calculate the minimum number of customers should this ticket be sold to.

which has the solution as:

Since every digit of the number varies from 0 to 9, there are 10⁵ numbers in total. As a result of it, according to the Pigeonhole Principle, the number of customers should be at least 10⁵ + 1 = 100001.

So, as far as I understand the question and try to relate it with the basics of the principle:

  1. The possible lottery tickets are holes here.
  2. Customers are pigeon here.
  3. We need to find the minimum pigeons (customers) required such that one hole (ticket) is occupied by more than one pigeon.

This evaluates to: $100001$ customer required such that two of them share the same ticket. But I cannot understand the solution as it tries to evaluate minimum number of customer required to make the game "justified" or "such that there is at least one winner".

  1. The machine can pickup any random number (of $5$ digits) and if a lottery number can be assigned only to a single customer, then $10^5$ is least number of customers required such that at least one wins. And in this case, there are no ticket left so $10^5+1$ ($1$ extra customer) is out of question.
  2. If more than one customer is allowed to pick up same tickets, then $10^5+1$ customers can also pickup the same ticket (obviously, very low probability) and there could be no winner.

I'm sure I'm missing something here.

Jean-Claude Arbaut
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Paras Lehana
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    I agree, the stated example is horribly written and seems incorrect for the reasons you stated. If I were you, I would ignore that example and look elsewhere on this site for other pigeon hole principle examples – JMoravitz Aug 20 '19 at 16:36
  • For some great applications of the pigeon hole principle, I recommend this list (though you should note that not all are particularly easy to prove). – JMoravitz Aug 20 '19 at 16:39
  • That's very confusing language. Is that the exact language from the article? It seems odd, because if there are multiple tickets with each number, then there is no way to ensure that somebody wins - they could all get number 12345, and nobody wins. And if there are 10,000 tickets, one per possible result, then you only need to sell 10,000 tickets to ensure somebody wins. – Thomas Andrews Aug 20 '19 at 16:44
  • do you know the variant of pigeon-hole principle? If you have more bullets than pigeons and you shoot and never miss...then exists one pigeon with at least 2 holes ;) – L F Aug 20 '19 at 22:23
  • @JMoravitz I had also moved on. Just doing my bit - didn't want to be icy over this. There's no related questions on Lotto and Pigeon Hole as well. Thanks for the list. :) – Paras Lehana Aug 21 '19 at 11:50
  • @LuisFelipe Great one. So, firing random bullets doesn't guarantee that a particular pigeon would get at least one hole. I mean all bullets can hole one pigeon. Hope you can relate. ;) – Paras Lehana Aug 21 '19 at 11:50
  • Yeah, that is the idea. The same for a set of 8 students and the birthday day (Monday, Sunday, ...) – L F Aug 21 '19 at 11:53
  • @ThomasAndrews Yes, the quoted text is taken from the paper as it is (link provided, page 8). If you want I can save it to my drive and then share here.

    Actually your numbers make sense and that's what I tried to describe. There are not much stuff about Lotto + Pigeon Hole either.

    I think, that's because, only distributions makes sense (no. of people having same ticket) and not probabilities (no. of people such that one wins) as far as Pigeon Hole is considered in Lotto Games.

    I'll drop the answer after waiting for a while.

     – Paras Lehana Aug 21 '19 at 11:54
  • I am reminded of a problem I was given in undergrad one time. A grocery store hosts a giveaway contest. Each time someone visits the store they are given a random whole number from $1$ to $100$, independently and uniformly selected. If their number matches someone who arrived earlier, then they win the prize and the game ends. (You can notice that there will be a guaranteed winner by the 101st person and the 1st person cannot win by pigeon hole) .... (continued) – JMoravitz Aug 21 '19 at 12:06
  • ... Now, the actual question is... Mr. A loves to win and so camps outside the store before it opens and can time when he enters the store in order as he pleases. What spot in line does he take in order to maximize his chances of winning? – JMoravitz Aug 21 '19 at 12:08
  • What I think is Pigeon Hole Principle doesn't actually maximize the chance of winning but actually guarantees winning. The example is Birthday Paradox, where 23 people gives 50% chances of same birthday while 367 will give 100% chance of same birthday (this is by pigeon hole).

    In your case, Mr. A can guarantee his winning (100% probability) by taking the 101 spot in the line, no?

     – Paras Lehana Aug 21 '19 at 12:37
  • No., and the connection to pigeonhole ends at the fact that there is a guaranteed winner by the 101st person. That is where the analogy ends unfortunately. It is only the first person to have matched an earlier person who wins. At that point, the game ends and no more numbers are given out. If he plans to take the 101st position, he would win only in the case that the previous 100 people all had different numbers, which occurs with probability $\frac{100!}{100^{100}}\approx 10^{-42}$. result – JMoravitz Aug 21 '19 at 12:53
  • Ah, okay. I didn't know that a number can be selected twice. Actually, it's a great example that is somewhat related to the question. If in my case, you can select the number twice, there could never be a justified game (100% probability that someone wins). And if you cannot, then there cannot be one extra customer than the number of tickets. – Paras Lehana Aug 21 '19 at 13:24

1 Answers1

2

Just so this question has an answer:

  • Yes, you are correct saying that if $10^5$ different numbers tickets are bought then the probability of somebody winning is $100\%$

  • Yes, you are correct saying that if $10^5+1$ or more tickets are bought then the pigeonhole principle means that at least two tickets have the same numbers, since there are only $10^5$ possible numbers

  • Yes, you are correct saying that if different tickets can have the same numbers and the purchasers choose the numbers or they are generated at random then it is possible that some numbers may not appear on any tickets and so the probability of there being at least one winning ticket may be less than $100\%$ no matter how many tickets are sold

  • There might be a different answer to this last point if Pitter (the ticket seller) decides which numbers appear on each ticket, but that could take you back to the first point of $10^5$ rather than $10^5+1$ as the minimum number of tickets needed to be sold to be able to ensure a $100\%$ probability of a winner

  • If $n$ tickets are sold, and the winning number is picked uniformly at random from the $10^5$ possibilities independently of the numbers on the sold tickets, then the expected number of winners is $\frac{n}{10^5}$; this reaches $1$ when $n=10^5$ but the expectation of at least one winner is not equivalent to there being a $100\%$ probability of at least one winner

  • If $n$ tickets are sold, with each sold ticket and the winning pick chosen independently uniformly at random from the $10^5$ possibilities, then the probability of at least one winner is $1-\left(1-\frac1{10^5}\right)^n \approx 1-e^{-n/10^5}$, which is about $50\%$ for $n=69315$, about $63.2\%$ for $n=10^5$, and about $99.995\%$ for $n=10^6$

Henry
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