12

In a lottery game, each ticket consists of 4 distinct numbers chosen from the set ${1, 2, \ldots, 15}$. The host randomly selects a secret ticket, and a winning ticket must have at least 2 numbers in common with the host's secret ticket. Players can choose to purchase any specific tickets. For example, if the host’s secret ticket is ${1, 5, 6, 13}$:

  • ${6, 10, 12, 13}$ is a winning ticket because it shares 2 numbers (6 and 13).
  • ${1, 3, 4, 9}$ is not a winning ticket because it only shares 1 number.
  1. Show that buying 75 tickets is sufficient to guarantee a winning ticket.
  2. Can we guarantee with at most 49, 24, 7 tickets respectively?

I have no idea how to start with. Can any show me some hints?

RobPratt
  • 52,643
ohana
  • 1,067
  • 2
    The way to start is to influence MathSE reviewers to react positively to your posted question. For what it's worth, virtually every MathSE posted question that I have seen, that followed this article on MathSE protocol has been upvoted rather than downvoted. I am not necessarily advocating this protocol. Instead, I am merely stating a fact: if you scrupulously follow the linked article, skipping/omitting nothing, you virtually guarantee a positive response. – user2661923 Oct 19 '24 at 06:00
  • Do u know the numbers of the secret ticket ? If so, lets say first 2 numbers are common with winning ticket. So you have $13 \times 12 \times {4 \choose 2} \times 2$ winning ticket numbers. Answer seems to be $13 \times 12 \times {4 \choose 2} \times 2$ possibilities for a winning ticket. –  Oct 19 '24 at 07:27

3 Answers3

25

We can answer the all the questions all at once; there exists a solution with only $7$ tickets.

The idea in this answer is due to trueblueanil. I realized that their answer was inefficient, and could be improved to $8$ tickets. In the course of writing this, I realized that $8$ can be improved to $7$.

Divide the numbers into four groups contiguous groups, as follows: $$ \begin{array}{ll} A = \{1, 2, 3, 4\}, & B = \{5, 6, 7, 8\}\\ C = \{9, 10, 11, 12\}& D = \{13, 14, 15\}. \end{array} $$ The first four chosen tickets will correspond to these four groups. Group D has only three numbers, so the fourth number can be anything arbitrary.

Which winning tickets are not covered by these four chosen groups? A winning ticket is covered as long as it shares at least two numbers with a chosen ticket, so the only uncovered tickets will have exactly one number from group A, exactly one number from group B, exactly one number from group C, and exactly one number from group D. In particular, every uncovered ticket contains a number group A and a number from group D. Therefore, these additional four tickets will cover all of the uncovered tickets: $$ \{1,13,14,15\},\quad \{2,13,14,15\},\quad \{3,13,14,15\},\quad \{4,13,14,15\}\tag{*} $$ To be clear; every uncovered ticket contains a number from group A and from group D. Suppose the uncovered ticket has the number $a$ from group A, and the number $d$ from group $D$. Then that ticket will be covered by $\{a,13,14,15\}$, because this ticket contains $a$, and contains all numbers $d\in D$.

However, there is a further improvement! I initially said that the group D ticket should contain three numbers plus some arbitrary fourth number. So, the group $D$ ticket could be $\{1,13,14,15\}$. But this takes care of one of the tickets in $(*)$, so we this means we only need three additional tickets to get all of $(*)$.

To be clear, here is the final list of seven tickets. $$ \{1, 2, 3, 4\},\quad \{5, 6, 7, 8\},\quad \{9, 10, 11, 12\}, \\\,\\ \{1,13,14,15\},\quad \{2,13,14,15\},\quad \{3,13,14,15\},\quad \{4,13,14,15\}. $$

Mike Earnest
  • 86,264
  • 3
    Nice answer. +1. Additionally, WLOG you don't have to define the sets in precisely this way. Providing the four sets cover all 15 numbers, you can split them up any way you want. Then your first three tickets should be the three sets of four, and the final four tickets should be the set of three augmented with each number from any particular one of the sets of four. – jonnybolton16 Oct 20 '24 at 07:44
  • For example, $A={1,5,9,13}$, $B={2,6,10,14}$, $C={3,7,11,15}$ and $D={4,8,12}$ with the chosen tickets being ${1,5,9,13}$, ${2,6,10,14}$, ${3,7,11,15}$, ${3,4,8,12}$, ${7,4,8,12}$, ${11,4,8,12}$ and ${15,4,8,12}$ – jonnybolton16 Oct 20 '24 at 07:46
  • Awesome answer, although this brings back bad memory of university tests where by solving the last question, the first questions automatically became moot and I was afraid of getting a bad grade because of it. – Stef Oct 21 '24 at 07:17
5

This is a covering problem. The elements are all $4$-combinations of the numbers $1, 2,\dots, 15$ i.e. all the tickets the host could select. A ticket covers all the elements that have intersection of size at least $2$ with it.

To make this more formal let's introduce the space we need to cover as

$$ X=\{ \space \{x_1,x_2,x_3,x_4\} \space | \space x_1,x_2,x_3,x_4 \text{ distinct numbers of } [15]\} $$

and the ticket $t = \{t_1, t_2, t_3, t_4\}$ covers all $x\in X$ such that $|t\cap x| \geq 2$.

In another words we must make sure that all the $\binom{15}{2}=105$ pairs of numbers are present in one of our tickets. UPDATE: As noted in the comments by C.C, this isn't necessary. For example the pair $1,5$ isn't present in any ticket in the solution given by Mike Earnest.

Exercise 1. is solved by noticing that we can put two pairs in each ticket, so we need only $\left\lceil \frac{105}{2} \right\rceil = 53$ tickets to do this.

For exercise 2. 41 tickets will suffice. Take the tickets $\{1,2,3,4\}, \{5,6,7,8\}, \{9,10,11,12\}$ and $\{1,13,14,15\}$. By this brute force Sage calculation:

X = Subsets(range(1,16),4)
ts = [set((1,2,3,4)), set((5,6,7,8)), set((9,10,11,12)), set((1,13,14,15))]
rest = [x for x in X if all(len(set(x).intersection(t))<2 for t in ts)]
#print (len(rest))
#print (rest)
restPairs = set()
for x in rest:
    for p in Subsets(x,2):
        restPairs.add(p)
print (len(restPairs))
#print (restPairs)

there are $73$ pairs left to cover after this. For those we only need $\left\lceil \frac{73}{2} \right\rceil = 37$ tickets. So $4+37=41$ in total.

For the lower bounds some sort of approximation is necessary. Or after taking the four tickets mentioned above (it's quite clear that it's best to take these (or take any element instead of $1$ in the last)) there are only $144$ tickets left to cover and we could make it an linear integer programming problem.

ploosu2
  • 12,868
  • 1
    turns out "we must make sure all $\binom{15}2=105$ pairs are present in one of our tickets" is sufficient but not necessary – C.C. Oct 20 '24 at 02:15
5

Part B gives a hint that $75$ may be an unnecessarily large number, so solving Part B wil automatically solve part A

Divide the tickets into $4$ groups,
$1-4,\quad 5-8, \quad 9-12,\quad 13-15$ named, say, $A,B,C,D$

Configuration of winning #s in these groups will be one of the following

$4000, \quad 3100, \quad 2200,\quad 2110,\quad 1111$

Choose Tickets A,B,C, and you may succeed, but to guarantee success, you have to cover the configuration $1111$

So going by the worst case scenario, $\mathtt{create}$ four more tickets by taking the #s $13,14,15,$ of Group D and adding $1,2,3,4$ of Group A turn by turn to create Tickets E,F,G H

With the $1111$ configuration one of these $7$ tickets have to succeed

Thus maximum # of tickets to be bought for the worst case scenario is $3+4 = \boxed{7}$

Note: Credit to Mike Earnest who has helped reduce the initial number of tickets I had computed.

true blue anil
  • 49,980
  • I understand your first four tickets are ${1,2,3,4}$, ${5,6,7,8}$, ${9,10,11,12}$, and ${13,14,15,x}$ ($x$ is arbitrary). But I have no idea what you mean by " all the tickets from the 4 # groups and try one by one from the small group of 3 #s", so I cannot determine from your answer what the remaining $36$ tickets are. Could you add more explanation, or at least list out like five of the tickets explicitly, as an example? – Mike Earnest Oct 19 '24 at 15:09
  • @MikeEarnest: I have edited, I believe it should now be quite clear – true blue anil Oct 19 '24 at 15:42
  • Huh, now you are overcounting! For the group A + group D case, only $4$ tickets are required; the tickets are ${1,13,14,15},{2,13,14,15},{3,13,14,15}$, and ${4,13,14,15}$. Similarly, the group B + group D and group C + group D are four tickets each, so this is only 16 tickets. – Mike Earnest Oct 19 '24 at 15:51
  • Furthermore, you only need the four initial tickets, and the group A + group D tickets, for this to work. So you have actually discovered an 8 ticket solution! – Mike Earnest Oct 19 '24 at 15:52
  • There are the four initial tickets, groups A through D. There are four tickets consisting of a single ticket from group A combined with all three tickets in group D. Four + four = eight. – Mike Earnest Oct 19 '24 at 16:27
  • I think if you try to literally list out all 11 tickets in your solution, you will realize that you are making a logical error. – Mike Earnest Oct 19 '24 at 16:37
  • I found that there exists a seven ticket solution, where four of the tickets are group A, group B, group C, and (group D + x) for some number x. You should try to find this yourself. In case you give up, see my answer for the solution. – Mike Earnest Oct 19 '24 at 17:08
  • 2
    This answer could be greatly improved by matching the words used to those in the question. For example, "divide the tickets into 4 groups" actually means "divide the numbers into 4 groups"; "cost=4#s" means "cost=4 tickets" and so on. – psmears Oct 19 '24 at 22:05
  • @psmears: I have edited to improve the wording, to make the idea clearer – true blue anil Oct 20 '24 at 03:06