Lottery Question, Show that 2 people will have chosen 2 of the same number

Question

A lottery game is played where a person must choose 6 unique numbers from 1 to 54 inclusive.

(a) Show that if 10 people play then at least 2 must have chosen at least one of the same numbers.

This one is pretty straight forward, with 10 people choosing 6 numbers thats 60 numbers chosen out of the total of 54 numbers. By the pigeon hole principle we can conclude at least 2 have the chosen the same number.

(b) Assume 100 people play this game, prove it is true that 2 people have at least 2 number in common.

This one I am stuck on, I am pretty sure we still need to use the pigeon hole principle but I am not sure in what capacity. I thought about representing this as a 54 x 100 grid and trying to show some number of columns collisions but with 100 rows its pretty daunting to do so. Hints over direct answers would be appreciated.

Answer 1

Imagine each person as having six of their own specific color ball unique to themselves.

Imagine a bucket for each of the $54$ numbers in the lottery.

A person choosing a particular lotto ticket corresponds to them placing one of their respective balls in each of the respective buckets corresponding to the numbers they chose.

There are $6\cdot 100 = 600$ balls that are placed into $54$ buckets.

So there is a bucket with at least $\lceil 600/54\rceil = 12$ balls in it. Note that each ball in this bucket must have come from someone else!

Continuing:

We do this again, giving all players their balls back, removing all players other than the ones who were in our known heavily filled first bucket, and having them repeat the game with the one fewer bucket and having them replace their five balls again.

$~$

There are $53$ buckets remaining in play

$~$

And there are at least $12\cdot 5 = 60$ remaining balls in play