How many powerball tickets will guarantee a 3/5 match?

Trying it figure this out, but looks like the obvious answer might not be right. According to the probabilities there is a 1 in 579.76 chance of matching 3 of 5 numbers in a Powerball ticket. So how many tickets will you need to guarantee a 3/5 match? Will you need 580 tickets?

Doesn't seem right when considering that to guarantee at least 1 number match you will need to buy 14 tickets as each ticket has 5 numbers. So 14 tickets can give you all possible 69 numbers.

However, probability of matching exactly 1 number would be around 0.28 using the formula ( 5 choose 1 * 64 choose 4 / 69 choose 5 ). Probability of matching at least 1 will be higher. So based on that you should need only 4 tickets or less to guarantee that at least one number matches.

Note: Powerball allows you to pick 5 numbers from 1 to 69. Probabilities given above may or may not consider bonus but the problem stays the same.

A crude upper bound for the number of tickets needed to guarantee a win is $18106$, as explained below . . .

Suppose a lottery is defined as a triple $(n,c,w)$ of integers, with $1\le w\le c\le n$, where

• $S=\{1,...,n\}$ is the set of numbers from which $c$ distinct numbers are chosen.$\\[4pt]$
• A ticket is an arbitrary choice of $c$ distinct numbers from $S$.$\\[4pt]$
• To win, you need to match only $w$ of the $c$ chosen numbers.

For the general $(n,c,w)$ lottery, let $g(n,c,w)$ be the least number of tickets required to guarantee at least one winning ticket.

In the context of the given question, we want to find an upper bound for $g(69,5,3)$.

Claim:

An upper bound for $g(n,c,w)$ is $f(n,c,w)$ where $f$ is defined recursively by $$ f(n,c,w)= \begin{cases} \text{if}\;n=c,\;\text{then}\\[4pt] \;\;\;\;\;1\\[4pt] \text{else if}\;w=1,\;\text{then}\\[4pt] \;\;\;\;\;\left\lceil{\large{\frac{n}{c}}}\right\rceil\\[4pt] \text{else}\\[4pt] \;\;\;\;\;f(n-1,c,w)+f(n-1,c-1,w-1)\\[4pt] \end{cases} $$ Implemented in Maple, we get $f(69,5,3)=18106$.

To justify the upper bound, the following ticket buying strategy, though not claimed to be optimal, suffices to guarantee a win, and requires buying exactly $f(n,c,w)$ tickets . . .

If $n=c$, there is only one possible draw, and only one legal ticket, which is therefore guaranteed to win. Thus, for this case, we get $f(n,c,w)=g(n,c,w)=1$.

If $w=1$, only one number needs to be matched, so an optimal covering can be had by buying tickets corresponding to $\left\lfloor{\large{\frac{n}{c}}}\right\rfloor$ pairwise disjoint $c$-element subsets of $S$, plus one more ticket if $n$ is not a multiple of $c$ (in order to cover the remainder). Thus, for this case, we get $f(n,c,w)=g(n,c,w)=\left\lceil{\large{\frac{n}{c}}}\right\rceil$.

Otherwise, recursively, use the same strategy to:

• Buy $f(n-1,c,w)$ blank tickets for the $(n,c,w)$ lottery, and mark them as if the goal was to guarantee a win for an $(n-1,c,w)$ lottery. Note: Those tickets will guarantee a win for the $(n,c,w)$ lottery if none of the drawn numbers is equal to $n$.$\\[4pt]$
• Buy $f(n-1,c-1,w-1)$ blank tickets for the $(n,c,w)$ lottery, choosing $n$ for one of the $c$ numbers, and for the other $c-1$ choices, mark them as if the goal was to guarantee a win for an $(n-1,c-1,w-1)$ lottery. Note: Those tickets will guarantee a win for the $(n,c,w)$ lottery if one of the drawn numbers is equal to $n$.