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

Question

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.

Answer 1

Guaranteed means $p = 1$. Like I said in my comment, if you buy one more than all possible $0, 1$ and $2$ number matches you are guaranteed at least a $3$ number match.

For $0$ number matches, there are $^{64}C_5 = 7624512$ ways to get that.

For $1$ number matches, there are $5$ ways to match that one number and for every one of those there $^{64}C_4 = 635376$ ways to not match the other $4$ making a total of $3176880$.

For $2$ number matches, there are $10$ ways to match $2$ numbers from $5$ and for every one of those there are $^{64}C_3 = 41664$ ways not to match the other $3$ making a total of $416640$.

The number of tickets to purchase to guarantee a $3$ number match is therefore:

$7624512 + 3176880 + 416640 + 1 = 11218033$

For an interpretation of this problem whereby guaranteed = likely, as in the same birthday problem, that is $23$ random people are required for a likely match of two people with the same birthday where $p > 0.5$.

So, the minimum number of tickets purchased for a probability $>0.5$ of $3$ out of $5$ matching numbers is the number of iterations of $1 - \frac{11218032}{11238512}\cdot \frac{11218031}{11238512}\cdot \frac{11218030}{11238511}...............\frac{11217652}{11238133} = .500001$. Hence $\approx 380$ tickets are needed to have a minisculely better than even chance of getting $3$ out of $5$ matching numbers.

Answer 2

There are ${69 \choose 3}=52394$ triplets of numbers. Each draw produces ${5 \choose 3}=10$ triplets. Each ticket you buy covers $10$ triplets. This means each ticket has $100$ chances to win. You would expect a good chance to have a winner if you buy $524$ tickets or so with the probability rising quickly around there. This is very close to your $580$ number. You need to choose your tickets to avoid repeating triplets. This is far from a guarantee. To get a guarantee you need to cover all but $9$ of the triplets, so need to buy at least $5240$ tickets. Probably you need to buy more because you wind up repeating triplets, but I don't know how to compute that.

Answer 3

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 $n$.