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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.

K Vij
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  • Its one more ticket than the sum of all the 0, 1, and 2 number matches. – Phil H Jun 22 '18 at 00:24
  • Each ticket contains only $5C3 = 10$ triples in it, so that is the most it can cover. Since there are $69C3 = 52394$ possible triples that might show up, and there are ten in each drawing, you get a lower bound of at least $53$ tickets required. But there is probably some necessary redundancy in how you spread out those triples on your tix. But I do not know the formula for this and it was very, very messy by hand. – John Samples Jun 22 '18 at 00:54
  • you mean lower bound of at least 5300? – K Vij Jun 22 '18 at 00:57
  • Not unless I am being an idiot, which is totally possible. You divide by 100 since there are ten triples to hit on the rollout, and each ticket has ten chances to hit. This is a very crude bound, though my intuition is that you probably don't need more than, say, 65. – John Samples Jun 22 '18 at 01:03
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    Bad answers get thumbs up, it can be frustrating. Especially if they post it as an answer, rather than a comment. – John Samples Jun 22 '18 at 01:05
  • Guaranteed means a probability p = 1. There are so many ways of not getting 3 matches that 42 or even 4200 makes no sense. – Phil H Jun 22 '18 at 01:12
  • Phil, if you buy every possible ticket then obv you will have one that hits three numbers. – John Samples Jun 22 '18 at 01:14
  • I know that John, but there is a big difference between a high probability, which seems to be the main discussion, compared to a guaranteed p = 1. – Phil H Jun 22 '18 at 01:22
  • @JohnSamples I think so too.. Although 53 also looks like a very low number too, since the probability of 3/5 match with a single ticket is 1 in 580.. – K Vij Jun 22 '18 at 01:23
  • @PhilH you are probably thinking in terms of probabilities but looks like the probabilities are not giving the right answer here.. we are looking for min number of tickets and 11436769 looks way too big... that many will guarantee a 5/5 match... – K Vij Jun 22 '18 at 01:25
  • Look like it might be an Open Problem as per this https://math.stackexchange.com/a/66503/571475 – K Vij Jun 22 '18 at 01:40
  • Also found this https://www.ccrwest.org/cover.html "La Jolla Covering Repository", so looking into that now – K Vij Jun 22 '18 at 01:40
  • You might be interested in "lottery wheeling". https://en.wikipedia.org/wiki/Lottery_wheeling – awkward Jun 22 '18 at 12:22
  • Jordan Ellenberg, among others, has written popular literature about this topic: "How not to be wrong" – David Diaz Jun 22 '18 at 20:20

3 Answers3

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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 $n$.
quasi
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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.

Ross Millikan
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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.

Phil H
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  • If you matched 2, then you already matched one. So you are double-counting. Among other things. – John Samples Jun 22 '18 at 01:19
  • Couldn't add up 4 numbers. 11218033 < 11238513 – Phil H Jun 22 '18 at 01:43
  • Still looks too big as there are only 52394 possible triplets – K Vij Jun 22 '18 at 01:50
  • But so many more 5 numbers that do not contain any triplets. – Phil H Jun 22 '18 at 01:52
  • I think it should be less then 52394 coz there will be replication once you start picking the next 3 numbers for each set in a list of all possible triplets.. – K Vij Jun 22 '18 at 02:06
  • There are 11218032 five digit numbers that do not contain a triple or more. How does buying 52394 tickets guarantee they aren't all from those numbers? – Phil H Jun 22 '18 at 02:11
  • The $52394$ triplets each do not have any duplicate numbers. I can guarantee a win with $52385$ tickets by marking one different triplet one every ticket, then marking two random numbers – Ross Millikan Jun 22 '18 at 03:49
  • For any $5$ winning numbers, there are $10$ triplets of interest. For each triplet, there will be $64 C 2\cdot 10 = 20160$ tickets.with a successful triplet. Add in the $4$ and $5$ number successes and the total is $20481$ tickets with at least one successful triplet and $11218032$ tickets with zero.successful triplets.If you buy $52385$ tickets the probability of getting at least a triplet is $1 - \frac{11218032}{11238613}^{52385} = <1$. The issue here seems to be the word guarantee which means 100% certain. For other interpretations of guarantee, is that p = .99 or .999 or .9999.....etc? . – Phil H Jun 22 '18 at 05:24