Sorry not related to poker just a question thats bugging me. Need help

[StuR]

Hi,

Sorry for using this forum I thought you guys would be some of the best to ask.

In the UK in our lottery there are a possible 49 numbers and six get chosen.

Whats the minimum number of tickets I would have to buy to GUARANTEE matching at least five numbers?

I have a solution that I think may be right but I'm not sure I just wanna see what you guys can come up with.

Once again sorry if this is a problem post in a poker forum.

StuR

[ninjaunderwear]

Where n = 49, the quantity of numbers from which you can choose and r = 6, the quantity to be chosen.
n! / (r! (n - r)!) = 13,983,816 total possible lottery tickets.

If the numbers drawn were 1,2,3,4,5,and 6, the tickets with 5 or more matching numbers would be the following seven:
12345x, 1234x6, 123x56, 12x456, 1x3456, x23456, and 123456. Of the tickets with 5 matching numbers, there are 43 combinations for the solitary wrong number on the ticket (49 total numbers - 6 correct numbers), for a grand total of 259 tickets with 5 or more matching numbers.

If you purchased 13,983,557 DIFFERENT tickets, all of these could still conceiveably have only 4 or fewer matching numbers. Purchasing one additional unique ticket would guarantee you 5 or 6 matching numbers. However, you would need to make sure none of your tickets have the same selections, a daunting task in itself.

So, in short, "No, it would not be possible to beat the lottery system."

[pzhon]

These are called covering designs or wheels. Many people have worked on these. Finding the optimum is a hard problem. You can find a lot of information on the web.

[SumZero]

[ QUOTE ]
If you purchased 13,983,557 DIFFERENT tickets, all of these could still conceiveably have only 4 or fewer matching numbers. Purchasing one additional unique ticket would guarantee you 5 or 6 matching numbers.

[/ QUOTE ]

Sorry, the question was what was the minimum number and I can guarentee that you match 5 with only 12,271,512 tickets. I'm not sure that my number is the minimum (and I don't think it is).

My method is pretty darn easy. There are 49 possible numbers and you pick 6 of them on each ticket and the winning draw picks six of them (I'm assuming the draws are without replacement - I.e., you can't have the winning 6 numbers be 1, 5, 7, 12, 1, 12). So my method is buy tickets that cover all the possible combinations of tickets as if there were only 48 numbers. That is 48C6. Then if the winning 6 numbers don't include 49 you have a ticket with all six (and many with 5), if they do include a 49 then you still have 48 tickets that match 5.

Um, why not 49C5 = 1,906,844

To guarantee at least 5 numbers in any N size set drawn from the set of 49, all you need to have is every possible 5 number combination of 49. So, 49C5

Because you have every possible 5 number combination of the original 49, any 6 number subset you choose out of this 49 will be covered.

I have no idea if this is the lowest number, but it seems likely.

[pzhon]

[ QUOTE ]
Um, why not 49C5 = 1,906,844

To guarantee at least 5 numbers in any N size set drawn from the set of 49, all you need to have is every possible 5 number combination of 49.

[/ QUOTE ]
Since each 6-tuple contains 6 5-tuples, you can hope to cover all 5-tuples with 49C5/6 tickets. I'd guess this is possible. Usually, when there is no number theoretic obstruction, such designs exist.

However, covering all 5-tuples is not necessary. You can omit some, just not all 6 5-tuples contained in some 6-tuple.

The total number of tickets sharing at least 5 numbers in common with a given ticket is 1+6*43 = 259. Ideally, you would like the 259 neighbors of each ticket in your wheel to be distinct, in which case you would need only 49C6/259 tickets. That's a lower bound for the number of tickets needed. Unfortunately, 49C6/259 isn't an integer, so you need some overlap. The minimal wheel could be very messy.

[StuR]

I understand the 49C5 would give you the total number of five number combinations but why the 259 again.

I thought the answer would be 49C5/6 because each ticket can hold 6 different five number combinations.

The problem my friend and I encountered was whether it was possible in this number (i.e. whether it was possible to group the combinations into batches of six that fit on one ticket) We thought this is where the problem would occur.

Is there anyway to tackle this? Can you explain the 259 calculation again.

I did something similar with there being 258 (1+6*43)winners (matching at least 5 numbers) for each possible outcome but then came derailed when I said choose one of these 258 for each possible set of 6 numbers = 13.9 million again!!!

Thanks for the input guys.

StuR

[StuR]

Just to check I sent this question to an any question answered service and the foll wing was the response

The odds of matching 5 numbers in the lotto draw is 1 in 55492, This means you need to buy 55492 different tickets to be guaranteed to match 5

It then directed me to an odds calculator that also said the odds of matching one number is 1 in 2.42

From his answer it would seem he/she thinks I would require 3 tickets to match one number.

This will only cover 18 numbers tops! SO I can just pick the "winning" numbers out of the remaining 31 possibilities to disprove this.

The logic fails for this so it definitely bought it into question - I'm trying to get my money back from the service!

StuR

[pzhon]

[ QUOTE ]

The odds of matching 5 numbers in the lotto draw is 1 in 55492, This means you need to buy 55492 different tickets to be guaranteed to match 5

[/ QUOTE ]
You need to buy at least 49C6/259 = 53991.6... or else the average number of tickets with at least 5 matches will be less than 1, which implies that there is a chance you won't have any 5 number winners.

This says nothing about the existence of a configuration of tickets ensuring that you would have at least one 5 number winner.

[pzhon]

[ QUOTE ]
I understand the 49C5 would give you the total number of five number combinations but why the 259 again.

[/ QUOTE ]
Suppose the winning ticket is 123456. There is one ticket matching all 6 numbers, 43 that match only 12345, 43 that match only 12346, etc. for a total of 1+6*43=259 tickets that match 5 or more numbers. Every ticket has a probability of 259/49C6 of matching at least 5 numbers.

[ QUOTE ]
I thought the answer would be 49C5/6 because each ticket can hold 6 different five number combinations.

[/ QUOTE ]
No. Suppose you cover each 5 number combination exactly once. Then suppose 123456 is drawn, and this is not one of your tickets. That means you have a ticket covering 12345, a ticket covering 12346, ..., 23456. You have 6 tickets matching 5 numbers.

[ QUOTE ]

The problem my friend and I encountered was whether it was possible in this number (i.e. whether it was possible to group the combinations into batches of six that fit on one ticket) We thought this is where the problem would occur.


[/ QUOTE ]
From what I remember of design theory, this type of design is not hard to construct, but the construction may be a bit messy.