“2006年10月，英國Bradford University 一班教授，合資買Lotto(六合彩)，結果獨中頭獎5百多萬英磅，相等於港幣8千多萬"

The writer of the article claimed that the professors had a “smart" way of buying the Lotto. I’d like to say something about this.

With basic training in applied mathematics, to ensure “sure-win" in playing Mark Six, we need to buy different tickets. Cost? Sorry to tell you that I had never bought any of Mark Six ticket before and had no idea of how much it costs. The article claimed that, instead of 13983816, it is required only 17 lottery tickets. It is of course questionable. The meaning of “all numbers appearing in that 17 tickets" means we can find a “lucky number" (out of 6 “lucky numbers") in one of the 17 tickets. But, there is no guarantee to have ALL 6 lucky numbers appearing in 1 ticket! In fact, it is extremely simple to ensure ONE lucky number in the tickets (no need 17 tickets), we need at most 9, the arrangement is as follows.

{01,02,03,04,05,06}

{07,08,09,10,11,12}

{13,14,15,16,17,18}

{19,20,21,22,23,24}

{25,26,27,28,29,30}

{31,32,33,34,35,36}

{37,38,39,40,41,42}

{43,44,45,46,47,48}

{43,44,45,46,47,49}

However, the following question may be meaningful and mathematical.

How many tickets we need to buy to ensure TWO lucky numbers appearing in one of the tickets? Or, what is the minimum number of tickets we need to ensure there must be TWO lucky numbers in one ticket and how to buy?

Consider the following 5 number sets.

= {01,02,03,04,05,06,07,08,09,10,11,12}

= {13,14,15,16,17,18,19,20,21,22,23,24}

= {25,26,27,28,29,30,31,32,33}

= {34,35,36,37,38,39,40,41,42,43}

= {44,45,46,47,48,49}

Now, no matter which 6 lucky numbers are, TWO of them must belong to one of the 5 number sets above. (Why? Pigeon hole principle)

OK, suppose the TWO lucky numbers are in which is a set of 12 numbers.

Then, all we need is 6 tickets to ‘capture’ those TWO lucky numbers. The arrangement is

{01,02,03,04,05,06}

{01,02,03,07,08,09}

{01,02,03,10,11,12}

{04,05,06,07,08,09}

{04,05,06,10,11,12}

{07,08,09,10,11,12}

Follow the pattern above, we can consider the cases when the TWO lucky numbers appearing in the number set . Hence, up to now, we have to buy tickets.

Now, suppose the TWO lucky numbers are in which is a set of 9 numbers.

Then, all we need is 3 tickets to ‘capture’ those TWO lucky numbers. The arrangement is

{25,26,27,28,29,30}

{25,26,27,31,32,33}

{28,29,30,31,32,33}

OK, up to now, there are tickets required. Go ahead.

Suppose the TWO lucky numbers are in which is a set of 10 numbers.

Then, we need 4 tickets to ‘capture’ those TWO lucky numbers. The arrangement is

{34,35,36,37,38,39}

{34,35,40,41,42,43}

{36,37,38,39,42,43}

{36,37,38,39,40,41}

OK, up to now, there are tickets. And finally, if the TWO lucky numbers are in , we need to buy one more ticket

{44,45,46,47,48,49}

Here, we are certain that TWO lucky numbers can be found in those 20 tickets mentioned above.

But, is the minimum number required? Can we serve the same purpose by buying less than tickets? My mathematics knowledge is so poor to come up with any further conclusion. Please share your idea or finding if you have time. This is actually a good topic to go further studies. As for example, how to ensure THREE lucky numbers there? Mark Six is more than just gambling or probabilities. Many years ago, I hope to have a topic of “mathematics in gambling" as a project for F.3 students, but I blamed myself immediately for this silly thought.

Looking back to that $188 book, I think I am able to write similar book by presenting pages of numbers (actually rubbish, non-sense, like what I usually do in forum and blog).