Quod Erat Demonstrandum

2007/09/14

[初中] Let me ‘teach’ you how to play Mark Six

Filed under: Junior Form Mathematics,Teaching — johnmayhk @ 10:34 下午

Reply to the question from 笨木:

“大學教授買六合彩獨中八千多萬" John Sir..我想問呢..你有冇睇過呢篇報導?mainly..個d大家教授話..買六合彩有方法..其中一個所謂的聰明..係話..買哂49個號碼..話d人買既只係49個數字既一部分 所以會有d數字唔中..我覺得呢個係白痴的..因為你買哂49個數字..抽出黎的一定係呢49個數字中間既數= =and..佢話賣既天書要188hkd一本..如果有傻仔買..當有100萬人買..(即6-7人中有一個人買)都有100萬x188..即係1億8千幾萬- -比起中六合彩仲要賺得多..John Sir你又點睇?By yahoo search, I come across the following

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 C_6^{49} = 13983816 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.

A_1 = {01,02,03,04,05,06,07,08,09,10,11,12}
A_2 = {13,14,15,16,17,18,19,20,21,22,23,24}
A_3 = {25,26,27,28,29,30,31,32,33}
A_4 = {34,35,36,37,38,39,40,41,42,43}
A_5 = {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 A_1 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 A_2. Hence, up to now, we have to buy 6\times 2 = 12 tickets.

Now, suppose the TWO lucky numbers are in A_3 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 12 + 3 = 15 tickets required. Go ahead.

Suppose the TWO lucky numbers are in A_4 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 15 + 4 = 19 tickets. And finally, if the TWO lucky numbers are in A_5, 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 20 the minimum number required? Can we serve the same purpose by buying less than 20 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).

廣告

9 則迴響 »

  1. Well, I have really read this but I still have got no idea what you’re talking about, but it’s funny that a ‘clever’ professor can get all the numbers. And I suppose I’m reading this blog again after I finish the University…

    迴響 由 Edmund — 2007/09/28 @ 4:41 下午 | 回應

  2. What I was writing is : if you follow my instruction to buy the 20 tickets, you MUST win at least 2 lucky numbers!

    After university? Too late ar. May be you visit my blog occasionally and share if you want to.

    迴響 由 johnmayhk — 2007/09/28 @ 5:34 下午 | 回應

  3. Hi sir, I am really interested in the detail for this topic.
    But I’m just too poor in understanding this part :

    “OK, suppose the TWO lucky numbers are in A_1 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}"

    Could you explain a little bit more on how to calculate and form the set that can ensure that it captures the two lucky numbers? Thank you very much.

    迴響 由 Chris — 2007/11/23 @ 11:52 上午 | 回應

  4. Suppose the two lucky numbers in the set
    {1,2,3,4,5,6,7,8,9,10,11,12}

    Now divide the set into 4 groups, say
    1,2,3 4,5,6 7,8,9 10,11,12

    We have the following 2 cases:
    (1) The two lucky numbers in the same group.
    (2) The two lucky numbers in different groups.

    For case (1), it is easily see the following tickets include the two lucky numbers

    {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}

    because each ticket consists numbers of same groups, e.g. the two lucky numbers are 7,8; then we can see 7,8 in the group “7,8,9″ and the ticket {1,2,3,7,8,9} consists of the group “7,8,9″ and hence contains the two lucky numbers 7,8.

    For case (2), it is not difficult to see the following tickets include the two lucky numbers.

    {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}

    since there are 4 groups, and each ticket contains 2 groups, hence the number of combination of possible tickets = C_{2}^{4} = 6, and ALL the 6 possible tickets are listed above. As for example, if the lucky numbers are 4,9. They are in different groups “4,5,6″ and “7,8,9″; and the ticket {4,5,6,7,8,9} consists of the groups “4,5,6″ and “7,8,9″ and hence it contains the two lucky numbers 4,9.

    迴響 由 johnmayhk — 2007/11/23 @ 1:19 下午 | 回應

  5. Thank you very much for your explanation.

    But I’m sorry to ask again if there’s any generic method on how to divide the groups?
    Say I am not confident on how do you determine for the case A4 (i.e. 10 numbers with 2 lucky numbers inside)
    I tried to divide them into 34,35 36,37 38,39 40,41,42,43
    so if the two numbers are in the same group or different group , the following will cover them all
    {34,35,36,37,38,39}
    {34,35,40,41,42,43}
    {36,37,40,41,42,43}
    {38,39,40,41,42,43}
    but I can’t get your combination. (or actually my combination is wrong?)
    Would you mind explaining a little bit on this? Thanks again.

    迴響 由 Chris — 2007/11/23 @ 2:57 下午 | 回應

  6. Yes, I actual use the method similar to what you’d mentioned. (But, in my previous post, I made mistakes and corrected now. Thank you Chris!)

    Pairing up 10 numbers in A_4 forming 5 groups, namely

    g1,g2,g3,g4,g5

    Your arrangement is as follows

    {g1,g2,g3} – – -> (12)(13)(23)
    {g1,g4,g5} – – -> (14)(15)(45)
    {g2,g4,g5} – – -> (24)(25)(45)
    {g3,g4,g5} – – -> (34)(35)(45)

    We see 12 small brackets contain ALL C_{2}^{5} = 10 combinations of selecting 2 groups out of 5 groups.

    The arrangement is not unique, for example

    {g1,g2,g3} – – -> (12)(13)(23)
    {g1,g4,g5} – – -> (14)(15)(45)
    {g2,g3,g5} – – -> (23)(25)(35)
    {g2,g3,g4} – – -> (23)(24)(34)

    10 combinations included.

    迴響 由 johnmayhk — 2007/11/24 @ 1:07 下午 | 回應

  7. It seems better to consider
    A_1 = \{1,2,3, \cdots, 10\}
    A_2 = \{11,12,13, \cdots, 20\}
    A_3 = \{21,22,23, \cdots, 30\}
    A_4 = \{31,32,33 \cdots, 40\}
    A_5 = \{41,32,33 \cdots, 49\}
    By your above consideration, we only need 4*4 + 3 = 19 tickets, which is one number smaller than 20 tickets.

    迴響 由 davistang628 — 2012/12/25 @ 8:54 上午 | 回應

  8. The problem is, to actually win the lottery, you must at least have 3 numbers matching (in 1 single lottery ticket)

    迴響 由 rudolfalex — 2014/05/31 @ 6:11 下午 | 回應

    • Payout table:
      $40 3 standard numbers
      $320 3 standard numbers + Special Number (4 numbers)
      $640 4 standard numbers
      $9600 4 standard numbers + Special Number (5 numbers)
      3rd prize 5 standard numbers
      2nd prize 5 standard numbers + Special Number (6 numbers)
      Jackpot 6 standard numbers

      迴響 由 rudolfalex — 2014/05/31 @ 6:15 下午 | 回應


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