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The Magic Professors

simonmaal

Critical Thinker
simonmaal
Joined
Sep 29, 2006
Messages
372
Take a look at the following story:

http://www.dailymail.co.uk/pages/live/articles/news/news.html?in_article_id=412386&in_page_id=1770

In an attempt to win the National Lottery, a syndicate of University professors and students devised a system that allowed them to 'rely less on chance'. Instead, they used two boxes, 49 pieces of paper and a mathematical system of divination. They won £5.3 million on Saturday night after matching all six numbers, and they and the press have attributed their win to the system rather than pure chance.

It just goes to show that intelligent people can sometimes be so stupid.
 
simonmaal said:
Take a look at the following story:

http://www.dailymail.co.uk/pages/live/articles/news/news.html?in_article_id=412386&in_page_id=1770

In an attempt to win the National Lottery, a syndicate of University professors and students devised a system that allowed them to 'rely less on chance'. Instead, they used two boxes, 49 pieces of paper and a mathematical system of divination. They won £5.3 million on Saturday night after matching all six numbers, and they and the press have attributed their win to the system rather than pure chance.

It just goes to show that intelligent people can sometimes be so stupid.
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i hope they don't lecture in maths....

....they're probably sports science or film studies profs :D
 
No ... I think what this shows is that Daily Mail journalists are stupid.

The syndicate used a sensible method of maximizing their odds. The journalist has spun this into an "unbeatable formulae".
 
andyandy said:
i hope they don't lecture in maths....
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Me too, I dread to think what kind of magical thinking they're trying to propose to their students in maths lectures!:yikes:

....they're probably sports science or film studies profs :D
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Whatever they are, I reckon a book and website will quickly follow in the wake of this. Anything that can help reduce the anxiety associated with an uncertain universe tends to become a million-seller, no matter how much it relies on superstition and claptrap.
 
simonmaal said:
Yes, stupid for believing such flim flam.

How can you possibly devise a system that can overcome the vagaries of lottery balls?
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OK, I'm no maths whiz, but doesn't a system that covers all 49 numbers have more chance of winning than a system that only covers, say, half of the numbers? That's what was done in this case.
 
tkingdoll said:
OK, I'm no maths whiz, but doesn't a system that covers all 49 numbers have more chance of winning than a system that only covers, say, half of the numbers? That's what was done in this case.
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In each case, we would be trying to say that the appearance of any given number could be predicted, rather than being the result of a purely random event. In other words, the system was one of divination, therefore belonging alongside numerology, tarot and I-ching rather than mathematics and physics. It is magical thinking dressed up as a mathematical system.
 
simonmaal said:
In each case, we would be trying to say that the appearance of any given number could be predicted, rather than being the result of a purely random event. In other words, the system was one of divination, therefore belonging alongside numerology, tarot and I-ching rather than mathematics and physics. It is magical thinking dressed up as a mathematical system.
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No, I still don't get it.
 
tkingdoll said:
No, I still don't get it.
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if you're looking at jackpot odds there's 49 numbers, 6 of which are chosen,
that gives you 49 choose 6 different combinations..which google informs me is
.....13 983 816........

however you order the numbers, whatever system you use, each individual set of numbers will have a 1/13983816 chance of being selected......
the only way to maximise your jackpot odds is to buy more tickets....:)

edit,
their "system" seems just be to of the type, buy one lottery ticket - numbers 1,2,3,4,5,6

buy another ticket don't choose 1,2,3,4,5 or 6 because they're already chosen on ticket 1.....

so system 1
ticket 1; 1,2,3,4,5,6
ticket 2; 7,8,9,10,11,12

system 2
ticket 1; 1,2,3,4,5,6
ticket 2; 1,2,3,4,5,6

system 1 spreads the chance of hitting a number on either ticket 1 or ticket 2, but does not effect the E(v)....

"Teaching maths through national lotteries" goes through the n choose r stuff......
http://www.cimt.plymouth.ac.uk/journal/ijnatlot.pdf
 
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tkingdoll said:
No, I still don't get it.
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That's because you're correct -- and simon wrong.

Here's the relevant quote from the article:

Instead of each member randomly choosing numbers or relying on birthdays and significant dates as they had done this time they came up with an unbeatable formulae.

All 49 numbers were written on pieces of paper and placed in one box. Each syndicate member in turn then picked out six numbers, until eight lines were filled, using 48 of the 49 numbers.
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People, in general, are lousy random number generators. If you want to prove this to yourself, find a reasonably sized group of people, ask everyone to write a number from one to ten on a slip of paper, and then see how many people don't pick either one or ten.

And many -- most? -- lottery participants rely on "significant dates" as one method of playing the lottery. This, of course, means that numbers below 32 are overrepresented and numbers above 32 are underrepresented, unless you were born on the 35th of February....

Picking a ticket uniformly at random from the entire set of possible tickets makes it much more likely that the team will pick from the under-represented part of the ticket space. This doesn't make it more likely that they'll win, but it does make it much more likely that if they win, they will win a larger prize (since they won't have to share it).

The other thing they're doing properly is making sure that they aren't doubling up on numbers, so they're covering more of the winning-space than they otherwise would. (This is Dr A's point, of course). Otherwise, if three of the team were born in July, you would probably see too many '7's appearing on the ticket.
 
Let's give a simple example, so you can see how it works. Imagine there are six balls, and two are picked. You get the jackpot for getting both right, and a lesser prize for getting one right. You intend to buy two tickets. How do you proceed?

You should pick two pairs of numbers with no overlap. Here's why.

The possible balls picked, with equal probability are:

{1,2} {1,3} {1,4} {1,5} {1,6} {2,3} {2,4} {2,5} {2,6} {3,4} {3,5} {3,6} {4,5} {4,6} {5,6}

If your tickets have identical numbers (e.g. {1,2} {1,2}) you have a 1/15 chance of the jackpot and a 9/15 chance of a minor prize.

If your tickets overlap by one number (e.g. {1,2} {2,3}) you have a 2/15 chance of the jackpot and a 12/15 chance of winning a minor prize.

If you numbers have no overlap (e.g. {1,2} {3,4}) then you still have a 2/15 chance of winning the jackpot, but you now have a 14/15 chance of winning a minor prize.

We may note that if only the jackpot was at stake, the only rule you'd need would be: don't buy two tickets with EXACTLY the same set of numbers. What you gain by avoiding overlapping numbers in your selection is an increase in your expected winnings by increasing your chances of winning one of the minor prizes.
 
Dr Adequate said:
Let's give a simple example, so you can see how it works. Imagine there are six balls, and two are picked. You get the jackpot for getting both right, and a lesser prize for getting one right. You intend to buy two tickets. How do you proceed?

You should pick two pairs of numbers with no overlap. Here's why.

The possible balls picked, with equal probability are:

{1,2} {1,3} {1,4} {1,5} {1,6} {2,3} {2,4} {2,5} {2,6} {3,4} {3,5} {3,6} {4,5} {4,6} {5,6}

If your tickets have identical numbers (e.g. {1,2} {1,2}) you have a 1/15 chance of the jackpot and a 9/15 chance of a minor prize.

If your tickets overlap by one number (e.g. {1,2} {2,3}) you have a 2/15 chance of the jackpot and a 12/15 chance of winning a minor prize.

If you numbers have no overlap (e.g. {1,2} {3,4}) then you still have a 2/15 chance of winning the jackpot, but you now have a 14/15 chance of winning a minor prize.

We may note that if only the jackpot was at stake, the only rule you'd need would be: don't buy two tickets with the same set of numbers. What you gain by avoiding overlapping numbers in your selection is an increase in your expected winnings by increasing your chances of winning one of the minor prizes.
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Good example. The one flaw in it is that you are not allowing for the possibility of winning two minor prizes (or, for that matter, two jackpots).

In particular, if your tickets have identical numbers, you have a 9/15 chance of winning "a minor price", but if you win one minor prize, you will actually win two minor prizes.

Under this more detailed analysis, the reason that spreading your numbers out is actually that you can't win multiple jackpots, since the jackpot is divided among all winning tickets. Winning the jackpot twice is equivalent to winning it once, and hence gives you no advantage.
 
drkitten said:
Under this more detailed analysis, the reason that spreading your numbers out is actually that you can't win multiple jackpots, since the jackpot is divided among all winning tickets. Winning the jackpot twice is equivalent to winning it once, and hence gives you no advantage.
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good point...i overlooked that :)
 
drkitten said:
Good example. The one flaw in it is that you are not allowing for the possibility of winning two minor prizes (or, for that matter, two jackpots).
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Prizes are split between winning ticket holders, so this only comes into play if someone outside the syndicate also wins.

(I did err by counting jackpot-winning tickets as winning minor prizes. Every time I do math on this forum, I get the darn numbers wrong, it's embarrassing.)

If the winnings were fixed sums, so that you could win two jackpots or two minor prizes, we'd have a different situation, but it would still be capable of optimization, and the solution would still be exactly the same.

If your tickets are {1,2} {1,2}

One jackpot (0/15) : {1,2}
Two jackpots (1/15) : {1,2}
One minor prize (0/15)
Two minor prizes (8/15) : {1,3} {1,4} {1,5} {1,6} {2,3} {2,4} {2,5} {2,6}
Lose (6/15) : {3,4} {3,5} {3,6} {4,5} {4,6} {5,6}

If your tickets are {1,2} {2,3}

One jackpot (2/15) : {1,2} {2,3}
Two jackpots (0/15)
One minor prize (7/15) : {1,3} {1,4} {1,5} {1,6} {3,4} {3,5} {3,6}
Two minor prizes (3/15) : {2,4} {2,5} {2,6}
Lose: (3/15) : {4,5} {4,6} {5,6}

If your tickets are {1,2} {3,4}

One jackpot (2/15) : {1,2} {3,4}
Two jackpots (0/15)
One minor prize (8/15) : {1,5} {1,6} {2,5} {2,6} {3,5} {3,6} {4,5} {4,6}
Two minor prizes (4/15) : {1,4} {1,3} {2,3} {2,4}
Lose (1/15) : {5,6}

Hence, whatever the cash values involved, {1,2} {3,4} is an optimum choice.

The actual situation in the National Lottery may be taken to be between the two extremes modeled. And as both of these have the same solution, I stand by the professors.
 
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Dr Adequate said:
Prizes are split between winning ticket holders, so this only comes into play if someone outside the syndicate also wins.

(I did err by counting jackpot-winning tickets as winning minor prizes. Every time I do math on this forum, I get the darn numbers wrong, it's embarrassing.)

If the winnings were fixed sums, so that you could win two jackpots or two minor prizes, we'd have a different situation, but it would still be capable of optimization, and the solution would still be exactly the same.

If your tickets are {1,2} {1,2}

One jackpot (0/15) : {1,2}
Two jackpots (1/15) : {1,2}
One minor prize (0/15)
Two minor prizes (8/15) : {1,3} {1,4} {1,5} {1,6} {2,3} {2,4} {2,5} {2,6}
Lose (6/15) : {3,4} {3,5} {3,6} {4,5} {4,6} {5,6}


If your tickets are {1,2} {3,4}

One jackpot (2/15) : {1,2} {3,4}
Two jackpots (0/15)
One minor prize (8/15) : {1,5} {1,6} {2,5} {2,6} {3,5} {3,6} {4,5} {4,6}
Two minor prizes (4/15) : {1,4} {1,3} {2,3} {2,4}
Lose (1/15) : {5,6}

Hence, whatever the cash values involved, {1,2} {3,4} is an optimum choice.

The actual situation in the National Lottery may be taken to be between the two extremes modeled. And as both of these have the same solution, I stand by the professors.
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yep, your E(v) would be the same for the two different ticket combis above if the prize was a fixed amount that was not divided amongst winning tickets but given to any tickets which hit.....

say with grand prize a
and minor prize b

tickets {1,2} {3,4}
over 15 rounds would have an E(v) of 2x1a/I] + 8b + 2x4b

and tickets {1,2} {1,2}
over 15 rounds would have an E(v) of 2a + 2x8b

both of which give you 2a + 16b

but with the grand prize not given as a fixed amount to any winners, but split between winners then

tickets {1,2} {3,4} have a 15 round E(v) of 2a + 16b
and tickets {1,2} {1,2} have a 15 round E(v) of a + 16b
 
If your tickets are {1,2} {2,3}

One jackpot (2/15) : {1,2} {2,3}
Two jackpots (0/15)
One minor prize (7/15) : {1,3} {1,4} {1,5} {1,6} {3,4} {3,5} {3,6}
Two minor prizes (3/15) : {2,4} {2,5} {2,6}
Lose: (3/15) : {4,5} {4,6} {5,6}
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but on this example, you've not taken into account the fact that [2,3] wins a jackpot and wins a minor prize
as does [1,2]

and [1,3] wins you two minor prizes

so with this adjusted we have a 15 round E(v) of 2a + 16b as before.....both for a fixed jackpot and for a split jackpot

so the only system to increase your E(v) is to not play the exactly the same numbers*......:)

* disregarding human lottery factors - like how people are drawn to the same numbers when picking - like "birthday" numbers etc. Avoiding these as DrKitten pointed out can increase your E(v) because you'll share the jackpot with less people......
 
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