Lottery syndicate - is this guy breaking any laws?

Actually I think we first need to establish if the larger, smaller or middle wins give a disporotionate share of the total winnings. So we know where we should be targetting.

Why do you think he would need to "freeze a particular slot"? The argument is not baloney, it is perfectly true that his system of 44 tickets will give a winning probability of 1/1.9M, the problem is that buying 44 tickets with *no system* can do very much better, i.e. give a winning prob of 44/14M.

Actually I think you will find buying 2 tickets does double your chances of winning.

It does double your chances of winning;-) The problem is the chances of losing!!

Can't be arsed to do the mathematics though;-)

CRF

He says: Only play with 5 balls, by this he means that *each* ticket will have a 7 fold chance. He considers that the freezing of a slot is a side effect of his 44 tickets. (but this is incorrect).

No it's not true at all. The only thing that is true is that playing a different game with only 5 numbers, there will be 1.9 mill different tickets.

He then thinks that you can convert the game into a 5 numbers game by a special selection of numbers which require 44 tickets, these 44 tickets would then

*each* have a 7 fold chance. I repeat: *each* ticket!

But if you look at the equations above, the missing factor is 44/6 ! The 44 comes from there is 44 choices left after choosing numbers of the first 5 balls. The number 6 comes because that 6'th ball can be in 6 different slots after the

5 balls have been placed. So he has achieved absolutely nothing; nothing more, nothing less. Presumably, it's one of those things which is easy to flog because the wins are so rare.

the

No it's n

I bought 100 random tickets one week many months ago. I was having a very bad week and wanted to give G-D a chance to let me escape to the sun. I won absolutely nothing whatsoever. I guess I deserved my suffering :-)

The maths is easy. The chances of losing reduce from 13999999 in 14M to 13999998 in 14M, i.e. they stay at 100.0000%, or to be one decimal point more precise, at 99.99999%, or two more, they reduce from

99.9999928% to 99.9999857%.

ISTM you might just as well buy 44 lucky dips.

Or if it's a syndicate, pick 7,8 etc numbers and cover all combinations of '6 from'.

[then]

[finally]

Yep - you're totally right Fat Freddy - Some members of the public are easily mislead on simple mathematics!

Luckily we have mathematicians like yourself (or your cat?) to set them straight! I think you should run for Chancellor when Brown becomes PM.

Complain to the Lottery that your tickets were not working :-)

Yes, his system offers nothing.

If you buy many tickets in one game, you can choose to either concentrate to numbers, or spread them over the board. In the first case, it becomes a small chance for a high win (high game), i.e. a 6-match may be accompanied by a

5-match on a neighbouring ticket. In the second case it becomes a better chance for the smaller wins (low game).

I did once use a system with 8 tickets of non-repeating numbers (low game). It gave me quite a few 3-matches, but nothing more. Nowadays, I just occasionally buy a single ticket because you're most likely to lose the investment anyway.

[...]

And you don't believe that?

I think that what he has missed is that there are six possible ways to choose your 5 numbers from the 6 drawn, e.g. if the actual numbers drawn are 1,2,3,4,5,6 your winning five could be:

1,2,3,4,5 1,2,3,4,6 1,2,3,5,6 1,2,4,5,6 1,3,4,5,6 2,3,4,5,6

Now, 49!/(5!(49-5)!) = 1906884 but there are six possible combinations of five balls in the 6 actually draw, so we need to divide by 6

1906884/6 = 317814

which by strange coincidence is equal to 13983816/44

Not a coincidence, because the difference between

49*48*47*46*45*44/(2*3*4*5*6) for a 6 numbers draw and 49*48*47*46*45/(2*3*4*5) for a 5 numbers draw is exactly the factor 44/6 !

There is a slight difference. By keeping 3 (or more) numbers the same in every ticket, it means that if those 3 numbers come up you will win something with every ticket. It does not of course increase the odds of winning, but it may give an extra sense of anticipation to know that you will win a reasonable amount by matching only 3 numbers. As such it may for some people afford a little extra fun.

Easier to demonstrate the fallacy by simply looking at the odds of winning with 44 random tickets, which would be 14 mill divided by 44, which is exactly the same odds as with having the 44 tickes following his "system"

He appears to be saying that having 44 random tickets would give you the same odds of winning as having just 1 ticket, and from that fallacy can easily demonstrate that buying 44 tickets gioves you a better chance of winning. Duh!

"Cynic" wrote

If that were true, then the odds of winning with 14M random tickets would be 14M divided by 14M = 100%. But you are *not* guaranteed to win the jackpot, if you buy 14M *random* tickets!

[You might have, randomly, missed that particular selection of 6 numbers...]

Which is perfecty true

Wrong

Whilst true, in this case popular opinion is correct and it is you who are wrong.