Since every combination of six numbers (however selected) 
has the same probability of winning, there is no difference 
between the odds for the two punters. In particular, the 
numbers don't need to be randomly selected by the 
punter from the full set of 49, as long as they are 
randomly selected in the lottery draw.

However, if one wishes to use one's knowledge of statistics 
to maximise the ratio of potential investment to outlay, 
the paper 'Using maximum entropy to double one's expected 
winnings in the UK National Lottery' by Cox et al (1998) in 
The Statistician 47, Part 4, pp. 629-641 is worth a read. 
Unfortunately, the greater the number of people who take up 
their advice, the less worth the advice is. In fact, if the 
numbers 1, 41, 43, 45, 46 and 49 ever do come up it would 
be interesting to see how many statisticians win the 
jackpot. I think I would have kept it quiet.

Andy Scally 

On Wed, 1 Sep 1999 14:14:09 +0000 
[log in to unmask] wrote:

> Dear All,
> 
> A colleague posed this question, and since I don't play 
> the lottery (being a statistician) I thought I'd pose it generally. 
> Apologies if it is well known. 
> For non-lottery players - You choose 6 
> numbers between 1 and 49 inclusive for one pound. If you match all 6 
> with those generated by a random number machine you win the jackpot. 
> Ignore payouts for less than 6 matches, and  the bonus ball. Two 
> punters each spend  5 pounds per week. Punter 1 has a bag with 49 
> numbered balls in it, and after choosing 6, replaces them in the bag 
> for the next go. Punter 2 has the same bag, but does not  replace the 
> balls before each new draw.
> 
> Who has the best chance of winning?.
> 
> One way of arguing goes thus:
> The chance of winning on one go is about 1/14000000.
> Thus  Punter 1's chance of winning on 5 goes is about 5/14000000.
> 
> Punter 2 has a chance of 1/14000000 of winning on the first go. On the
> second, if any of the winning numbers appeared in the first set, then
> the chance of winning with the second and subsequent sets is zero.
> Thus the chance of winning with the second set is less than
> 1/14000000. Similarly with the 3rd, 4th and 5th. Thus the overall
> chance of winning is less than 5/14000000 and so Punter 1 has a better
> chance.
> 
> 
> 
> However, another way of looking at it is that the above imposes an
> artificial ordering on the numbers. The lottery  has no interest in
> how or in what order the sets were generated - all it sees are five
> distinct sets of 6 numbers, each with a 1/14000000 chance of winning.
> Since there is a small, but finite chance of  Punter 1 drawing the
> same set of 6 numbers more than once in the 5 draws, thus Punter 2 has
> a (slightly)  better chance. Of course in reality Punter 1 would not
> enter the same set of numbers twice, so the two methods are equal. 
> 
> I favour the latter but my colleague (being of little faith) would
> like a more expert opinion.
> 
> Mike
> 
> 
> Mike Campbell
> Professor of Medical Statistics with an Interest in Primary Care
> Institute of Primary Care
> Community Sciences Building
> Northern General Hospital
> Sheffield S5 7AU
> Tel 0114 271 5919
> FAX 0114 242 2136
> email [log in to unmask]

----------------------
Andrew John Scally
[log in to unmask]



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