any mathematicians about?? Lottery Odds

[sundial]

And what are the odds on a person (e.g. me!!) getting two numbers for four weeks on the trot?

:evil: :twisted:

3249 to 1 ... ie 7.55?

Cheers

dunk

[gingerboy]

There is a higher chance that you will be seriously injured in a road accident than win the lottery jackpot ... They are both extreme chances at the extreme tail ends of the distribution ... probably in a considerably less than a "permilletile" part of the ogive.

Cheers

dunk

I was knocked down by an on-call fireman when I was 6, does that mean that, I, statistically, have a better chance of winning the lottery as I was fortunate enough to be hit by a car travelling at high speed and survive? :lol:

On reflection though, the 3 weeks in intensive care I have no recollection of and the further 6 weeks in hospital and then having to sleep on the floor for 3 months were really no fun so I think I deserve to win even if it was only a tenner..... best start buying lottery tickets as I don't play at the moment. Complete waste of money as far as I am concerned.

Cheers

Paul

[redmonaco]

Is it better to have 7 tickets in a 14m -1

or

1 ticket in a 2m to 1 ?

Neither.

I think you must have missed my point. It's exactly the same odds.

I understand that it seems the same but can't get over this bit:

if you have 7 tickets in a 14m to 1 shot then you have 7 chances of winning and 13,999,997 chances of losing.

if you have 1 ticket in a 2m to 1 shot then you have one chance of winning and 1,999,999 chances of losing.

So therefore you have 12,000,000 less chances of losing in the smaller lottery.

Which makes me believe that the second way is far better.???

I'm hoping that someone can explain why they are equaly as good bets even though they dont appear to be...

[Pugwash]

I understand that it seems the same but can't get over this bit:

if you have 7 tickets in a 14m to 1 shot then you have 7 chances of winning and 13,999,997 chances of losing.

if you have 1 ticket in a 2m to 1 shot then you have one chance of winning and 1,999,999 chances of losing.

So therefore you have 12,000,000 less chances of losing in the smaller lottery.

Which makes me believe that the second way is far better.???

I'm hoping that someone can explain why they are equaly as good bets even though they dont appear to be...

Wow, I don't know where to start. :)

[GraniteQuarry]

I understand that it seems the same but can't get over this bit:

if you have 7 tickets in a 14m to 1 shot then you have 7 chances of winning and 13,999,997 chances of losing.

if you have 1 ticket in a 2m to 1 shot then you have one chance of winning and 1,999,999 chances of losing.

So therefore you have 12,000,000 less chances of losing in the smaller lottery.

Which makes me believe that the second way is far better.???

I'm hoping that someone can explain why they are equaly as good bets even though they dont appear to be...

Yes, but you're talking about buying SEVEN tickets in the 2m lottery so multiply the 1 in 1,999,999 chances of losing by seven - same overall position as the one ticket in the 14m lottery.

:)

[Glamdring]

I usually find probability to be counter-intuitive so I just give up. The odds of my getting the answer correct are about 14m:1 against...

[JasonG]

:sigh:

:roll:

[jools]

Nobody mention the three door question :blackeye:

[darrenw]

Is it better to have 7 tickets in a 14m -1

or

1 ticket in a 2m to 1 ?

Neither.

I think you must have missed my point. It's exactly the same odds.

I understand that it seems the same but can't get over this bit:

if you have 7 tickets in a 14m to 1 shot then you have 7 chances of winning and 13,999,997 chances of losing.

if you have 1 ticket in a 2m to 1 shot then you have one chance of winning and 1,999,999 chances of losing.


So therefore you have 12,000,000 less chances of losing in the smaller lottery.

Which makes me believe that the second way is far better.???

I'm hoping that someone can explain why they are equaly as good bets even though they dont appear to be...

Sorry Pugwash, but you're wrong. They are completly different odds; 7 chances in 14m, compared to 1 chance in 2m - big, big difference (although let's be fair, you don't stand much chance either way).

Redmonaco, your workings are correct - you do stand a much better chance of winning if you buy 1 ticket in 2m.

However, you're missing an important factor, which is how much you stand to win. Let's assume that in both cases tickets cost £1 and that all the money is paid to the winning ticket holder. So if you win in the 2m draw you get £2m, but if you win in the 14m draw you get £14m.

If you want the best odds for winning, bet on something that gives you odds of 1 in 2 (footie, boxing, black/red in Roulette etc) just don't expect to retire on your winnings :wink:

The beauty of the lottery is that you could retire on an investment of £1, but it's probably best to have a plan B in life :D

So the question isn't just about the odds, it's about how much you stand to win.

[Pugwash]

Sorry Pugwash, but you're wrong. They are completly different odds; 7 chances in 14m, compared to 1 chance in 2m - big, big difference (although let's be fair, you don't stand much chance either way).

... and this is why Casinos are still in business.

7:14,000,000 is identical to 1:2,000,000.

[darrenw]

Sorry Pugwash, but you're wrong. They are completly different odds; 7 chances in 14m, compared to 1 chance in 2m - big, big difference (although let's be fair, you don't stand much chance either way).

... and this is why Casinos are still in business.

7:14,000,000 is identical to 1:2,000,000.

Oh dear, it's just not. You can't just divide 14m by 7 (you're confusing division with subtraction).

Redmonaco summed up the odds perfectly with his expanded workings, as follows...

if you have 7 tickets in a 14m to 1 shot then you have 7 chances of winning and 13,999,997 chances of losing.
if you have 1 ticket in a 2m to 1 shot then you have one chance of winning and 1,999,999 chances of losing.
So therefore you have 12,000,000 less chances of losing in the smaller lottery.

[JasonG]

:sigh:

:roll:

If you buy one ticket then the odds of picking the jackpot numbers are 1 in 14000000 (approx).

If you buy seven tickets with different sets of numbers (not necessarily all different numbers, just at least one different) then each ticket has odds of 1 in 14000000 (approx) of being the drawn jackpot numbers; you now have odds of 7 in 14000000 of being the holder of the/a winning ticket. 7 in 14000000 (approx) is the same as 1 in 2000000 (approx), end of.

It's simple to see...just decide to keep increasing the number of (different) tickets you buy until you get to 13983816; you will then have a chance of 13983816 in 13983816, ie: one.

To combine the odds of uncorrelated, discrete events you ADD the odds. To blithely state that you cannot evaluate 7/14000000 to get 1/2000000 is simply wrong.

As I previously explained, to buy seven tickets covering seven numbers, "to permute six of seven", gives you seven chances of winning the jackpot so you have a one in two million chance of any ticket winning. However, the returns are improved since any non-jackpot wins are likely to be repeated.

Okay?

I know that the maths around probabilities are remarkably confusing and difficult to understand but I would recommend that anyone who is interested should set up their own test lottery with a spreadsheet or pencil and paper (and probably a calculator which has nPr and nCr) and try out smaller systems. You'll get the idea pretty quickly.

Tra

Jason

[darrenw]

OK Jason, I should requalify, I was talking about winning the jackpot, but I accept that it's far more complex if you're considering all winning combinations, regardless of payout size.

[Pugwash]

Oh dear, it's just not. You can't just divide 14m by 7 (you're confusing division with subtraction).

Redmonaco summed up the odds perfectly with his expanded workings, as follows...

if you have 7 tickets in a 14m to 1 shot then you have 7 chances of winning and 13,999,997 chances of losing.
if you have 1 ticket in a 2m to 1 shot then you have one chance of winning and 1,999,999 chances of losing.
So therefore you have 12,000,000 less chances of losing in the smaller lottery.

Let's do it with smaller numbers.

My small lottery has 14 chances of winning and next-door's small lottery has 2. The top prize is the same and the cost of seven tickets in my lottery is the cost of one next door. Would you buy 7 tickets in mine or one ticket next door?

With mine, you have 7 losing numbers. Next door you have one.

Both lotteries have a 50:50 chance as you have 7 chances of winning and 7 chances of losing in mine and 1 chance of winning and one chance of losing next door. Same odds.

We both decide to have bigger lotteries. My medium lottery has 140 tickets and next door has 20. You buy 7 tickets at mine and one next door. You have 133 chances of losing at mine and only 19 next door. You will win my lottery seven times out of the 140 draws and next door you will win once every 20 draws. If you do next door's lottery seven times, you'll most likely lose 133 times. Both are 1-in-19.

If you extrapolate this to 14 million and 2 million, it's the same.

[redmonaco]

I think I understand the workings now, many thanks to those that replied.

It was not so much about the National Lottery (I only do 1 lucky dip per week!!) but it was that that sparked the initial quandry for me.

The thing with this math of course is it assumes equal distribution. The sand or ping-pong ball analogy points to that there is equal distribution throughout and so, in this case, there would be one ball amongst every 'set' of 2,000,000 balls (or grains of sand). Of course though, the universe is not like that and it is quite possible (dont know what the formula is for that one... :shock: ) that all seven balls, or grains of sand, are grouped together at the bottom of the barrel.

After all, a 14,000,000 to 1 shot can come up the first time, or after 14,000,000 times or more...

I see the laws of 'chaos' saw fit for an English couple to winn £25million on the Euro lottery this week with odds of 76,000,000 to 1!!!!!!!!!!!!!!!

Which introduces another factor in to the original argument, can Chaos and probability be combined???

[rfrazier]

Which introduces another factor in to the original argument, can Chaos and probability be combined???

Sure. Take a chaotic system to be a deterministic one described by simple equations, but where slight differences in initial conditions can have significant effects. Now these systems can produce short term disorder, but long term order. This long term order can be described probabilistically. Indeed, one might take the basis of probability to be an underlying chaos.

Best wishes,
Bob

PS All this talk of the lottery got me dreaming of riches beyond my wildest dreams (well, actually not, as some of my dreams are exceedingly wild). So, I bought not 1, but 2 lottery tickets, thereby doubling my chance of winning. I am pleased to report that my two sets of lucky dip numbers didn't have any numbers in common. I am less pleased to report that out of those 14 numbers, only one came up in the lottery.
RLF

[Pugwash]

A simple way to look at it is a roulette table. If you put 6 numbers on (1-6, for instance), you have 31 losing options, if you put six chips on a group of 6 numbers (first six, for instance) on you have 5 groups that you'll lose on, plus the zero. 6 losing options. However, it's exactly the same odds, same wins, same losses. It's the exact same play, just described differently.

Does that make sense now?