Well, cynic is somewhat right to a high degree, since 44 is much less than 14mill.
And yes, you're also right since the probability of winning when buying 14M random tickets is not assured, but is ~ 1-1/e ~ 0.632 . I love nitpicking :-)
I was *more* right than Cynic!
"johannes" wrote
;-)
That is not how the maths of probability works at all.
Having a *probability* of 100% does not mean that you will win every time, just that the average number of wins over a large number of attempts will be equal to the number of attempts. That includes the situations where more than one ticket wins in a draw.
Yes, I am aware that the mechanics of the lottery mean that if you win multiple times the prize is divided by the number of wins, but that is not relevant wrt the probability of winning.
If you wish to exclude the double-win situation, you need to introduce the additional rule that no two random selections may be the same. The 100% average will then become a 100% certainty.
"Cynic" wrote
I know - that's why I said: " **IF** that were true ..."
"Cynic" wrote
You must need to brush-up on your basic probability. It means *exactly* that - a probability of 1.0 (aka "100%") *is* a certainty.
"Cynic" wrote
You are confusing averages with probabilities. Think a little about what you are saying!
"Cynic" wrote
Exactly. The probability of winning is only ever 1.0 (100%) if you have covered all 14M-odd possible selections of 6 numbers. Only in that case is the probability 100% and then it *is* a certainty that you will share the jackpot.
"Cynic" wrote
With that additional rule, it would **NOT** be random!!
Incorrect, I'm afraid! Try statistics, combinations and permutations.
CRF
>
That's my problem too - I always pick the correct numbers, but the lottery machine never agrees with me. :)
On a more relevant note - in 12 years, I've had 4 numbers come up only once (plus once as part of a syndicate).
If you want I will sell you my old tickets for 1/2 price. Your chances are winning are not much different plus you'd have saved 50.
I also had 4 numbers, but only once, although it happened on a single ticket play. It only netted £65. In fact the ~45% payout pot is structured such that more money is concentrated on 3-macthes and 6-matches (jackpot) at the expense of a relative hole in the middle for 4,5,5B payouts. Presumably, the 6-matches is the big carrot, and the 3-matches keeps players sucked in.
yes it does.
Quite correct;-) The only certainty that I know of is that in the future "I'll pop me clogs"
True;-)
True! Or win outright;-)
Well done;-)
CRF
It is *not* incorrect.
With one ticket your chance of getting all six numbers right is
1 in 14M and therefore the chance of losing is 13999999 in 14M.Of course, in this context, by "losing" I meant "not getting all six numbers right". I know you can also win lesser prizes by getting fewer than 6 right, but I was disregarding those.
Also I was using 14M only as an approximation to 13983816 and so for 13999999 please read 13983815 etc.
With two (different) tickets the chances of winning double to
2 in 14M and so the chances of losing reduce to 13999998 in 14M (or 13983814 in 13983816), which is a reduction from "almost certain" to "almost certain".
It can be random - it depends on how you define picking random tickets.
If you want 13M tickets and just pick random 6 from 49 multiple times you will almost always end up with duplicates.
However, if you pick 13M tickets from the set of 14M tickets then you will not end up with duplicates.
The four bridge hands are random but never have a duplicate card. Cutting a pack 13 times for each hand can (and almost always will) end up with duplicates. (I'm assuming here that you don't remove the card, just write down what was cut)
Tim.
That applies to finite sets. (Most games involve finite sets). If you have infinite sets, such as e.g. all real numbers in the interval [0,100], then it becomes more complicated.
Picking a real number randomly in [0,100], the probability of hitting one particular number, e.g. 12.345 is zero!
So the probability of not hitting 12.345 is one! Although it would still be possible to hit that number. So in this case, probability one does not mean certainty.
As long as you sell me the half that goes to 'good causes'.
If you play a finite number of times, it means your probability of winning every time is 1. In general (but not in this example) it still might be possible for you to lose.
A mathematician would say you win 'almost surely'.
E.g. the probability of not picking an irrational number if you pick at random from all the real numbers in the range [0,1] is 1. So you pick an irrational 'almost surely'. But it's still possible that you might pick a rational. The probability is exactly zero, but it's within the rules of the game.
Peter
And in this case, were you buy 14m random tickets, your probability of winning is about 63%.
Not necessarily. A drug dealer could buy all 14m tickets with cash at different shops. He is then guaranteed to win the jackpot, which of course will be a lot less than £14m, but he has converted his dirty drugs cash into clean lottery winnings.
"johannes" wrote
Yep.
"johannes" wrote
Agreed!
"johannes" wrote
How would you go about devising a system to do that, though?
For instance - would it be a digital system (such as using a digital computer), or an analogue one (such as sticking a pin into a numberline)?
I'm not sure that a digital system could cope with an infinite number of possible outcomes. So it might need to be an analogue system...
"johannes" wrote
Yep.
"johannes" wrote
Yep.
"johannes" wrote
Is it, though - really? ;-( How would you *know* ("for sure") that you'd picked
12.345000000..., and not (say) 12.345000000100000... or 12.344999999900000..., with an analogue system?I mean, there you are with your pin stuck in the numberline - how do you discover what number it actually hit? You might think it looks quite close to 12.345. And when you look closer it's still definitely within 12.344999 and 12.345001. When you look even closer it's still definitely within 12.344999999 and 12.345000001 ... BUT when you look many times closer still, you see it's actually 12.345000000000000312...
Or if it was a digital system (so you *can* know you picked exactly 12.3450000..), then how did you ensure that all members of the infinite set had a chance of being picked?
"johannes" wrote
Instead of thinking that the probability of not hitting 12.345 being one means that it's certain not to hit it, think of it as meaning that you are certain not to *know* ("for sure") you picked it!
"Peter Robinson" wrote
Ah, but perhaps *I'm* a mathematician!
"Peter Robinson" wrote
Please see my reply to johannes' post.
"Jonathan Bryce" wrote
To get the probability of 100% mentioned, you'd need to cover *all* 14M-odd possibilities with your
14M-odd tickets. In that case, you *would* win.BeanSmart website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.