The other flaw in Punoqllads’ argument is that he’s only considering single-number overlaps, but overlaps don’t matter until you get to four-number overlaps (since the lotto doesn’t give a prize for 1, 2, or 3 matching).
BTW, in my previous post I didn’t mean for the word “scheme” to come across in any perjorative sense. I probably should have used something like “plan”. Sorry about that.
Chronos, great signature line. Did you get the name Chronos from that Johnny Bravo episode with the angry bear?
Africanus, my loss is NOT guaranteed. With those two tickets I will always win. My neighbor who picks two tickets randomly is NOT guaranteed to win. So what’s his probability of winning? Can someone with a probability of 100% of winning still lose?
Y’all are forgetting I’m looking for the probability of the SET of two tickets, because you keep bringing up the probability of each individual ticket.
Oh, and the ratio of my neighbor’s wins to losses is not 50-50, it’s 11/20.
You will always walk away from a lottery drawing with one winning ticket and one losing ticket. If walking away with at least one winning ticket regardless of how many losing tickets you have is your goal, then you will obtain that goal.
If the lottery draws three numbers without replacement from 1 through 6, and you have two tickets with 1,2,3 and 4,5,6, respectively, then one will always win and one will always lose.
I’m not saying that you won’t have at least one winning ticket, I’m saying that you’ll have one winning ticket and one losing ticket. You simply can’t have both tickets win with the way it is set up. That’s all.
How did you calculate your neighbor’s odds?
Rusalka, your loss is guaranteed. One of your tickets will always lose. You will always have exactly one winner and one loser. You have paid twice as much as someone who has only purchased one ticket, and have twice the expected number of winning tickets as them. They have 1/2 of a winning ticket, and you have 1 winning ticket.
Your neighbor who, say, picks one number to use on both of them, but picks the other four numbers randomly, on the other hand, is not guaranteed to have one winner each time and is not guaranteed to have a loser each time. He might have zero, one, or two winning tickets.
The chance that he will have n winners is:
0: choose(3,1)*choose(3,2)2/3choose(2,1)*choose(1,1)/(choose(6,3)*choose(3,2))
1: choose(3,3)/choose(6,3) + choose(3,2)*choose(3,1)2/3choose(1,0)*choose(2,2)/(choose(6,3)*choose(3,2)) + choose(3,2)*choose(3,1)1/3choose(1,1)*choose(2,1)/(choose(6,3)*choose(3,2)) + choose(3,2)*choose(3,1)1/3choose(1,0)*choose(2,2)/(choose(6,3)*choose(3,2)) + choose(3,1)*choose(3,2)2/3choose(2,2)*choose(1,0)/(choose(6,3)*choose(3,2)) + choose(3,1)*choose(3,2)1/3choose(2,2)*choose(1,0)/(choose(6,3)*choose(3,2)) + choose(3,1)*choose(3,2)1/3choose(2,1)*choose(1,1)/(choose(6,3)*choose(3,2)) + choose(3,0)/choose(6,3)
2: choose(3,2)*choose(3,1)2/3choose(1,1)*choose(2,1)/(choose(6,3)*choose(3,2))
which simplify to
0: 1/5
1: 3/5
2: 1/5
So, your neighbor will win nothing 1 time out of 5, and will have one winning ticket 3 times out of 5. But, 1 time out of 5, your neighbor will have 2 winning tickets, for a total expected value of 1 win every time, just like you. Your varience is lower, however, because you will win once every time, while your neighbor has a chance of winning more than once every time, but also less each time, too.
If your neighbor picks every number randomly the odds are:
0: (choose(3,0)*choose(3,3)+choose(3,1)choose(3,2))[sup]2[/sup]/choose(6,3)[sup]2[/sup]
1: 2(choose(3,0)*choose(3,3)+choose(3,1)choose(3,2))(choose(3,2)*choose(3,1) + choose(3,3)*choose(3,0))/choose(6,3)[sup]2[/sup]
2: (choose(3,2)*choose(3,1)+choose(3,3)*choose(3,0))[sup]2[/sup]/choose(6,3)[sup]2[/sup]
Which simplify to:
0: 1/4
1: 1/2
2: 1/4
Again, still an expected return of one winning ticket each time.
Chronos, as the above system has overlaps, you can see that the expected value remains the same even in those situations.
The point I’m trying to make is that I’ve heard lots of smart people say that they have some way of “beating” a lottery. They have some system of increasing their odds of winning a prize over what just random guesses would get them. But they haven’t worked out the math, and because the sample size is so low, compared to their odds of acutally winning something, that they can’t reasonably say that their win/loss ratio shows whether or not their system works. But if they did work out the math, they would find that all their system does is alter the expected variance, not the expected return.
The only way to increase your odds for winning a lottery is to have inside information on which numbers are going to win. The only real way to get an edge without cheating is to reduce the odds that someone else picks the same number as you, such as only picking numbers 32 and larger so that all of the people who use dates, times, or names to generate their tickets’ numbers won’t ever pick your numbers.
Which, incidently, is something that the master also says.
Please don’t use the term “variance”, it’s confusing because we’re not sampling (or at least I wasn’t). Also, I don’t understand the convention you use to prove the math. (but that’s o.k.)
However, one key phrase clarified the whole issue for me.
“1 time out of 5, your neighbor will have 2 winning tickets, for a total expected value of 1 win every time, just like you. Your varience is lower, however, because you will win once every time, while your neighbor has a chance of winning more than once every time, but also less each time, too.”
That winning more than once within a set was the missing piece of the puzzle for me, but again you are forgetting something: The vagaries of the Lotto. I’d still rather pick the sure bet, because even if my neighbor’s set of two randomly chosen tickets wins twice in one game, HE DOESN’T GET A FULL SECOND PRIZE. At most he’ll get a some fraction more (if the pot were 300 dollars and one other person wins at the same time, he’d get 200 dollars instead of the 150 he’d get with one winning ticket)
So his second winning ticket is just a blown chance, (which would make his total chances really less than 5/5 according to your calculations) whereas each of my ticket sets guarantees 1 win, so if I play this game over and over, I would come out ahead… Right?
So according to your scenario over 5 games with no other winners:
My neighbor
loses once = 0
wins three times = $900
wins twice in one game = + $300
Total winnings = $1200
I on the other hand
win every time = 5 x $300
Total winnings = $1500
With two lotto winners it would be:
neighbor:
loses once = $0
wins three times = $450
wins twice in one game = $200
Total winnings = $650
I on the other hand:
win every time = 5x $150 =
Total winnings = $750
Of course this is just one round, but overall I come out ahead.
You don’t need sampling to have a variance, you can just have a mathematical model.
The choose() function I use is ususally written as one number above another, inside parentheses. For two numbers, a & b, choose(a,b) = a!/(b! * (a-b)!). The “!” is factorial notation, defined recursively over all non-negative integers where, if n > 1, n! = n * (n-1)! and if n = 1 or n = 0, then n! = 1. So, 5! = 5 * 4 * 3 * 2 = 60. The choose function is used to determine how many ways you can choose b different items from a set of a items where the order you pick them in doesn’t matter.
So, I divided the set of numbers into two sets, winning numbers and non-winning numbers. If you call those numbers w and n, respectively, then the odds of selecting s numbers and getting exactly k winning numbers is choose(w,k) * choose(n, s - k) / choose(w + n, s).
Well, I don’t think I was “forgetting something”. All lottos I’ve ever heard of behave almost nothing like what you’ve been talking about. Normally, the jackpot is divided up evenly between all winning tickets, and consists itself of a lump sum plus a fraction of the cost of all tickets purchased since the last jackpot was won.
But, for what you’ve described as your lotto, yes, if all you care about is one win for each set, then making sure you don’t repeat numbers will increase your chance of winning something in a given set.
Hi Everyone
Assuming that there are 6 numbers drawn from 42 and for this example you need all 6 to get a prize.
Say person A has 7 tickets all different numbers no duplicates
Person B has 7 tickets, each ticket has the number 1 on it and then 5 unique unduplicated numbers.
Person A after the first number has been drawn (which he definatly has on one ticket) has a 5/41 x 4/40 x 3/39 x 2/38 x 1/37 chance of winning, this is 1 in 749398.
Person B needs the number 1 to come up. Odds of 6/42.
So if that comes up he then needs the other 5 numbers to come up. This is the same as above in exampleA ( 1 in 749398 ). However he has 7 tickets with these odds ( 6/42 x 1/749398 ) so its those odds x 7. Which is 1 in 749398.
Therefore the odds of winning the prize are equal in both cases.
I would assume this is the same for all examples. eg more prizes for less numbers etc. This would be why we have never heard it stated that in multiple tickets you increase your chance with non-duplicated numbers.
Let me know what you think everyone.
Andy UK
p.s. Feel free to point out any mistakes.
Just to add I did a calculation for same scenario except 2 numbers wins and 2 numbers are drawn not six, they still pick six numbers etc. The odds of winning compared to each other were the same. The maths for 6 numbers being drawn is a bit messy and I’m too lazy to do it at the moment. The only way I can see it being different is possibly a scenario where in one you are 7 times less likely to win but if you do you win 7 times for that 1 result. So I conclude it doesn’t matter anyway.
Andy