The probability of matching a given set of winning numbers would be lower or higher, depending on the numbers included in that winning set, but the expected value of total winnings for any ticket bought before the drawing would be the same.
So, big whoop, it doesn't matter.
But for the group whose numbers were supposed to have had random numbers on their tickets, it should matter.
Since the winning numbers are picked randomly, those whose numbers were not picked randomly must have a less chance of winning - especially since 2 of the winning numbers were not picked as often as they ideally should.
What were the numbers drawn for the period of time when the glitch was in action?
The example above: 3 - 8 - 10 - 12 - 13 - 18 - 23 is from the period where the generating RNG had a glitch.
Hang on: there are TWO random number generators to talk about. The first one (let's call it the "vendor") is making up an auto-ticket to sell to someone. The second one (let's call it the "drawing") is picking a single number which is published in the newspaper.
Exactly. The problem is not with the drawn numbers in the lottery, but with the numbers picked by the RNG for those who just wanted a pre-filled out ticket.
Suppose the drawing machine picks a number A, uniformly, from 1-40. If the vending machine is malfunctioning, and spits out the number B = 1 every time, then the odds of A=B are 1/40.
Suppose the vending machine is semi-malfunctioning, and spits out B=1 or B=2, but never anything else. You have a 1/80 chance of getting B=1 and A=1 (win!), plus a 1/80 chance of getting B=2 and A=2, for a total chance of winning of 1/40. You have a 1/80 chance of getting A=1 and B=2; ditto for B=1 and A=2, and a 38/40 chance of getting A > 2. So your overall odds of winning are ... 1/40.
Suppose the vending machine is not malfunctioning, and picks B uniformly from 1-40, then your odds of winning are again 1/40.
The only way the odds change are if *both* the vending machine *and* the drawing machine are biased. If the drawing machine always picks A=1, but the vending machine always picks B=40, then you never win. If the drawing and vending machines both pick A < 10 and B <10, then your odds are 1/10 rather than 1/40.
There's also the issue of splitting the prize. In a lotto with 1e6 independent players picking from 1e8 number-combinations, you have a 1e-8 chance of winning the prize, and (if you win) a 1e-2 chance of sharing it with someone. In a lotto with 1e6 players and 1e7 number-combinations (if the vending-machine skipped 90% of the choice space), you still have a 1e8 chance of winning the prize, but now a 10% chance of sharing it with someone.
Sure, but this isn't a case where the numbers 1-9 weren't picked at all. Those numbers were picked less often than the rest (10-36).
However you arrive at the numbers on your ticket, the chance it will win is the same.
Even if the ticket printer in the shops always printed 1234567, your chance of winning would be no different from a truly random system. If you did win, however, you would be more likely to end up sharing the prize with other people who used the same machine.
Why will it be the same?
What you're forgetting (presumably) is that there are two random things happening here.
But here you say that
If you mean that even when generating the number that actually wins their RNG is skewed then there's a problem. If the only effect is that automated filled-out coupons didn't select the numbers 1-9 as often as it should, then your chances remain the same.
Here's a simple thought experiment to prove that: if the winning number is chosen perfectly randomly, then if picking 1-9 less often decreases your odds of winning from random chance, necessarily picking 1-9 more often increases your odds, which is ridiculous.
Why is it ridiculous? If you have a RNG that in 50% of the time picks 1-7, why aren't your odds of winning increasing by playing 1-7?
But we're talking retrospectively, surely? So in the period of the glitch, you were less likely to get a ticket with the numbers 1-9 on it. However, the winning tickets during that period might have had 1s, 2s, 3s, to 9s on them. So if you'd bought you're ticket via the glitched system, surely you were at a disadvantage?
That's what I'm thinking, yes.
Your chances of getting the winning ticket = same.
Why?
Once drawn, however, if it contains a number you were less likely to have had, surely you can argue that your chances of getting that specific winning ticket were not the same?
Someone please correct me, I do NOT find probability in any way intuitive!!!!
That's why I am asking for some math here.
At the point at which the tickets were brought they had an equal probability of winning. That's all you can ever say, judging any selection method on additional knowledge from after the event will skewing your reasoning.
By the same logic, you could make the claim that picking 1 2 3 4 5 6 biased the lottery against you after they didn't come up.
Why do they have an equal probability, if some of the numbers on the coupons are not chosen as often as the rest?
PonderingTurtle was referring to the method for drawing the winning number.
OK. But we are talking about the method for drawing the numbers on the automated tickets.
Yep.
However, their hard luck would have more sympathy if they hadn't picked the numbers themselves, but left it to a biased machine.
Which they did, in this case.
I suppose that if people are paying you to indulge their fantasies about how "luck" works, then that's what they are paying for. For example, if you went to a casino to play roulette, and the croupier told you "Sorry, the wheel is broken today---instead, I'll be reading off numbers from this pseudorandom list I generated this morning" --- well, a lot of people would walk away. They wanted to enjoy their incorrect notions about how the wheel behaves.
Yeah: But in this case, the issue is that the numbers picked for them which they thought were picked randomly, wasn't. Some numbers were picked less often than others.
In your example, the list generated this morning would have 1-9 less often than 10-36. If people were expecting that all numbers were picked evenly random, they would be cheated.
Perfect: 42,072,307,200:1 =
1st no = 36:1 * 2nd no = 35:1 * 3rd no = 34:1 * 4th no = 33:1 * 5th no = 32:1 * 6th no = 31:1 * 7th no =30:1 =
36*35*34*33*32*31*30=42,072,307,200
Glitched: (assuming a consistent glitch!) 44,476,439,040:1 = a guess based on your original assessment of the glitch.
No math? Just a guess?
Well, more than that. They are paying to have those notions entertained. Buying the lucky dip ticket is a contract, the terms of which presumably state that the numbers are generated randomly with equal probability.
Exactly. That's the idea of having the automated system: You expect to get an evenly distributed set of numbers picked for you, when in fact you didn't.
Meh, I'm clutching at straws and for no real reason as I hate the lottery with a passion, but it does seem to me that the game people paid for is not the game they got, regardless of the fact that their chances of winning were the same.
Well, that's my question. Why is it the same?
As per my coin toss analogy - if you were told that you were buying a lucky dip ticket that might have heads or tails on it, that's the game you pay for. If it turns out that all the lucky dip tickets had tails on, then that's not the contract you entered into.
The same applies, when the lucky dip tickets had less tails than heads on it.
Having said all this, I daresay the small print of the lottery machine says something like "we are not responsible if it's skewed".
It doesn't.
Lets say you only need to get one number correct for the lottery.
What are the chances that the final draw will be a number from 1 - 9?
9/36
What are the chances of you buying a ticket which contains a 1- 9?
0
This would mean that you have a 9 out of 36 probability of it being impossible to win.
When the draw is a number above 9, your chances increase because your selection pool was smaller, every 27/36 draws you have a 1 in 27 chance of winning instead of 1 in 36.
This then mitigates the 9/36 times that you have a 0% chance, So therefore it does not make any difference if you choose your own number or get a electronically generated one. (maybe
)
The same principle would apply with more numbers.
But it isn't a case of the numbers 1-9 never being picked by the generator. It's a case of the numbers 1-9 being picked less often than 10-36.
Well, I don't know if there is an explicit contract. The OP says that the system is automatic and that they are now claiming their generator didn't affect your chances of winning. If that's the extent of their promise they've lived up to it.
Wait, wait: All players aren't using the automated system. Only some of them are.
The "lottery picker" will always have odds of 42,072,307,200:1 for any sequence of 7 numbers of 36, but the "random generator", if glitched, will not. So for the example given, ( 3 - 8 - 10 - 12 - 13 - 18 - 23), the odds are 42,072,307,200*(44,476,439,040/42,072,307,200):1 for you if you chose the lucky dip.
That's how I see it, too. I just want to know how you got to the 44 billion number.
That's not quite what happened in this case, but actually your odds of winning are the same.
There are (27 choose 7)=888030 ways for your flawed drawing to come out, but you will be picking one of (36 choose 7)=8347680 values. If you want to break it down consider case 1: you are lucky and don't pick 1-9 and case 2: you pick 1-9 and are lost from the beginning.
Odds of case 1 occurring:
(888030 / 8347680)
Odds of case 2 occurring:
(8347680-888030)/8347680
Odds of winning in case 1:
1 / 888030
Odds of winning in case 2:
0
Total odds of winning:
(888030 / 8347680) * 1/888030 + (8347680-888030)/8347680 * 0 = 1/8347680
If you had some way of knowing that 1-9 were not being drawn you would dramatically increase your chances of winning of course, which is why that's a much more important problem.
What if 1-9 were not being drawn as often as 10-36?
I think what Teek was getting at is that the ticket is a form of contract, not only guaranteeing specific odds of winning, but specific odds of payout possibilities.
Yes: Those who choose the generated numbers expect that the numbers are picked evenly random, because the balls are also picked randomly.
If there are ten numbers in the drawing, my odds are 10:1, but I'm also entering into this knowing that not only are everyone else's odds equal, but the odd of hitting the same numbers and sharing the pot are also equal.
So, 10 people are assigned random numbers, making everyone's odds 10:1. However, we don't use the numbers 1, 2, or 3 in our random generation, so everyone has an increased chance of having the same number as someone else. This is definately not part of the contract.
Say we only use 6-10. Of the ten people, all of them share a number with one other person, meaning thier odds are the same, but their payout is halved.
More simple; if we flip a coin, with two contestnts ranomly assigned heads or tails, we can reasonable expect that we have a 2:1 chance of winning, and a 2:1 chance of splitting the pot. So, person A and person B in a fair coin toss will have eight possible outcomes, where A is A, B is B, and F is the Flip, and we're looking at A's winnings, betting 2 dollars each time:
A B F
H H H Win half (1/2)
H H T Lose (0/2)
H T H Win Full (2/2)
H T T Lose (0/2)
T H H Lose (0/2)
T H T Win Full (2/2)
T T H Lose (0/2)
T T T Win half (1/2)
So, for every 16 dollars bet, A should win back 6. These are the odds A was entering the contract for.
Now, our random generator only assigns Tails and not Heads. Again playing 8 games and betting 2 dollars, A can be expected to win...
A B F
T T H Lose (0/2)
T T T Win Half (1/2)
T T H Lose (0/2)
T T T Win Half (1/2)
T T H Lose (0/0)
T T T Win Half (1/2)
T T H Lose (0/0)
T T T Win Half (1/2)
This time, A only won 4 dollars for every 16 dollars spent. A won just as frequently (four wins out of 8 flips) but those wins alway shared the pot. This was most definately not the contract we entered.
Was that what you were getting at?
Not quite. The numbers 1-9 are picked less often, they are not never picked.
The main assumption I made is that the people who pick their own numbers pick evenly distributed combinations. As andyandy pointed out this isn't a very good assumption on my part. People do pick low numbers more often, because they play birth dates, lucky numbers and such. While any combination of numbers is equally likely to win (assuming the draw itself is fair), you want to choose a combo that no one else has picked so you don't share the pot.
So if self pickers were good at actually not favouring some numbers over others, a slight bias in the automated system would actually increase your chance of sharing a pot. However, because people tend to be biased towards lower numbers, the fact that the automated system is biased decreases your chance of sharing your pot with some of the self-pickers.
Yes, that's a good point.
