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Lotto: Statistics question

CFLarsen

Penultimate Amazing
C
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Aug 3, 2001
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As in many other countries, Denmark has a national lotto. Here, you have to guess 7 correct numbers out of 36 possible.

To make it easy, there is an automated system that will select the numbers for you. You basically press a button, and the system spits out 7 numbers out of 36. It's easy, and it doesn't alter your chances of winning.

However: They have recently discovered that their random number generator had a glitch. In some drawings, the number generator for the automated filled-out coupons didn't select the numbers 1-9 as often as it should.

They claim that this didn't influence your chances of winning. Is that right?

With a perfect system, the numbers 1-36 should each be selected 1 out of 36 times. I don't know exactly how less often the numbers 1-9 were selected, but let's assume that, with a glitched system, the numbers 1-9 were only selected 1 out of 37 times, and the rest of the numbers, 10-36, were each selected 1 out of 36 times.

The winning chance for the perfect system (1:36 for all 36 numbers) is (36!)/(36-7)! = 8,347,680.

What is the winning chance for the perfect system, if the winning set of numbers were:
3 - 8 - 10 - 12 - 13 - 18 - 23

What is the winning chance for the glitched system, if the winning set of numbers were:
3 - 8 - 10 - 12 - 13 - 18 - 23

3 and 8 were not selected on the automated coupons as often as they should have been. Therefore, the chances of getting all 7 right must be lower, if you had chosen an automated filled-out coupon.

Right?
 
Well assuming that their method of randomly generating the winning numbers was valid, any set of numbers should be equaly likely to win as any other specific set of numbers.
 
What is the winning chance for the glitched system, if the winning set of numbers were:
3 - 8 - 10 - 12 - 13 - 18 - 23

3 and 8 were not selected on the automated coupons as often as they should have been. Therefore, the chances of getting all 7 right must be lower, if you had chosen an automated filled-out coupon.

Right?
Click to expand...
But, if the the winning numbers didn't contain the numbers 1-9, then you would have higher odds of winning. It follows that if 1-9 were picked less often that expected, then 10-36 were picked more often than expected.

In your example:
let's assume that, with a glitched system, the numbers 1-9 were only selected 1 out of 37 times, and the rest of the numbers, 10-36, were each selected 1 out of 36 times
Click to expand...
You have 9 numbers with 1 in 37 odds, and 27 with 1 in 36 odds. 9*(1/37) + 27*(1/36) = 0.993. As long as the lotto draws are independent, and each number has an equal chance of coming up, then any ticket has the same odds of winning.

The problem that arises from this, is that while your odds of winning are the same, your odds of sharing the pot will likely increase. (ETA: this last statement does make some assumptions, but I am in to much of a hurry to point them out right now).

Walt
 
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CFLarsen said:
As in many other countries, Denmark has a national lotto. Here, you have to guess 7 correct numbers out of 36 possible.

To make it easy, there is an automated system that will select the numbers for you. You basically press a button, and the system spits out 7 numbers out of 36. It's easy, and it doesn't alter your chances of winning.

However: They have recently discovered that their random number generator had a glitch. In some drawings, the number generator for the automated filled-out coupons didn't select the numbers 1-9 as often as it should.

They claim that this didn't influence your chances of winning. Is that right?

With a perfect system, the numbers 1-36 should each be selected 1 out of 36 times. I don't know exactly how less often the numbers 1-9 were selected, but let's assume that, with a glitched system, the numbers 1-9 were only selected 1 out of 37 times, and the rest of the numbers, 10-36, were each selected 1 out of 36 times.

The winning chance for the perfect system (1:36 for all 36 numbers) is (36!)/(36-7)! = 8,347,680.

What is the winning chance for the perfect system, if the winning set of numbers were:
3 - 8 - 10 - 12 - 13 - 18 - 23

What is the winning chance for the glitched system, if the winning set of numbers were:
3 - 8 - 10 - 12 - 13 - 18 - 23

3 and 8 were not selected on the automated coupons as often as they should have been. Therefore, the chances of getting all 7 right must be lower, if you had chosen an automated filled-out coupon.

Right?
Click to expand...

I get:

Perfect: 42,072,307,200:1

Glitched: (assuming a consistent glitch!) 44,476,439,040:1
 
Whatever method was used to choose the numbers has absolutely no effect on the odds of winning. You could choose 1,2,3,4,5,6,7, you could throw darts and only choose numbers between 1 and 20, you could restrict yourself to prime numbers, fibonaci sequences, no matter - your probability of winning is completely unaffected.

It would make a small difference to your expected value (EV) winnings by skewing the number of people with specific numbers - but as this is a greater than 9 skew, and seeing as 1-9 are the most overly chosen numbers and should be normally avoided in fair lotteries when considering EV, then even this may be neglible.....

But the odds of winning are still (36 choose 7) = 8 347 680

I expect a CF argument, but this is incontrovertible.
 
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With a perfect system, the numbers 1-36 should each be selected 1 out of 36 times. I don't know exactly how less often the numbers 1-9 were selected, but let's assume that, with a glitched system, the numbers 1-9 were only selected 1 out of 37 times, and the rest of the numbers, 10-36, were each selected 1 out of 36 times.
Click to expand...
Not a good assumption, since 9/37 + 27/36 < 1.

It wouldn't affect your chance of winning. Hell, printing every ticket with a series of 1s wouldn't affect your chance of winning. It only affects your chances of matching a given set of winning results, which of course you don't have when the numbers are generated.

It would affect the distribution of winning tickets, however.
 
In my new lotto one number will be chosen uniformly from 1-2. You may press a button to get a number but it will give you a 1 with probability 0.4 and a 2 with probability 0.6. Your chance of winning if you choose randomly is:

P(L1)P(G1) + P(L2)P(G2) = 0.5*0.5 + 0.5*0.5 = 0.5

And the button's chances are:

P(L1)P(B1) + P(L2)P(B2) = 0.5*0.4 + 0.5*0.6 = 0.5

ETA: Language question. Why did I use "is" for your "chance" of winning and "are" for the button's "chances"?
 
ponderingturtle said:
Well assuming that their method of randomly generating the winning numbers was valid, any set of numbers should be equaly likely to win as any other specific set of numbers.
Click to expand...

But that's the problem: Their method is not valid, as they discovered that some of the numbers (1-9) came out less often than the rest (10-36).

Walter Wayne said:
But, if the the winning numbers didn't contain the numbers 1-9, then you would have higher odds of winning. It follows that if 1-9 were picked less often that expected, then 10-36 were picked more often than expected.
Click to expand...

Yes, that's what I figured. It wouldn't matter, if the winning numbers were all above 9. But, since two numbers were below 10, the chances would be lower.

Walter Wayne said:
In your example:

You have 9 numbers with 1 in 37 odds, and 27 with 1 in 36 odds. 9*(1/37) + 27*(1/36) = 0.993. As long as the lotto draws are independent, and each number has an equal chance of coming up, then any ticket has the same odds of winning.

The problem that arises from this, is that while your odds of winning are the same, your odds of sharing the pot will likely increase. (ETA: this last statement does make some assumptions, but I am in to much of a hurry to point them out right now).
Click to expand...

I'd like to hear them, once you get back.

Macoy said:
I get:

Perfect: 42,072,307,200:1

Glitched: (assuming a consistent glitch!) 44,476,439,040:1
Click to expand...

It appears the glitch was consistent. Can you show how you got to those numbers?

andyandy said:
Whatever method was used to choose the numbers has absolutely no effect on the odds of winning. You could choose 1,2,3,4,5,6,7, you could throw darts and only choose numbers between 1 and 20, you could restrict yourself to prime numbers, fibonaci sequences, no matter - your probability of winning is completely unaffected.

It would make a small difference to your expected value (EV) winnings by skewing the number of people with specific numbers - but as this is a greater than 9 skew, and seeing as 1-9 are the most overly chosen numbers and should be normally avoided in fair lotteries when considering EV, then even this may be neglible.....
Click to expand...

I don't understand this. The point is that you (in this case, the randomizer) would not choose all numbers equally often. It would seem to me that "you" don't choose 1,2,3,4,5,6,7 or throw darts or whatever. "You" throw darts but only hitting the top numbers.

mumblethrax said:
Not a good assumption, since 9/37 + 27/36 < 1.

It wouldn't affect your chance of winning. Hell, printing every ticket with a series of 1s wouldn't affect your chance of winning. It only affects your chances of matching a given set of winning results, which of course you don't have when the numbers are generated.

It would affect the distribution of winning tickets, however.
Click to expand...

But that's precisely it, isn't it? If the first group of numbers (1-9) were selected less often than expected, the value would be less than 1. Therefore, if your automated filled-out coupon contained any number from 1-9, you have a less chance than if they came out as often as the numbers from 10-36.
 
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.
 
What were the numbers drawn for the period of time when the glitch was in action?
 
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.

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.
 
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.
 
What you're forgetting (presumably) is that there are two random things happening here.

However: They have recently discovered that their random number generator had a glitch. In some drawings, the number generator for the automated filled-out coupons didn't select the numbers 1-9 as often as it should.

They claim that this didn't influence your chances of winning. Is that right?
Click to expand...

But here you say that
But that's the problem: Their method is not valid, as they discovered that some of the numbers (1-9) came out less often than the rest (10-36).
Click to expand...

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.
 
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? Your chances of getting the winning ticket = same. 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!!!!
 
tkingdoll said:
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? Your chances of getting the winning ticket = same. 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!!!!
Click to expand...

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.
 
tkingdoll said:
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?
Click to expand...
If you're going to apply that logic, then every loser in every drawing can claim they had zero percent chance of winning in retrospect because they didn't have the winning combination.
 
RecoveringYuppy said:
If you're going to apply that logic, then every loser in every drawing can claim they had zero percent chance of winning in retrospect because they didn't have the winning combination.
Click to expand...

Yep.

However, their hard luck would have more sympathy if they hadn't picked the numbers themselves, but left it to a biased machine.

ETA: I should clarify that I'm thinking of this in terms of consumer rights/contracts, and also if the 1-9 weren't in the picture at all. I think in the real life situation, anyone daft enough to do the lottery shouldn't be moaning about probability.

It's as if I said "I am going to toss a coin, and you can buy a ticket which will randomly say heads or tails". You buy the ticket, but you aren't aware that they all say 'tails'. I toss the coin. Now, if it comes up heads, everyone loses. If it comes up tails, everyone wins. So it's fair from that point of view, but that's not the contract you entered into and not what you paid for.

Am I making sense or not? It's nearly 1am.
 
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tkingdoll said:
It's as if I said "I am going to toss a coin, and you can buy a ticket which will randomly say heads or tails". You buy the ticket, but you aren't aware that they all say 'tails'. I toss the coin. Now, if it comes up heads, everyone loses. If it comes up tails, everyone wins. So it's fair from that point of view, but that's not the contract you entered into and not what you paid for.

Am I making sense or not? It's nearly 1am.
Click to expand...

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.
 
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