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Lotto: Statistics question

sthomson said:
I admit, I can't answer this one. Anyone who's better at statistics than I am?
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With dice it's very simple to play enough games to have a high chance of winning.

For the glitch to affect you in the lottery, you would have to buy thousands of tickets. Before that, the tickets you bought would still be sufficiently different from each other.

And on average, it all evens out. Take the dice example given just now: With a broken dice, you would win *big* one in every 6 games.
 
sthomson said:
OK, I think I see what you're saying. You're saying that if I buy 10 biased random lottery tickets, and someone else buys 10 non-biased random lottery tickets, they have a much better chance of winning because their tickets will cover more of the possible numbers than mine will.

I admit, I can't answer this one. Anyone who's better at statistics than I am?
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It doesn't matter whether the numbers that are picked by the players are biased or not. For an INDIVIDUAL drawing, each sequence is equally likely.

It is when you look at where the winning numbers come from over the long term, that there will be a bias away from the winning numbers coming from the biased group. See post 127. Again, for any single drawing it doesn't make a difference.
 
Rasmus said:
With dice it's very simple to play enough games to have a high chance of winning.

For the glitch to affect you in the lottery, you would have to buy thousands of tickets. Before that, the tickets you bought would still be sufficiently different from each other.

And on average, it all evens out. Take the dice example given just now: With a broken dice, you would win *big* one in every 6 games.
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As you would with a normal die.
 
sthomson said:
OnlyTellsTruths said:
Say you have 3 people and 3 dice (normal 6 sided). The second two people try to match what the first person rolls, they each get 10 tries. Say you are one of these second people and your dice is broken (gum on it) so that it always displays 3.
The other guy gets 10 rolls to try and match, but you don't.
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OK, I think I see what you're saying. You're saying that if I buy 10 biased random lottery tickets, and someone else buys 10 non-biased random lottery tickets, they have a much better chance of winning because their tickets will cover more of the possible numbers than mine will.

I admit, I can't answer this one. Anyone who's better at statistics than I am?
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Exactly.
 
sthomson said:
OK, I think I see what you're saying. You're saying that if I buy 10 biased random lottery tickets, and someone else buys 10 non-biased random lottery tickets, they have a much better chance of winning because their tickets will cover more of the possible numbers than mine will.

I admit, I can't answer this one. Anyone who's better at statistics than I am?
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This does not affect odds of winning in the sense in which you're implying (see below for the disclaimer) :).

Say person 1 has two lottery tickets in the OP lottery

1,2,3,4,5,6,7 and 2,3,4,5,6,7,8

given that there are (36 choose 7) = 8 347 680 unique different combinations, and he has chosen 2 distinct combinations he has a 2/8,347,680 chance of winning.

say person 2 has two lottery tickets in the OP lottery

1,2,3,4,5,6,7 and 8,9,10,11,12,13,14

given that there are still (36 choose 7) = 8 347 680 unique different combinations, and he has also chosen 2 distinct combinations he has a 2/8,347,680 chance of winning.

Those 8,347,780 combinations are all unique - and each one has the same odds of being chosen.


*It is worth noting that there would be an absolutely minutely greater chance of the glitched RNG throwing up exactly the same numbers. And if you're sitting there with two duplicate tickets, then you only have one distinct (1/8,347,680) chance of winning. This may or may not affect your EV dependent on whether there is a fixed payout for every winner, or a shared pot for all winners.

This scenario would happen by chance with a non-glitched machine 1/(8 347 680x8 347 680) = 6.96837614 × 10-13
And say, the glitched RNG was so glitched that it completely missed out 1-9 - so the combinations were (27 choose 7) = 888 030, the chance of this RNG throwing up identical tickets would be

1/888 030x888 030 = 1/788,597,280,900 = 1.26807437 × 10-12
So this would be about 90 times more likely than with a completely random RNG, but still vanishingly, incredibly unlikely.*
 
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sthomson said:
OK, I think I see what you're saying. You're saying that if I buy 10 biased random lottery tickets, and someone else buys 10 non-biased random lottery tickets, they have a much better chance of winning because their tickets will cover more of the possible numbers than mine will.

I admit, I can't answer this one. Anyone who's better at statistics than I am?
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It the bias was so bad that you got the same number 10 times in a row, then your chance of winning would be reduced. But if the bias allows you to get 10 different numbers, you have just as much chance of winning as someone who has "non-biased" tickets. You have 10 of the possible numbers, and so does the other guy.
 
You're still not answering the same question I am, you must account for multiple tickets.

The person that buys the non-biased tickets can buy multiple tickets that cover a broader range than the person buying the same number of biased tickets. The increased odds are in the total range not each specific sequence.

I'll clarify the extreme example I used even further (from 6 sided dice to the even easier to comprehend coin-flip).

You have 3 people and 3 two-sided coins. The second two people try to match what the first person FLIPS ONCE, they each get 10 tries. Say you are one of these second people and your coin is biased (gum on it) so that it always displays tails.

You flip tails on all 10 tries, you would then only match the first persons flip 50% of the time. (multiple biased tickets)

The other guy gets 10 flips with an unbiased coin (both sides available). There is way less than 1% chance of him not matching the first persons flip in ten tries. (multiple non-biased tickets)

Beausoleil is, unlike forgetting about the multiple tickets completely, confusing it with the other ingredient of multiple numbers picked.


I'm well aware of how people think number sequences matter (they don't see how 10 head flips isn't any different than any other particular sequence of 10 flips, odds wise), please stop confusing that with this case that clearly introduces other factors. 10 head flips has the same very low probability of all other pre-ordained sequences of 10 flips........... unless the coin has gum on the tails side, then 10 heads goes from way less than 1% to 100%.
 
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Say we stick with the 10 single digit numbers being available for 5 slots

Think of the entire chart of numbers from 00000 to 99999 (or, for no repeats, from 01234 to 98765 minus all numbers containing repeating digits).

Now say someone is using a machine that doesn’t allow for ½ of the allowed digits to be used. That would remove 1/2 of allowed numbers in your chart, it doesn’t matter which 5 digits it is (that are removed) or how many total allowed numbers you have (repeating or non). It’s already been discussed how any specific sequence has the same odds as any other, but don’t forget when you increase the number of possible digits allowed, and/or the length of the sequence, the separation in odds increases (lower percentage).

Everyone is hung up on if both groups just chose one number from their available range (same odds), and not taking into consideration that they are allowed multiple guesses within their differing sets for multiple days (not same odds as a single guess basis).
The person with all digits equally available for his multiple guesses will always have the possibility of having the correct answer be somewhere in his available source range. Likewise, the person with a limited source range not only has to guess the number each day, but have the number each day be available to his limited source range.
 
Re-reading my last post I'd like to note how key it is in seperating the number of allowed digits per number in a sequence from the number of numbers (slots) that are in the sequrence.

For instance, the way in which the differences between 1111(binary) and 11111(hex) are distinguished is not the same way in which the difference between 1111(binary) and 11111(binary) are different. The first involves changing both the number of slots and the number of digits allowed per slot, while the latter only involves changing the number of slots.
 
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Dare I quadruple post with semi-derail? I dare!

The need for more precise and non-conflicting terminology rising with respect to the complexity of the topic (see above references to the physics bet topic) is just one reason I’m always open to being any amount of wrong on something (which is ingrained to be hard for us humans to do). (I would hope of course I’m not completely wrong, though I’m sure I’ll find something I said wrong or didn’t mean to say when I re-read the topic tomorrow……. not even taking grammEr into account.)

Though both issues (terminology and humility) are eons apart, we should always remember to try the best we can/be optimistic.
 
As alluded to by Ramsus (and someone else I think, sorry I missed that) the number of possible outcomes removed by the bias is relative to the number of possible outcomes purchased.

We would need to know A: ‘exactly how many tickets were purchased from the station that had bias’ and B: ‘exactly how many possible outcomes were not available because of the bias’.

As far as I see it it’s something like: If the percentage of total outcomes removed by the bias in the machine overcomes the percentage of total outcomes that machine sells then the relative odds of winning start to decrease noticeably. Add people using that same machine for multiple daily drawings (spreading the bet) and you get even worse relative odds.

Bill Gates doing what he feels like it on a whim scenarios aside, I’m sure some places sell a lot of tickets to a lot of people on a consistent basis.
 
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Lothian said:
I am not dismissing the question. I am questioning why you are asking it.
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Because you think I have to have sinister motives. It simply cannot be because I merely wanted to know.

Lothian said:
We are not talking about Sylvia Browne asking questions about black holes. I am questioning why Phil Platt is asking if Planet X will hit us.
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No, no. You pointed to the difference between those who ask questions: Woos and non-woos. We should treat them differently, based on who they are.

A similar case, if you'll allow me.

Randi receives a claim of a paranormal phenomenon.

Should he question the claim, based on who the claimant is?

Lothian said:
the hell put you in charge of what question I can ask? :)
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But that's what you are doing: Casting aspersions on me because I ask the question.

Lothian said:
Ok let me give an example. See your quote about Sylvia Browne above. If next week you started asking lots of questions along the lines of; what is wrong with Sylvia Browne? Why can’t we speak to the other side? What reasons are there for calling her the lowest form of life ? Would I be weird to question why you have changed your tack.
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How would that be changing my tack? And how does "native" enter the picture?

Lothian said:
Same here. I can’t understand why you had a good understanding of maths and now it has deserted you. Are you playing devils advocate or has my memory tricked me ?
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I have a good understanding of math. I was asking a statistics question.
 
OnlyTellsTruths said:
You're still not answering the same question I am, you must account for multiple tickets.

The person that buys the non-biased tickets can buy multiple tickets that cover a broader range than the person buying the same number of biased tickets. The increased odds are in the total range not each specific sequence.

I'll clarify the extreme example I used even further (from 6 sided dice to the even easier to comprehend coin-flip).
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Onlytellsthetruth - you keep confusing the issue by repeatedly changing your scenarios - the original situation and subsiduary discussion has been completely resolved - there is no further clarity necessary. Each new scenario you introduce brings with it different considerations - which you are not being rigourous enough in your descriptions to address accurately.
The range is another red herring for the lotto example. Chooosing two lotto tickets 1,2,3,4,5,6,7 and 2,3,4,5,6,7,8 gives you precisely the same odds of winning the jackpot as choosing 1,2,3,4,5,6,7 and 8,9,10,11,12,13,14. Just because the latter two tickets have a greater spread does not affect your chances of striking the jackpot as both are unique single combinations out of the (7 choose 36) possibilities.

Everyone is hung up on if both groups just chose one number from their available range (same odds), and not taking into consideration that they are allowed multiple guesses within their differing sets for multiple days (not same odds as a single guess basis).
The person with all digits equally available for his multiple guesses will always have the possibility of having the correct answer be somewhere in his available source range. Likewise, the person with a limited source range not only has to guess the number each day, but have the number each day be available to his limited source range.
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*sigh*

no. Say the draw is on Friday, and you buy 1 ticket on monday, 1 ticket on tuesday, 1 on wednesday and 1 on thursday. You use a glitched RNG which only selects between 1 and 18.

Someone else buys 1 ticket on monday, 1 ticket on tuesday, 1 on wednesday and 1 on thursday. He uses a non-glitched RNG which selects between 1 and 36.

The only factor to consider is the miniscule chance of duplicate numbers being generated - that aside there is absolutely no difference in the odds of hitting the jackpot.
 
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andyandy said:
No. Say the draw is on Friday, and you buy 1 ticket on monday, 1 ticket on tuesday, 1 on wednesday and 1 on thursday. You use a glitched RNG which only selects between 1 and 18.

Someone else buys 1 ticket on monday, 1 ticket on tuesday, 1 on wednesday and 1 on thursday. He uses a non-glitched RNG which selects between 1 and 36.

The only factor to consider is the miniscule chance of duplicate numbers being generated - that aside there is absolutely no difference in the odds of hitting the jackpot.
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Yes. I will also say that the first person's RNG may be glitched in that it only selects numbers from 1 to 18, but at the same time, it may be based on a form of linear congruential generator or block cipher and having the property of not generating the same result until it cycles.

The OP and this thread has considered the probability of a single ticket hitting the jackpot. If the drawing is fair, this probability is fixed and cannot be altered. Other interpretations of "probability of winning" can of course be changed: for example, the probability of a person hitting the jackpot can be trivially increased by playing more tickets. The probability of winning a lesser price (such as matching fewer numbers) while playing multiple tickets can also be altered, perhaps notably, by not using any RNG for individual tickets (since that would be quite likely to repeat some numbers in different output sets), but instead choosing some method that would make sure that your multiple tickets repeat as few numbers (and groups of numbers) as possible.

As andyandy has said, these are different situations and therefore require different considerations. The original case is clear though.
 
OnlyTellsTruths said:
You're still not answering the same question I am, you must account for multiple tickets.

The person that buys the non-biased tickets can buy multiple tickets that cover a broader range than the person buying the same number of biased tickets. The increased odds are in the total range not each specific sequence.

I'll clarify the extreme example I used even further (from 6 sided dice to the even easier to comprehend coin-flip).

You have 3 people and 3 two-sided coins. The second two people try to match what the first person FLIPS ONCE, they each get 10 tries. Say you are one of these second people and your coin is biased (gum on it) so that it always displays tails.

You flip tails on all 10 tries, you would then only match the first persons flip 50% of the time. (multiple biased tickets)

The other guy gets 10 flips with an unbiased coin (both sides available). There is way less than 1% chance of him not matching the first persons flip in ten tries. (multiple non-biased tickets)

Beausoleil is, unlike forgetting about the multiple tickets completely, confusing it with the other ingredient of multiple numbers picked.


I'm well aware of how people think number sequences matter (they don't see how 10 head flips isn't any different than any other particular sequence of 10 flips, odds wise), please stop confusing that with this case that clearly introduces other factors. 10 head flips has the same very low probability of all other pre-ordained sequences of 10 flips........... unless the coin has gum on the tails side, then 10 heads goes from way less than 1% to 100%.
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The problem with this is that the number of flips is less than the number of options, so the "biased" person always repeats his attempt to match (always gets tails). Many people have pointed out that the bias only matters insofar as it increases your chance of a duplicate among your attempts to match.

As long as all the attempts to match are different from one another, it doesn't make any difference how they are generated or how restricted the set is. In the case of the coin toss above, they can't be different from one another by definition so it is a different situation. In the case of the lottery, the chance of the biased machine generating two identical numbers out of any feasible number of lottery ticket purchases is not significantly higher than the chance with the unbiased machine.

Suppose two people each buy 100 tickets. One uses a biased machine that always gives exactly the same 100 consecutive numbers (eg 012345 012346 012347...). The other gets 100 truly randomly selected numbers. The first person is more likely to win by a very small amount, because the second one might get a duplicate ticket. If the randomly selected numbers are all different, the chances of each person winning are the same.

I'm thinking of a 10 digit number. You get 100 chances to guess what it is. As long as you provide 100 distinct numbers (not all the same number as in the coin toss example above) it matters not at all how you generate them.
 
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