Winning proximity

[nimzov]

So who is closer, the woman who pushed in, or the guy on the other side of the country who got the ticket that was one digit different to the winning number, or the guy who stayed too long in bed and missed his place in the queue and would have bought the winning ticket, or the guy who bought a ticket in another shop that would have been the winner if the ticket distributor's van loader had tossed the ticket block into a different stack, etc. ?

There are potentially millions of ways someone else could have bought the winning ticket. Are they all 'closer' than some guy buying a losing ticket in a shop in another city?
Perhaps there is there some universal hierarchy of 'closeness' we can agree on?

Quite right. In my opinion, there is a continuum in the closeness of tickets to the winning ticket. As a hierarchy, I have suggested the time-proximity as a candidate.

Or is it just that we like to hear stories that appeal to our non-rational and instinctive (or intuitive) misunderstandings of probability and causality? After all, why do people actually do the lottery at all?

To be clear, I am not proposing any kind of "system" or martingale here.

 

[nimzov]

Probability only works in one direction--forward. I think that's where you're getting mixed up.

I know this trap and really tried to avoid it in my examples and analogies. Maybe I failed.

 

[stilicho]

The dart is trown randomly by whatever physical mechanism you like. The outcome is a random letter. It is binary in the sense that the owner of a ticket wins by having the right letter or does not win. I think that the analogy of spatial-proximity if the dart lottery an time-proximity of the lottery in the OP, as in the 4th choice of the poll is correct.

Is this the same then for the lottery numbers selected a nano-second after the winning ticket was purchased? (In your universe, that is, since I don't think the timing of the purchase is in any way relevant to "closeness").

 

[nimzov]

Is this the same then for the lottery numbers selected a nano-second after the winning ticket was purchased? (In your universe, that is, since I don't think the timing of the purchase is in any way relevant to "closeness").

Yes just before and just after are equaly close. Like the letter to the left of "i" is equally close (spatially) to "i" as the letter to the right of the letter "i". And yes a continuum of closeness related to time-proximity, is what I suggest.

Spatial closeness is relevant for the darts throwing, and time closeness is relevant for the choice #4 in the OP lottery example ?

In your universe, that is,....

What do you mean by that ?

 

[bruto]

"I still say that closeness in this case is irrelevant, because winning or losing is essentially a binary operation, " bruto

===

Coming first in a race is also a binary result. Does not mean you cannot measure the closeness to winning.

In the context of the dart, here is a rudimentary definition.

In the dart example a definition of closer could use the size of the random variation or fluctuation in the trajectory of the dart to hit letter "j" compare to hitting the size of the random variation or fluctuation letter u.

As the dart goes along its trajectory, the probability of hitting a particular letter changes. A definition of closeness could use that fact. And in the example given, if you own ticket "j" you came closer to winning than if you own ticket "u".

Of course in a race you can measure the closeness to winning and if you're close to winning as the race progresses, your chance of winning is dependent in a large degree on that closeness. In addition, of course, closeness in one race, even if you lose, can help in developing a strategy for a subsequent race, because a race depends on skill. In many races, the closeness to winning is deemed sufficiently important that it is rewarded with a medal for second or third. But at the point where near-wins are not counted in the rules, you lose completely. Your name does not go up on the board. You are eliminated from the finals. Someone else gets the medals. There may be many ways to think about and measure your closeness to winning, but when considered in the context of prizes, medals, podium position, rankings and so forth, you're no better off than the person who came last.

I simply don't see any benefit in any kind of relative closeness you might care to identify in a lottery number. No indications of greater luck, no trends, no material with which to strategize. It may be interesting in some aesthetic way, or good for a laugh or a curse when you tear up the ticket, but it seems to me that analyzing losing numbers is an empty exercise.

 

[nimzov]

Of course in a race you can measure the closeness to winning and if you're close to winning as the race progresses, your chance of winning is dependent in a large degree on that closeness. In addition, of course, closeness in one race, even if you lose, can help in developing a strategy for a subsequent race, because a race depends on skill. In many races, the closeness to winning is deemed sufficiently important that it is rewarded with a medal for second or third. But at the point where near-wins are not counted in the rules, you lose completely. Your name does not go up on the board. You are eliminated from the finals. Someone else gets the medals. There may be many ways to think about and measure your closeness to winning, but when considered in the context of prizes, medals, podium position, rankings and so forth, you're no better off than the person who came last.

I simply don't see any benefit in any kind of relative closeness you might care to identify in a lottery number. No indications of greater luck, no trends, no material with which to strategize. It may be interesting in some aesthetic way, or good for a laugh or a curse when you tear up the ticket, but it seems to me that analyzing losing numbers is an empty exercise.

Thanks very interesting message.

One point. In a lottery you win some prize if you hit 3 numbers, more money for 4 numbers etc.

So there already is a kind of "closeness" defined in the game and you are rewarded proportionally to how close your ticket is to the winning combination.

Does that mean that no other definition of closeness is relevant. I don't know.

 

[bruto]

Thanks very interesting message.

One point. In a lottery you win some prize if you hit 3 numbers, more money for 4 numbers etc.

So there already is a kind of "closeness" defined in the game and you are rewarded proportionally to how close your ticket is to the winning combination.

Does that mean that no other definition of closeness is relevant. I don't know.

Yes, you can get a certain closeness from the quantity of matching numbers, and that is why I believe the fifth of your original examples comes closest to closeness, so to speak. But this closeness has to do with the set of correct numbers, not with the qualities of those that aren't correct.

Or how about a silly example from a race: four people enter a winner-take all foot race. The winner runs fast and finishes the race. Another runs very slowly but finishes the race well back. The third runs very fast but collapses of a heart attack in the last quarter. The fourth runs the wrong way around the course and is disqualified though he finishes as fast as the winner. Which of the three came closest to winning? Each did in some sense depending on how you view closeness, but their loss was complete.

cheers.....I'm shutting down to watch a movie, no more for a while.

 

[nimzov]

But this closeness has to do with the set of correct numbers, not with the qualities of those that aren't correct.

Good point.

Or how about a silly example from a race: four people enter a winner-take all foot race. The winner runs fast and finishes the race. Another runs very slowly but finishes the race well back. The third runs very fast but collapses of a heart attack in the last quarter. The fourth runs the wrong way around the course and is disqualified though he finishes as fast as the winner. Which of the three came closest to winning? Each did in some sense depending on how you view closeness, but their loss was complete.

cheers.....I'm shutting down to watch a movie, no more for a while.

I think only two racers respected the rules of the game by not running in the wrong direction and finishing the race. So I guess I would easily go with the slow racer.

Thanks anyway for your input. Appreciated.

 

[theprestige]

Please look at it in the context of the spatial example in my previous message.

Of course the rules says you do not win anything if you do not have the letter "i" on your ticket. Nonetheless...

Are all letters equal losers ?

Are all letters infinitely far from the letter "i" ?

Did ticket "j" came closer to winning than ticket "u" ?

My answers would be: No, No, Yes.

Your answers are based on the relative positions of the letters in a conventional English transcription of the Alphabet. When writing down a, b, c, ... z, of course "j" is closer to "i" than "u" is.

But the winning letter isn't chosen based on its position in the alphabet. Its quality of "winning" is based on something else entirely. And other letters gain no proximity at all to the winning quality by virtue of their place in the alphabet.

Likewise with other lotteries. The winning number is not chosen based on its position in time. Other numbers therefore gain no proximity to winning based on their relative positions in time.

The winning number is not chosen based on its position on a number line. Other numbers therefore gain no proximity to winning based on their relative positions on a number line.

The winning number is not chosen based on its membership in a particular set of numbers, or its amenability to certain kinds of mathematical transformations. Other numbers therefore gain no proximity to winning based on their membership in certain sets, or their amenability to certain kinds of mathematical transformations.

The only kind of "closeness" I can think of, in games of chance, is statistical proximity. If the results are weighted to a particular set or range of numbers, then numbers in that set or range are closer to being selected as the winning number. But that's in the future. Once the selection is in the past, one number won, and all the rest... did not.

"Did not win" is, in lottery terms, an absolute condition. There are no shades or variations of it. Unrelated characteristics such as relative position on a number line or the point in time when it was chosen do not make it any less of a loser, or shift it any closer to the winner.

I mean, that's the whole point of the lottery's random drawing: It's impossible for any number to be any closer to winning than any other. Before the drawing, all numbers have an equal chance of being the winner: They're all equally close to winning. After the drawing, all numbers but one have no chance at all of winning. They're all equally far away from the winner.

 

[JAStewart]

Equal - you have the same chance of getting 1-2-3-4-5-6 as you do getting 1-23-75-23-19964234-1

 

[High River]

Suppose there is a mega-huge-lottery where they draw 6 numbers from 100.
The winning combination came out and is:

12 - 14 - 30 - 46 - 52 - 68

Suppose also you have 5 tickets for this lottery with the following combination:

Ticket 1 : 11 - 13 - 29 - 45 - 51 - 67
Ticket 2 : 21 - 41 - 03 - 64 - 25 - 86
Ticket 3 : 06 - 07 - 15 - 23 - 26 - 34
Ticket 4 : 07 - 27 - 39 - 69 - 90 - 95
Ticket 5 : 12 - 55 - 61 - 83 - 85 - 89

We observe that:

In ticket 1, there is an offset of -1 of the winning combination
In ticket 2, we have the reversed digit of the winning combination
In ticket 3, we have half of every number in the winning combination
In ticket 4, (this ticket was chosen randomly by the terminal selling the ticket, but studying the listing of the randomly generated combination by the computer we find that this combination came out right before the winning combination (say if computed 1 millisecond later the computer would have printed the winning combination)
In ticket 5, Number 12 is the only number in the winning combination

Question is: With which ticket (1-2-3-4-5) did you came the closest to winning the lottery ?

If the lottery is working like this:
Grand Prize:6 numbers fixed
First Prize:5 numbers fixed
Second Prize:4 numbers fixed
Third Prize:3 numbers fixed
Forth Prize:2 numbers fixed
Fifth Prize:1 number fixed
Then all the combinations make a metric space.

Definition:

[latex]$$$
D := \{x \mid x \in Z \cap [1,100] \} \\
C := \{x \mid x \subset D \wedge |x|=6 \} \\
d(x,y) :=6-|x\cap y| \\
$$$[/latex]​

then

[latex]$$$
1.(\forall x,y\in C \,) (d(x,y) \ge 0) \\
2.(\forall x\in C \,) (d(x,y) = 0 \leftrightarrow x=y ) \\
3.(\forall x,y\in C \,) (d(x,y) = d(y,x)) \\
4.(\forall x,y,z\in C \,) (d(x,z) \le d(x,y)+d(y,z)) \\
$$$[/latex]​

proof:
1.

[latex]$$$
(\forall x,y\in C \,) (d(x,y)=6-|x\cap y|) \\
but, x\cap y \subset x \\
so,0 \le|x\cap y|\le |x|=6 \\
\Rightarrow d(x,y)=6-|x\cap y| \ge 6-|x|=0 \\
\Rightarrow (\forall x,y\in C \,) (d(x,y) \ge 0) \Box \\
$$$[/latex]​

2.

[latex]$$$
(\forall x,y\in C \,) (d(x,y)=6-|x\cap y|) \\
\Rightarrow d(x,y)=0 \leftrightarrow 6-|x\cap y|=0 \\
\Rightarrow d(x,y)=0 \leftrightarrow 6=|x\cap y| \\
but, x\cap y \subset x \\
x\cap y \subset y \\
|x|=|y|=6 \\
\Rightarrow 6=|x\cap y|\rightarrow(|x|=|x\cap y|\wedge |x\cap y|=|y|)
$$$[/latex]​

6 is a finite number

[latex]$$$
\Rightarrow (|x|=|x\cap y| \leftrightarrow x\cap y=x) \wedge (|x\cap y|=|y| \leftrightarrow x\cap y=y) \\
\Rightarrow |x\cap y|=6 \rightarrow (x\cap y=x \wedge x\cap y=y) \\
\Rightarrow |x\cap y|=6 \rightarrow (x=y) \\
$$$[/latex]​

On the other hand,

[latex]$$$
x=y\rightarrow (x\cap y=x \wedge x\cap y=y) \rightarrow |x\cap y|=|x|=|y|=6 \\
\Rightarrow x=y\rightarrow|x\cap y|=6 \\
so,|x\cap y|=6\leftrightarrow x=y \\
so, (\forall x,y\in C \,) (d(x,y)=0 \leftrightarrow x=y) \Box \\
$$$[/latex]​

3.

[latex]$$$
(\forall x,y\in C \,) (d(x,y)=6-|x\cap y|) \\
\Rightarrow d(y,x)=6-|y\cap x| \\
y\cap x=x \cap y \\
\Rightarrow |y\cap x|=|x \cap y| \\
\Rightarrow 6-|y\cap x|= 6-|x \cap y| \\
\Rightarrow d(y,x)=6-|y\cap x|=6-|x \cap y|=d(x,y) \\
so,(\forall x,y\in C \,) (d(x,y) = d(y,x)) \Box\\
$$$[/latex]​

4.

[latex]$$$
(\forall x,y\in C \,) (d(x,y)=6-|x\cap y|) \\
\Rightarrow d(x,z)=6-|x\cap z|,d(x,y)+d(y,z)=12-|x\cap y|-|y\cap z|\\
\Rightarrow d(x,y)+d(y,z)-d(x,z)=6-|x\cap y|-|y\cap z|+|x\cap z|=|y|-|x\cap y|-|y\cap z|+|x\cap z|\\
$$$[/latex]​

Definition:

[latex]$$$
\bar{(x)}:=C\backslash x \\
O:=\bar{x} \cap\bar{y}\cap\bar{z} \\
I:=\bar{x}\cap\bar{y}\cap z \\
II:=\bar{x}\cap y\cap\bar{z} \\
III:=\bar{x}\cap y\cap z \\
IV:=x\cap\bar{y}\cap\bar{z} \\
V:=x\cap\bar{y}\cap z \\
VI:=x\cap y\cap\bar{z} \\
VII:=x\cap y\cap z \\
then,|x\cap y|=|VI|+|VII|,|y\cap z|=|III|+|VII|,|z\cap x|=|V|+|VII|,6=|y|=|II|+|III|+|VI|+|VII| \\
so,|y|-|x\cap y|-|y\cap z|+|x\cap z|=|II|+|V| \ge 0 \\
\Rightarrow d(x,y)+d(y,z) \ge d(x,z) \\
\Rightarrow (\forall x,y,z\in C \,) (d(x,z) \le d(x,y)+d(y,z)) \Box \\
$$$[/latex]​

So,

[latex]$$$
(C,d)
$$$[/latex]​

is a metric space., and it is suitable for the lottery rules.

Thus the distance between those five tickets and the winning combination is

[latex]$$$
d(1,w)=0;\\
d(2,w)=0;\\
d(3,w)=0;\\
d(4,w)=0;\\
d(5,w)=1;\\
$$$[/latex]​

So the 5th ticket is the closest one.

 

[Vorticity]

"Number One, engage Universal Translator..."

If the lottery is working like this:
Grand Prize:6 numbers fixed
First Prize:5 numbers fixed
Second Prize:4 numbers fixed
Third Prize:3 numbers fixed
Forth Prize:2 numbers fixed
Fifth Prize:1 number fixed
Then all the combinations make a metric space.

The set of lottery combinations constitute a set of mathematical objects upon which some notion of non-negative "distance" can be defined.

Definition:
[latex]$$$
D := \{x \mid x \in Z \cap [1,100] \}
$$$[/latex]

Define D as the set of all integers between 1 and 100, inclusive.

[latex]$$$
C := \{x \mid x \subset D \wedge |x|=6 \}
$$$[/latex]

Define C as the set of all subsets of D that contain precisely 6 elements. C is the set of all possible lottery combinations.

[latex]$$$
d(x,y) :=6 - \left|x\cap y\right|
$$$[/latex]

Define the distance between two lottery combinations x and y as the 6 minus the number of numbers x and y have in common. E.g. if they are the same, then the distance between them is 0. If they share no numbers, the distance between them is 6.

A number of conditions on the distance function are now asserted, which establish that it is, in fact, an "actual" distance function in the mathematical sense:

then

[latex]$$$
1.(\forall x,y\in C \,) (d(x,y) \ge 0) \\
2.(\forall x\in C \,) (d(x,y) = 0 \leftrightarrow x=y ) \\
3.(\forall x,y\in C \,) (d(x,y) = d(y,x)) \\
4.(\forall x,y,z\in C \,) (d(x,z) \le d(x,y)+d(y,z))
$$$[/latex]​

1 = The distance between two lottery combinations is always greater than or equal to zero.
2 = If the distance between two lottery combinations is zero, then they are necessarily the same combination.
3 = The distance between combo x and combo y is the same as the distance between combo y and combo x.
4 = The "triangle inequality": The distance between combo x and combo z is necessarily less than or equal to the sum of the distance between x and y and the distance between y and z.

<Correct proofs of assertions 1-4 omitted...>

Thus the distance between those five tickets and the winning combination is

[latex]$$$
d(1,w)=0;\\
d(2,w)=0;\\
d(3,w)=0;\\
d(4,w)=0;\\
d(5,w)=1;\\
$$$[/latex]​

So the 5th ticket is the closest one.

Oops. I think this should read:

[latex]$$$
d(1,w)=6;\\
d(2,w)=6;\\
d(3,w)=6;\\
d(4,w)=6;\\
d(5,w)=5;\\
$$$[/latex]​

Otherwise, good show!

Although you've only answered the OP question for this choice of metric...

 

[Gord_in_Toronto]

I am using only two definitions, spatial and temporal,

And? You say that as though it means something.

 

[High River]

Oops. I think this should read:

[latex]$$$
d(1,w)=6;\\
d(2,w)=6;\\
d(3,w)=6;\\
d(4,w)=6;\\
d(5,w)=5;\\
$$$[/latex]​

Otherwise, good show!

Although you've only answered the OP question for this choice of metric...

Thank you.