@ Steve

[TurboGenius]

I posted this for The General (aka countless names and aliases) on the other forum -
Maybe you want to take a shot at it.
It's roulette related, don't worry.

======================

This might force you to go against everything you believe in lol.

Here's the question -

3 horses are in a race - they all start at the same time - except...
they each run randomly.
Meaning - their speed isn't predictable at all, it can change every step of the race.
A horse could run fast, slow, not move.....who knows ? But each horse runs the
race in this manner randomly.

One horse wins the race (eventually of course) - where do the other two horses finish ?

Here's some possible answers, or use your own...
1) They all finish equally at the same moment
2) One horse finishes and the other two trail behind about 1/3 of the way back
3) One horse finishes and the 2nd place horse is around 1/3 of the way back, the last place horse is about 1/3 of the way back from the 2nd place horse
4) Some other answer
5) I don't know anything about how random works, stop bothering me

I have a follow up question, if you decide to answer this one... and it just might change your
mind about everything you think you believe about random.


===========================

Take a shot at it - maybe you'll have a better reply to it than he "did" (so far).

[denzie]

Lets hope they leave the spaceships out. 

4) we cant know if they 1/3 behind...they might be right on his tail?


Just like hotties...one might start fast but freezes for a long time....so it get passed  by the others

[maestro]

3 horses are in a race - they all start at the same time

is race track straight line or circle...

[Joe]

Go to link:://rextester.com/l/pascal_online_compiler, remove the default code in the editor and replace it with the code below :

Code Select Expand

var
  h1, h2, h3, dist, i  : byte;
 
begin
  randomize;
  while (h1 < 24) and (h2 < 24) and (h3 < 24) do begin
    dist := random(3);
    case dist of
      0 :  inc(h1);
      1 :  inc(h2);
      2 :  inc(h3);
    end;
  end;
  writeln;write(h1:3,' ');
  for i := 1 to h1 do write('*');
  writeln;write(h2:3,' ');
  for i := 1 to h2 do write('*');
  writeln;write(h3:3,' ');
  for i := 1 to h3 do write('*');
  writeln;
end.

Then press F8 or click the "run it button" over and over to see how far back from the winner the other 2 "horses" are. Are there any patterns?

A real time simulation would be better because then you could see things like how often the horse which first takes the lead remains in the lead.

[vladir]

4)All we know is the other 2 horses are between the starting point and the end of the race(-1). But we can't predict how far they have gone.

[TurboGenius]

4)All we know is the other 2 horses are between the starting point and the end of the race(-1). But we can't predict how far they have gone.

But we can. On average the two losing horses will be around the same places when the winner crosses the finish line. I was hoping Steve (or the General) would tell me where... understanding random would make the answer simple.

[Steve]

Ok I'll play.

One horse wins the race (eventually of course) - where do the other two horses finish ?

Possibilities I see are:

1. One or more horses finished equal first

2. Half way down the track, the two losing horse starting humping. Maybe still are. They were disqualified.

3. One got hit by a low-flying plane and propelled it past the finish line first, but was disqualified.

4. One got really, really excited about law enforcement.



The most probable, given history of horse races is the other two horses finished the race, and came in second and third place respectively, at different times. We don't know times. We can estimate based on past performance but this is not an exact answer. Only precise timings will give an exact answer.

If you give more specific parameters, I could give a more specific response.

I don't see your point. You have given undefined variables and asked for the possibilities. Based on what you've said of possibilities in roulette, you're probably going to say it correlates to the expectation that a 3-repeater comes after 2-repeater, and so on.

[Steve]

And if a horse did that to me, I would punch it in its dick. It's just not on. Could you imagine the jokes from fellow officers...

"Hey Officer Smith, how is your doughnut this morning?"

Lots of potential..

[TurboGenius]

I don't see your point. You have given undefined variables and asked for the possibilities. Based on what you've said of possibilities in roulette, you're probably going to say it correlates to the expectation that a 3-repeater comes after 2-repeater, and so on.

It has nothing to do with repeaters at all - and nothing to do with horse racing.
Just random and how it's predictable.
I didn't ask when or how the other two horses would cross the finish line.
I asked when the winner wins - where are the other two horses typically (or "on average") on the track. It's a math calculation and easy to figure out if someone understands random.
If someone doesn't - then I can't expect them to know the answer.
It can however be used in roulette (or course) because just like my question above, we have random outcomes in which each horse could be a set of locations that appear completely randomly and independent from the last spin - yet are completely predictable.

[Steve]

Just random and how it's predictable.

Random basically means unpredictable. So first you need to change your definitions.

But to be accurate, nothing is ever completely unpredictable or without cause. Everything is just cause and effect. In the context of RNG, usually the numbers are from algorithms which sometimes are predictable. Usually they aren't predictable, at least with any method we know.

But that's another story. Your approaches don't change the odds because repeaters dont change the odds. Proper testing proves this. I've done my own tests and besides my tests, it is well known.

I asked when the winner wins - where are the other two horses typically (or "on average") on the track

Somewhere between the start and finish line.

It can however be used in roulette (or course) because just like my question above, we have random outcomes in which each horse could be a set of locations that appear completely randomly and independent from the last spin - yet are completely predictable.

No you cant, because you are talking about a very large number of possibilities. And in the end you are still left with the same accuracy.

You previously mentioned things like 4 repeats comes after 3 repeats and so on..... as if it had any meaning. But a simple test proves the odds don't change after 3 or 4 repeaters. I explained it already. Anyone can do the test and see they are stuck with 1 in 37.

Let's not dance around the issue. You can only make a reasonable guess about where the horse are, but it is very different to roulette in "predicting random/RNG" because you dont have any reference points. You need reference points and correlation. Exactly what are you getting at?

[maestro]

if you get distance from winner to second horse and distance from second horse to last horse then sum of distances should come to 0.38888 of whole track...just a guess

[TurboGenius]

Go to link:://rextester.com/l/pascal_online_compiler, remove the default code in the editor and replace it with the code below :

That is perfect. I would give you 10 extra likes if I could !
Easier than doing it each time with the calculation by hand as I was doing
just to verify the exact value.
When I compared mine to yours one value was exact and the other was
off less than .04 so they would have balanced out during more samples
as I was using averages.
Nicely done.
Cheers.

[TurboGenius]

Let's not dance around the issue. You can only make a reasonable guess about where the horse are, but it is very different to roulette in "predicting random/RNG" because you dont have any reference points. You need reference points and correlation.

No, the horses will be "on average" in specific places when the winning horse crosses the finish line. Perfect ? Nope, doesn't need to be.
And did you know - you can make "reference points" in a random game ? You can.

[Steve]

No, the horses will be "on average" in specific places when the winning horse crosses the finish line. Perfect ?

You mean like a bell curve? That doesnt work in roulette. It's no different to waiting for 55% more reds than blacks to spin, then betting red... because we expect things to average out.

And did you know - you can make "reference points" in a random game ? You can.

Any reference point that has no predictability in the equation will nullify an increase in accuracy.

So exactly what is your point? That there will be a long term "balance" that we can depend on?

You still arent changing the odds at all.

[psimoes]

The first is the one who knows a shortcut.
The third, by running a few metres before quitting, thinks is the true winner.
The second runs a few metres, goes home and returns 18 hours later.