Guessing game (with real edge)

[Drazen]

Hello

I haven't posted in a while, so I decided to come with a fun random experiment.

I will ask you to write down 2 different random numbers (it can be any random number positive or negative like 0.001, Pi, +100  -100, +/- million etc... You can use any RNG (true or pseudo)  hey you can even for example use last two decimals of bitcoin price or any stock price on any day you want. How cool is that?

Once you picked your random numbers, write them on a card and turn them face down. Let's call those cards A and  Z. 

Now I will come to the table,  overturn only one of the cards and I am going to tell you with over 50% of chance me being correct wether upturned one will contain bigger number than the one that is faced down. You would have thought that I have only 50:50 chance of being correct, but I will beg to differ.

So how am I going to do that? I will introduce the third random number for myself which will help me to raise the odds in my favour. Let's call that number “d”. That random number can be generated in any way I like as long as it follows a normal distribution.

After I generated my number “d”, I will have to follow a specific simple algorithm and decide upon it.

So if your first card (let's say I decided to turn A) will have bigger number written on it than my randomly generated number “d” I will say, A card will have the bigger number here (between your cards A and Z), and If A will have smaller number than my number “d”, I will say no no, Z is the bigger number there.


I hope I didn't create too much confusion here, but let's try to show it like this:

We will write 3 numbers. First one is your number A, in the middle will be my number “d” and the third one will be your number Z

A        d        Z
82        -98        63

So let's imagine your numbers are 82 and 63 which I cant see. So I came and I turned number A (82) then I generated my number “d” which is -98. Now I look at my algorithm and I say: So number A is bigger than my number “d” so I am guessing that your (still) unturned number Z will be smaller than your first number A. And my guess was right here!


Does that mean I have a “dependance” here? Dependence in single random numbers? Priyanka said that dependence creates bias? Is that what she meant?

How much over 50% correct guesses are we talking about? Can that be proven? Play and find out.

Best

Drazen

[Person S]

Hi, I think the emphasis is on probability.
We wrote 2 numbers in advance (more / less), and now imagine that “D” is a dice with many faces. Do not be this extra option, the odds will remain 50/50. But by including the bone in the selection, we create an additional event, and this somehow affects the result. It would be nice to hear this from experienced mathematicians.
In roulette, the next event may repeat the previous one, for example, “D” predicts that the next number will be greater. But here the problem number may remain the same ...

[Drazen]

Few more examples how you can play with it and test. Its really easy and simple. All numbers are from random.org btw


   A    d    Z
-85   -82   98
138   41   -1
174   -99   5

So if A is bigger than d we say A is bigger (than Z)

If A is smaller than d, then Z is bigger (than A)

You can limit min and max values and see how it goes with different values.

Hope its fun and you might be amazed with this simple trick.

And if you think it makes any difference that they are predetermined here, generate one by one 

Cheers

[Clf7]

Did you Test on rx Software? Or it cant be tested there?...So we can know If it is working Long term

[Drazen]

Thank you for your interest Clf7

I havent tested it in Rx as I unfortunately dont have any coding skills. It can be done manually though.

But if someone is interested to check this through a simulation I would encourage them to do so and show us the results here.

Cheers

[quos]

What is the idea to test it in RX?

1st spin = A and
2nd spin = d ?

Thank you!

[Clf7]

I dont know how, i am not into this

[Drazen]

What is the idea to test it in RX?

1st spin = A and
2nd spin = d ?

Thank you!

Yes, and you will need the third one for Z

[Drazen]

Short test

So let's say we took random numbers between 1-1000. Numbers are taken from random.org

478  196    463 W
465  844    589 W
228    533   10  L
73    371   713 W
61    822   406 W
276    353   102 L
989    198   661 W
280    316   203 L
941    337   192 W
452    920   924 W
293    676   519 W
996    100   962 W
891    246   38  W
470    897   383 L
812    527   655 W
154    876   888 W
503    649   530 W
975    527   631 W
337    869   294 L
303    356   156 L
458    5           710 L
810    966   753 L
817    281   345 W
258    108   727 L
763    569   970 L
989    852   472 W
361    31       780 L
542    583   229 L
521    390   839 L
324    134   455 L
12    187   863 W
8    278   884 W
831    376   696 W
274    719   775 W
352    221   789 L
62    730   110 W
706    534   317 W

22W and 15L

That's 59.46% win rate, right?

You can try different number ranges and observe the difference. If 60% winning rate here doesnt amuse you enough, maybe you can observe, experiment and find a way to get even higher percentage? (khm khm, yes for the case like this there is way)

Enjoy

[falkor2k15]

Is that based on erdos?

[ati]

I saw a numberphile video about this a while ago.

[Drazen]

Is that based on erdos?

No, this is not Erdosz theorem.

Ati is right, but I initially saw about it many years ago when RRBB (under different nick) pointed to it in American Scientist magazine back in 2011

[Person S]

Yes, it looks beautiful, but really nonsense.
Roulette example
Spin A-32, D-14. Now we need to cover the field from 0 to 31. This is a big bet ... We can not choose what we like. For example, we are waiting for the trigger A-15 / D - less than 15. And we rely on such events, but he will not play. We need to play in the stream. Can a series of small victories outweigh a major loss? I think no.

[falkor2k15]

Yes, it looks beautiful, but really nonsense.
Roulette example
Spin A-32, D-14. Now we need to cover the field from 0 to 31. This is a big bet ... We can not choose what we like. For example, we are waiting for the trigger A-15 / D - less than 15. And we rely on such events, but he will not play. We need to play in the stream. Can a series of small victories outweigh a major loss? I think no.

That's right - same problem as defining element in cycles or 2 dozens being more likely than 1 dozen - risk/reward ratio stays the same - but at least it's dependency (statistical) in some form. Not that I've yet found a way to use statistical dependency to gain real edge.

[precogmiles]

So how do you use this to make a prediction on the next number?