Optimum Stopping Theory

[daveylibra]

Hi, I'm still grappling with this!

To simplify, let k=10. Then int (k/e) = 4.
At the 4th spin we note the highest number.
Then suppose at the 6th spin a higher number occurs, eg 30.
Are we saying that, on AVERAGE,  30 will be the highest number after k spins 33% of the time?
Or that, on average, 30 will be the highest number after k spins, AT LEAST 33% of the time?
It seems from the video that the former is the case.
If so, we can say that 30 WILL NOT be the highest number 67% of the time.
ie a number between 31-36 will appear between spin 7-10, 67% of the time.

Another consideration is that a roulette number can appear twice or more, I'm not sure
that this is allowed in the "googol game /secretary puzzle"

But it is very interesting, if anyone cares to share their thoughts?

[huskerdu]

If we play live roulette wheel, the roulette table is a fake representation of the wheel. The numbers are dispersed all over the wheel, so the table with numbers is a virtual reality.
Also if you play RNG the RNG machine is dealing with digits. It doesn't realize small or big numbers. They are all digits.
So if we bet all numbers from 1-30 is the same as if we bet any 30 random numbers. We have exactly the same propability

[daveylibra]

Hi HuskerDu

Sometimes maths is not intuitive. Check the amazing puzzle 1 here -

link:s://johncarlosbaez.wordpress.com/2015/07/20/the-game-of-googol/

[huskerdu]

Sometimes maths is not intuitive. Check the amazing puzzle 1 here -

link:s://johncarlosbaez.wordpress.com/2015/07/20/the-game-of-googol/

Very interesting!
Thanks!

Talking about maths, maths is a solid truth.
And randomness is also under the rules of maths.
Randomness creates unbeliable outcomes on the short run that maths cannot forsee.
But on the long run maths rules.
let me tell you an example similar to the puzzle of the link you wrote:
Someone takes 100 slips of paper, and on each slip he writes numbers from 1 to 100 and turn them backword mixed on the table..
Every time he will say a random number from 1 to 100 (e.g. 58) and he will tell you to chose one of the papers and if this papper has not the number that he randomlly  said to you, you  win 1 euro. if it happens that the paper that you chose has the number that he randomlly said,  you lose 99 euros.
Are you willing to play?
You will think, ofcourse, because i have only 1% chances to choose the paper that he said.
You chose a random paper and it is a different number that he said and you win.
It's easy you think, you put back on the table the paper mix the again  and you ask him to continue saying to you a different number
He sais a random number  from 1 to 100 then you chose a paper. Again you win!
And you win and you win and you win.....
But some momment it happens that the number that he said is the number of the paper you chosse! And this moment you lose 99 euros!
Mathematically this moment will come on the 100th play. And the previous  99 euros you have earned will be lost.
Randomness says that this lost may come after 200 or 200 plays, but also may come from the first play.
You don't know.
If you are lucky you will have 200 winning sessions and you will stop, taking your 200 euros that you earned and run away and don't play again.
if you are unlucky maybe the lose will appear at the 20th play. until then you will have earned 19 euros but you will lose 99 euros.
And if you want to play this game for years every day here comes maths linking with real world :
In 100.000.000  times of play you will be even!
You will win 99.000.000 times 1 euro and you will lose 1.000.000 times 99 euros!
For 100 plays the 1/% of lose  is theoritical.
But in 100.000.000 plays , math theory becomes reality and it will happen.
Ofcourse we will not have 1 lose every 100 plays, but sum it up we will have 1.000.000 loses in 100.000.000 plays. No matter when those loses will come
maybe we play for 99.000.000 times not having any lose and earn 99.000.000 euros. But the next 1.000.000 plays will be loses and we lose all our earnings. That is the connection between maths and randomness

[cht]

If you chose random numbers the chance of your numbers hitting is according to probability distribution itlr. Carpet or wheel format are random selections. Probability and statistics fail.

What if I told you roulette spins or prng is naturally imbalanced that it won't return 50:50 for select EC as example. The key is this imbalance is naturally occurring no matter who the dealer or whatever wheel and ball. Condition A occurs more often than condition B everytime.

I hv posted the reason. I have shown the sample game. Yet people can't figure it out. It's easy yet it's difficult even after you've seen the actual game played that's clearly a mechanical selection process that exploits this naturally occuring imbalance.

[cht]

Nobody cares if he's a c*nt or a troll.

He's a sly scammer in disguise, "invest in ME!"

That precog guy is a fake scammer.

Check out the account for real science based precog as I hv described.

[CarpeDiem]

Sometimes maths is not intuitive. Check the amazing puzzle 1 here -

link:s://johncarlosbaez.wordpress.com/2015/07/20/the-game-of-googol/

Hi

There you have it, in plain sight
link:s://:.quantamagazine.org/solution-information-from-randomness-20150722

Applying this to roulette, where numbers are arranged from 0-36, the optimum choice is  G=18, place 1 unit on zero and 37 units on high(19-36).
Worst deviation, 0 doesn't show up, no big deal. You now have a game with a p>0.50 EC.

[CarpeDiem]

"Okay, it's almost time to wrap up the first Insights puzzle. Let me summarize the important insights, and point out the mistaken notions that may still be preventing some people from seeing the solution.

The Pure Math Question

The main puzzle is an existence problem in pure math â€" does there exist a strategy that guarantees that you - the player - can guess the larger number as your final choice after seeing one of the two numbers, no matter what two numbers the host selects?

The answer is yes, and this fact has been rigorously proved here for the unbounded case. The barriers to seeing the solution are:

1)   The pair of numbers A and B have already already chosen for you by the host using an unknown 'black box' process. There is no requirement that they come from a uniform distribution. In fact, a uniform distribution over the entire real line is impossible.

2)   There are an infinite number of winning strategies, all of which involve sometimes switching your guess to the other hand based on the number seen. One way this can be done involves generating an arbitrary guide number G having a non-zero probability of being present in any interval on the real line. There are an infinite number of distributions that have this property, and they can be based on trigonometric ratios, exponentials, Gaussian curves or even by simple coin-tossing. The guide number G must also be generated independent of A and B.

3)   The solution only proves that the probability of winning is more that 50 percent. The actual probability does not need to be calculated, and cannot be calculated without knowing the host's number generating algorithm. The exact (non-zero) probability of G lying between any given interval can however be calculated, based on the actual distribution the player uses.

4)   For any qualifying distribution selected by the player, the probability of winning may be extremely close to 50% for a given pair of numbers, but it is always higher than 50% by a finite amount.

5)   The infinity of the real line has the same cardinality as the infinity of any interval within it (aleph-one). So for any pair of numbers A and B on the real number line, we can find a pair of numbers A1 and B1 within any bounded interval such that the probability of a guide number lying between A and B under a given distribution on the real line is exactly equal to the probability of a guide number lying between A1 and B1 under a uniform distribution.

6)   As a consequence of the above, there is not much of a difference in the practicality of achieving success between the unbounded and bounded cases. This surprising fact is explained in detail below.

What about winning in real life?

I think one of the reasons that some people have difficulty with the unbounded case is that the infinity of the real line seems so vast, that it seems impossible to win in a real-world game. This is certainly true in an adversarial game, where the host is actually trying to prevent the player from winning. But as a consequence of point 5 above, this is also true in the bounded case, as we saw in bonus problem 3. In the adversarial case, whether bounded or unbounded, the host can choose two numbers that are so close together, say 10^-30 apart in the bounded case, that the player's chances of winning are reduced to very close to 50%. Only cosmic beings playing at the rate of a googolplex games per second and having all eternity can realize the win in such a case. However, this does not undermine the fact that the "pure math" winning strategy does exist even in the adversarial case. "What's the use of this?" a hard-nosed realist may ask. The pure mathematician will say "It's the principle of the thing" or as G. H. Hardy put it, "what is useful above all is technique, and mathematical technique is taught mainly through pure mathematics."

The utility of this technique bears full flower in the non-adversarial case, where the host is indifferent, as for example in a game show setting. When the host's pair of numbers are selected at random from any reasonable distribution over the entire real number line without explicit intent to cause the player to fail, the win in the unbounded case is as easy as in the bounded case. I carried out a simulation of 100,000 cases where the host's numbers A and B were drawn from the entire real number line using the function y=ln(x/(1-x)) with x being a uniformly distributed random variable between 0 and 1, where the player used G=tan(Ï€(x-0.5)) to generate G. The player still won 62.7% of the time. In fact, because the median of both these distributions are centered around zero, the "median" strategy suggested by the very first commenter, Sarah L. of using 0 as the guide number succeeds, as expected, 75% of the time.

Besides having a common median, another problem with the above distributions is that they have a sharp peak very close to zero (neither of the two generated numbers above 10 in my simulation!). To correct for these problems, I changed the generating distributions to y=(10^15)^ ln(x/(1-x)) with a 70% bias towards negative numbers for the generating distribution and G=10^(tan(Ï€(x-0.5))) with a 70% bias towards positive numbers for G. Now I obtained some hugely negative and positive numbers (absolute values of 10^150 or more in both cases). The player still won 65% of the time. The "dynamic median" strategy which I had suggested as the best for a repeated game won 75% of the time even though the median fluctuated wildly initially.

So it is just as easy to win in the unbounded case with a non-adversarial opponent as it in the bounded case. Why should this be? Well, any distribution that has appreciable probability density over a wide range on the real line will often generate a pair of numbers that are far apart, and there is a good probability that a guide number generated by almost any distribution will be able to lie between them.

I offer the following corrections to hitherto uncorrected comments made by people, not as criticism but to allow them and others to overcome false impressions. As they say in reminder letters, please ignore these messages if you have already corrected these impressions ☺.

Walt Donovan said (and this was echoed by Max Rockbin, Mark B, Matthew Kosak and others):
"Once you bound the range of A and B, though, then the probability of finding A<g<b becomes="" becomes="" greater="" than="" zero="" and="" non-infinitesimal,="" and="" you="" can="" then,="" and="" only="" then,="" achieve="">50% chance of success."

This is not true as we saw above. What is more important than bounded vs. unbounded is adversarial vs. non-adversarial. The results are the same in the bounded and unbounded cases.

Walt also said:
As the problem was originally stated, A and B are unbounded real numbers. Thus, the probability that one can find a G between A and B is an infinitesimal epsilon, greater than zero but smaller than any positive number. And the chance of success is 50% + epsilon, which converts to 50% as a real number.

This is not correct. There is no such thing as an "infinitesimal epsilon, greater than zero but smaller than any positive number." For any A and B, the epsilon is an actual positive number, with an infinity of smaller numbers still present between it and 0. The chance of success always remains measurably greater than 50%.

@Pauli
Benford's law is interesting and would have been relevant if the numbers were restricted to numbers with a fixed number of digits. This is not the case in this problem.

@sagef
I noticed in your code that you had used RANDBETWEEN(1,10). Didn't you mean RANDBETWEEN(0,10)? The former will give you success only 74.69% instead of 75% ☺

@Giovanni
Thanks! I agree, it is definitely a most fascinating problem. I think we are easily deceived not so much by statistical concepts, but by infinities. Most of the mental difficulties here relate to this.

The three mantras that need to be chanted ad infinitum ☺ are:
1.   Om â€" There are no uniform distributions on the real number line.
2.   Om â€" The probability of a number falling between A and B is not (B-A)/infinity. That quantity is undefined.
3.   Om â€" The actual probability of a number falling between A and B depends on the distribution and can certainly be finite."
P.M. Author

[precogmiles]

Hi

There you have it, in plain sight
link:s://:.quantamagazine.org/solution-information-from-randomness-20150722

Applying this to roulette, where numbers are arranged from 0-36, the optimum choice is  G=18, place 1 unit on zero and 37 units on high(19-36).
Worst deviation, 0 doesn't show up, no big deal. You now have a game with a p>0.50 EC.

So you bet 38 to win 36 back?

[CarpeDiem]

So you bet 38 to win 36 back?

With a p> 0.5, the focus is on the EC bet, not the single number 0. I meant a 18 units on EC and 1 unit on zero in a 37 spin cycle. You would bet 19 units to get a 18 units profit on a win and -19 on a loss, considering zero.
Trying to simplify a strategy to see the real value of a math certainty.
You can of course remove the zero altogether and stick to the method presented above, by picking G=randomly.
That would of course mean betting in a lot of cases on >18 individual numbers, good luck with that. The p of success would still be >50%.
Your choice.

[daveylibra]

Lets look at Bonus Question 1 from the link -

There are 2 differences between this and roulette.
One being that numbers can be repeated in roulette.
The other being that roulette numbers are integers.
I think that questions such as these allow any number whatsoever, eg 7.4218

The problem, as you say, is that although we win 75% of the time, this is offset
by the fact that we might bet > 27 numbers to achieve this  (0.75 x 36)
Its no good to bet when, say, A is close to 18, as this is placing conditions on the rules.

Can I go back to Euler's number at the start of the thread?

To simplify, let k=10. Then int (k/e) = 4.
At the 4th spin we note the highest number.
Then suppose at the 6th spin a higher number occurs, eg 30.
Are we saying that, on AVERAGE,  30 will be the highest number after k spins 33% of the time?
Or that, on average, 30 will be the highest number after k spins, AT LEAST 33% of the time?
It seems from the video that the former is the case.
If so, we can say that 30 WILL NOT be the highest number 67% of the time.
ie a number between 31-36 will appear between spin 7-10, 67% of the time.

I think I can answer my own question - the 33% can vary, and is subject to the value of the highest number.