I can Predict your response with '90%' accuracy.

[SidYuca]

I Enjoyed this thread and thought that I'd contribute.

1. Multiply 2 single digits to form a product
2. Continue to multiply the resulting product by a single digit of your choice until said product is 7 digits (or more in length)
3. Circle one of the non-zero digits of the product (for me to guess) and post the remaining digits in any order.
4. I will post your circled digit..

Two questions: How? and give a mathematically valid reason for % accuracy.

 

[CajunCenturion]

@kwbMitel - I emailed you the complete set of numbers last Tuesday.

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[kwbMitel]

You Did?

Hmmm, I'll have to check my junk folder.

Did that list include the quantities of factors like in Karluks sample previously? This is what I really need to calculate what I am trying to do.

**********************************************
What's most important is that you realise ... There is no spoon.

 

[SidYuca]

Come on karluk! Mercy... I was proposing a rough method of getting a number closer to 90% than 67% and your banging me for a rather rather big discrepancy of 3%. PLUS attributing it to combining green balls and not the following.

I took 8 picks working on an average of 5 to get 7 digits numbers when in reality 5^8 is 390,625 certainly a rather rather big discrepancy from 7 digits and 5^9 being close to 2,000,000 and missing a sampling. Even taking the true average of 5.5 ^8 is a 6 digit number. Could it be that my "rather big discrepancy" came from that and not combining the 3's and 6's.

I believe this crude approach was more than sufficient to justify my simulated statistical approach which the problem statement was based upon. After nearly 100 posts you were the one to provide a 'mathematical solution' based upon 8 picks.

 

[karluk]

I was proposing a rough method of getting a number closer to 90% than 67%

I don't think anyone involved in this thread has ever doubted that the probability is closer to 90% than 67%, assuming that you are allowed to force the clue generator into using your method of multiplying together randomly selected single digits. The problem, of course, is that you are NOT allowed to force him into doing anything of the kind, based on the way the problem is stated. As I clearly said when I first pointed out the 67% figure:

The actual percentage of numbers that meet these criteria is about 67.0%. To get your 90+ percent figure, you have to make unwarranted assumptions about how the numbers are selected. (Emphasis added.)

The "unwarranted assumptions", of course, is your implicit assumption that you can force the clue generator into using your clue generation technique, without an explicit rule that forces him to do it your way. You never declared such a rule, so the odds are NOT 90%.

Even taking the true average of 5.5 ^8 is a 6 digit number. Could it be that my "rather big discrepancy" came from that and not combining the 3's and 6's.

No. Your black ball/green ball model consistently produces excessively high estimates of the percentage of n-digit numbers that are divisible by nine, compared with the actual percentage of such numbers. The error persists for higher values of n.

After nearly 100 posts you were the one to provide a 'mathematical solution' based upon 8 picks.

I'm not sure I was the first. CajunCenturion calculated a 93.75% probability without explaining his methods. I would be interested in seeing how he derived this number, since it doesn't appear to be consistent with my formula.

 

[karluk]

Sorry, "n" is the number of single digit numbers multiplied together to form a product. In my previous post, I incorrectly stated that it was the length of the product. Sorry for the confusion.

 

[kwbMitel]

@karluk - as yet I do not think you have responded to my critisisms of your method. Possibilities vs. Probabilities. You are the skilled player who can manipulate the odds by methods beyond the norm. This does not change the odds as they exist but only biases them in your favour. I do not disagree for one moment your claim that 67% of the numbers are divisible by nine. I do dispute your obvious claim that each number has equal weighting with respect to probability. The odds of blackjack do not change when a card counter is playing. The odds of Craps do not change when the dice thrower can manipulate his throw. The odds do not change in this case by knowing all the possible outcomes and choosing selectively. You've proven the odds can be beaten, not that the odds are different.

**********************************************
What's most important is that you realise ... There is no spoon.

 

[karluk]

This does not change the odds as they exist

Even though I made the wisecrack in other thread about being a "smart-aleck student named karluk", I was dead serious about the general principle I stated there that probabilities are undefined in the absence of a probability density function that determines them. The game in this thread has NEVER, EVER had such a function defined on the seven digit numbers being produced by the person generating clues. Hence, when you talk about "odds as they exist", you are missing the point that the odds simply didn't exist for this game. Take a look at any introductory book on probability theory. Every single "compute the odds" problem is accompanied by either the actual probability density function, or enough information to determine what it should be. This thread has been sadly lacking the same basic information. Furthermore, for almost all of the thread, I was under the impression that this was an intentional omission by SidYuca, not an inadvertent oversight. In the absence of a probability density function, I naturally interpreted the phrase "single digit of your choice" as an open invitation to the clue generator to pick a distribution that maximized his edge. So to me a phrase like "odds as they exist" means "odds as they exist after the clue generator has decided what they should be".

The odds of blackjack do not change when a card counter is playing. The odds of Craps do not change...

You are talking about two games that have well-defined rules that mandate random number generation and how the randomization device works. In that respect they are both different from this "guess the missing digit game". It's ridiculous to talk about things like the "odds do not change in this case" when there are no odds, not even the slightest suggestion that the clue generator should use any sort of randomization at all, let alone randomization that leads to a huge advantage for the guesser.

 

[kwbMitel]

Thanks Karluk, I have a need to understand. I continue to calculate the odds based on random numbers.

**********************************************
What's most important is that you realise ... There is no spoon.

 

[SidYuca]

kwbMeitel said:
"I do not disagree for one moment your claim that 67% of the numbers are divisible by nine."

kwbMeitel, this statement is confusing. It would be more accurate to say That:

"Of all the UNIQUE prime factorizations I generated from a combination of the primes 2,3,5,and 7 67% of them (i.e. 882) were divisible by 9."

Those 882 can be selected a great deal MANY more times selecting from the product of digits 2 through 9 digits

11111320 is represented as a single entry of the 882 numbers disguised as its unique prime factorization of "2^3 * 4^3 * 5^1 * 7^3" with a factor of 9 and represented ONE time in the 'good' pool to be later compared to Good/(Good+bad)

But how many times can it be selected from the products of digits (2-9)? Too many to write out BUT I did earlier with the unique factorization of 36... i.e. 2^3 * 3^2 with digits 2-9?
4*9
2*2*9
4*3*3
2*2*3*3
6*6
2*3*6
2*3*2*3
2*3*2*3

 

[kwbMitel]

But how many times can it be selected from the products of digits (2-9)?

This is the angle I was investigating when Karluk responded most recently. I can accept Karluk's reasoning that without rules absolutly stating the number selection process then any selection process is valid. 67% is correct for his selection process.

I've calculated all of the different ways that the 882 numbers can be generated. 1111320 can be generating 16 ways.

There are a total of 27366 ways to generate numbers that are divisible by nine. 4677 ways are not. This = 85.4%

This number is still not a representation of random numbers though as the distribution is not even - Usage Shown Below.

2 105478
3 60920
4 45655
5 37556
6 32148
7 28825
8 26117
9 23914


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What's most important is that you realise ... There is no spoon.

 

[SidYuca]

KwbMitel,

I believe that the numbers do not have to be in a random distribution or order to pick/make one random. No need to study sand distribution at the beach before making a random i.e. unbiased selection.

Those 27366 which have factors of 9 were 7 digits in length and made only from products of digits 2-9? If so would you share code?

 

[kwbMitel]

Actually, randomization is absolutely necessary to achieve 90% or better. The distribution of number usage although not needing to be equal, needs to be within a margin of error.

With random numbers it is far more likely to get 8388608 by one of the 55 methods that is not 2^23. 2^23 does not have the same probability as many other numbers simply because it uses more digits to get there. If all combinations used the same quantity of digits then it would not matter but the fewest is 8 and the most is 23. Somewhere, there is a formula that can do this but I don't know it and I would have some difficulty deriving it

I wish I could write code to do this but it's too complicated for me. I resort to massive spreadsheets with numerous formulas to generate numbers such as these. It takes me much longer than most people but I get the job done. The advantage for me is that the a save all the numbers and methods. What I can give you is the .csv of all the numbers and their factors (not just the prime factors but all of them).

http://www.mediafire.com/?em9ammjkkvdkwrt

**********************************************
What's most important is that you realise ... There is no spoon.

 

[kwbMitel]

Another thing about the distribution of numbers.

Of the 882 there are 65 that are so unlikely so as not to occur even after 3,000,000 trials

In the same 3,000,000 the bottom 441 only account for 66,000 occurances or .02% 215 of the 291 numbers that are not divisible by 9 are in this group.

All of the top 140 are divisible by 9 and combined they occur 78% of the time.

To say that your puzzle is weighted (biased) towards numbers that are divisible by nine is an understatement.



**********************************************
What's most important is that you realise ... There is no spoon.

 

[SidYuca]

kwbMitel,
re. the 2 above statements. They may just account for the problem statement. In a hat with 9 red counters and one white counter I can predict with 90% etc... I base my refusal to continue on my 5th ammendment rights.