Having 360 numbered balls in a box

[Steven G]

Oh, this is very math related.

Plus, I'm only answering things I'm fluent in, just like you said.

I prefer that you say 'just like I asked' instead of just like I said. I am not your boss or a moderator on this forum so i can't tell you what to do.

 

[Cubist]

I've worked out the difference between the methods used by @pka and myself.

Consider this simplified scenario:- 3 balls in the box; 3 picks; need to have numbers 1 and 2 for success. The lines labelled "<->" are successful...

       pka     |   my method
               |
       ***||   |   111 
       **|*|  <->  112 121 211 
       **||*   |   113 131 311 
       *|**|  <->  122 212 221 
       *|*|*  <->  123 132 213 231 312 321
       *||**   |   133 313 331
       |***|   |   222 
       |**|*   |   223 232 322
       |*|**   |   233 323 332
       ||***   |   333

RESULT  3/10      12/27 = 4/9

Which method is correct? I feel 80% towards my method - since it captures the fact that some ball combinations can be picked in more ways than one and are therefore more likely?

 

[pka]

If a camp ground has three cabins and three campers show up to a weekend visit then how many ways can the cabins be occupied?
\(\dbinom{3+3-1}{3}=10\). This in known as an occupancy question. We do not care who is where only the distribution of the people or occupants.
On the other hand, How many solutions are there to \(X+Y+Z=9\) where each variable a positive integer(i.e. at least one).
The answer is \(\dbinom{6+3-1 }{6}\) the number of ways to place six ones into three slots.

 

[Dr.Peterson]

If a camp ground has three cabins and three campers show up to a weekend visit then how many ways can the cabins be occupied?
\(\dbinom{3+3-1}{3}=10\). This in known as an occupancy question. We do not care who is where only the distribution of the people or occupants.
On the other hand, How many solutions are there to \(X+Y+Z=9\) where each variable a positive integer(i.e. at least one).
The answer is \(\dbinom{6+3-1 }{6}\) the number of ways to place six ones into three slots.

But the problem given to us is not a mere occupancy problem; it is a probability problem.

What is the probability that there is someone in each of the first two cabins, if each person randomly chose a cabin? That's what you have to answer.

 

[pka]

But the problem given to us is not a mere occupancy problem; it is a probability problem.

We just disagree on that point. I think that it is just a mere occupancy problem.

 

[Odylio]

I'm curious about the source of this problem... please let us know!

You may find the origin of this question a bit shocking perhaps.
...Or maybe not. ?

First of all, thank you very, very much for all your effort, and your answer !
Appreciate it...

But since you asked, it has to do with the book of Ezekiel in the Old Testament Bible.
This particular book contains a total of 13 dates..., 4 of which are:
- the 1st day of the 1st month (or 11)
- the 1st day of the 5th month (or 15)
- the 5th day of the 4th month (or 54)
- the 10th day of the 7th month (or 107)

And we reckon a Biblical calendar to have 360 days.

Without going into too much detail...
In another study of some recent church-history these 4 dates play a crucial part,
in a structure where they form the 4 main waymarks within a single year.
Some very significant events took place on these 4 particular dates.
And only afterwards did we look into a possible connection with the book of Ezekiel.

Studying the book of Ezekiel with a somewhat "arithmetic" approach for some 2 years now, brought forward some very interesting patterns that defy any mathematical probability.
That is to say, the likelihood that these patterns that were discovered are mere coincidence..., is near zero.

The answer to my question on this forum seems to add more weight to this supposition.

 

[tkhunny]

1) The probability, whatever it is, doesn't change if you start with ANY four values. The fact that you managed to construct them in an arbitrary manner does not add any weight to your supposition. Pasting together numeric dates and numeric months? What even is that? It is nothing.

2) History is filled with meaningless numerology that has been used to mislead unsuspecting victims. The likelihood that these patterns are "discovered" is relatively high for that very reason - even if they are totally meaningless and 100% coincidental. Confirmation bias is a difficult thing to assess in one's own mind.

 

[Cubist]

3432*(5*t^9-90*t^8+780*t^7-5460*t^6+37044*t^5-186900*t^4+540150*t^3-645660*t^2-279394*t+978385) / t^13 where t=360
≈ 0.00000097178529996
or around 1 in 1,029,034

I have a slight correction to the above, it ought to be:-

3432*(t-2)*(5*t^8-80*t^7+620*t^6-2960*t^5+9389*t^4-20072*t^3+28066*t^2-23368*t+8855) / t^13
≈ 0.00000097179055874
or around 1 in 1,029,028

 

[Cubist]

I've done further research into this problem. The following method seems more elegant than my previous one. It uses Stirling numbers of the second kind:-

Number of ways to match is [math] \sum_{i=0}^{p-m}{S(p-i, m) m! \binom{p}{i} \left(t-m\right)^i} [/math]
Where
t = 360, the total number of balls
p = 13, the pick size
m = 4, the number of balls to match

The method is based on an answer in this post. To apply it to this problem I considered the number of ways to pick a matching set of balls, but where "i" of the picked balls don't match. This can then be summed over all the possible values of i to get the final answer.

Breaking the expression down:-

[math] S(p-i,m) m! [/math] gives the number of ways that the "p-i" matching balls can be distributed among the "m" matches with at least one ball for each match (in any order, thus the factorial)

[math] \binom{p}{i} [/math] the ways that we can choose the matching balls from the number picked, since choose(p,i)=choose(p,p-i)

[math] (t-m)^i [/math] the ways that we can choose the set of unmatched balls

Here's some python code that performs the calculation:-

Spoiler: reveal code

from sympy.functions.combinatorial.numbers import stirling
from sympy import binomial
from math import factorial

m=4   # Number of matches
t=360 # Number of balls in the bag
p=13  # Number of picks

r=0
for i in range(p-m+1):
    r += stirling(p-i, m) * factorial(m) * binomial(p,i) * (t-m)**i

print("Probability is", r, "/", t**p)
print("Or approx 1 in", float(t**p) / r)