For different settings of the probability-to-be-green, we get different modes in the stalagmite. For example, when the probability-to-be-green is 50% and we're working with 9-blocks, we get a mode at both "4" and "5." By "mode," we mean that there will be more of that type of 9-block as compared to any other 9-block in the stalagmite. Let's call 9-blocks that have exactly 4 green squares "4zees," or in short, "4-z."
This riddle looks at the relation between the setting of probability-to-be-green and the mode. For instance, what would the mode be when the probability is 13%? Will the stalagmite max at the 1-z column? At the 2-z column?
Our riddle was: What is the range of probabilities at which a 2-z is
the most prevalent in a randomly-generated 9-block? To look at this question,
one must first determine a system for finding when a given n-z should be most prevalent and only then apply this system to the specific case of the 2-z. For instance, one could
safely assume that a 9-block with no green squares would be most prevalent if the
probability of getting green is 0; similarly, a block with all green squares would be
most prevalent with a probability of 1. Thus, the range of probabilities for each n-z is somewhere
between the extremes "0" and "1." This may seem trivial, but is useful in framing a range for the probability values.
The n-z and prevalence
For a particular n-z to be the mode, the probability of getting it must be greater than any other n-z. Note that the higher the
probability-to-be-green, the higher the probability of getting a "greener" n-z. In other
words, there is a better chance of randomly getting a 4-z at a probability of 50% than at a probability
of 10%. This is even more so for a 5-z, because the 5-z is "farther to the right."
Also, this means that the range of probabilities for which 5-z "rule" as the modes of an experimental outcome is immediately to the right of the respective range for 4-z. We see here a direct relationship between the range of n-z and the range of probabilities.
In general, one might say that if an n-z "rules" at some probability range that ends at X, then (n+1)-z will rule at a probability that begins at X. For instance, the 2-z will begin "ruling" where 1-z "concedes."
Multiplying it out
The point at which the 2-z becomes more probable than the
1-z is when the probability of getting each is equal.� The probability of an n-z =
(number of combinations) * (probability of getting the n-z), so:
For # of 1-z = # of 2-z (with x = probability of success):
9 * (x)1 * (1 - x)8 = 36 * (x)2
* (1 - x)7
9 * (x)1 * (1 - x)8 = 36
* (x)2 * (1 - x)7
9 * (1 � x) = 36 * x��
=>� �9 = 45 * x��
=> ��x = 1 / 5 = 2 / 10 = 0.2 = 20%
Thus, the count of 1-z and 2-z would be expected to be the
same at the probability 20%. This means that the beginning of the range
of prevalence of 2-z is at the probability of 20%.
For # of 2-z = # of 3-z (with x = probability of success):
36 * (x)2 * (1 - x)7 = 84 * (x)3
* (1 - x)6
36 * (x)2 * (1 - x)7 = 84
* (x)3 * (1 - x)6
36 * (1 � x) = 84 * x��
=>� �36 = 120 * x��
=> ��x = 3 / 10 = 0.3 = 30%
Thus, the count of 2-z and 3-z would be expected to be the
same at the probability 30%. This means that the end of the range
of prevalence of 2-z is at the probability of 30%.
Thus, the answer to the riddle regarding the range of probabilities
of greenness in which a 2-z is most prevalent is from 2/10 to 3/10 =
the probability of 20% - 30%, not inclusive, since we expect the # 2-z to equal the # 3-z
at the probability of 30%.
Now, to take it a step further, let us look at the general case; it uses
the �combination� function (nCr), which determines the number of combinations
there are of a permutation of greens and blues,
n C r = n! / ((n-r)!* r!) :
For # of (n)-z = # of (n+1)-z (with x = probability of
success, k = number of squares in the block - (k-block), not necessarily a
9-block):
k C n * (x) (n - 1) * (1 � x) (k + 1 - n)
= k C (n+1) * (x) (n) * (1-x) (k - n)
k C n * (x) (n - 1) * (1 � x) (k +
1 - n) = k C (n+1) * (x) (n) * (1-x) (k -
n)
(k! / (n! * (k � n)!) * (1 � x) = (k! / ((n + 1)! * (k - 1 �
n)!)* x
(k! / (n! * (k � n)!) * (1 � x) = (k!
/ ((n + 1)! * (k - 1 � n)!)* x
(1 � x) / (k � n) = x / (n + 1)
(n + 1) * (1 � x) = (k � n) * x
n + 1 = (k � n + n + 1) * x�
=>� x = (n + 1) / (k + 1)
Thus, the formula proves that the point at which the n-z is expected
to be most dominant between the probability of
n / (k + 1) to (n + 1) / (k + 1).
To check our work, we wrote a "Brute-Force" model that finds the
probability at which a certain n-z is most prevalent. This model finds
where the probability of a given n-z is highest in the range of 0% - 100%. To use it,
select the size of the side of the block (a side of 2 would produce the results for
a 4-block, side of 3, 9-block), then press [Setup]. Next, choose the n-z for which
you are looking, and press [Go]. When this process stops, the 'probability' slider will indicate the probability at which the
n-z beging to be prevalent:
CM ProbLab support model: 9-Block Stalagmite Summation
Don't see nothin'?
As may be seen from the model
simulation, the probability at which the dominant 9-block switches from 1 to 2
is 20%+, from 2 to 3 is 30%+, etc.�
Earlier, we established that the
formula for the range of the 2-z was from 2/10 to 3/10, or in general, the
formula for the range of a n-z is from n / (k + 1) to (n + 1) / (k + 1).
One could have expected ~ / k, instead of ~ / (k + 1), since one could have thought that for 9-blocks (k = 9),
we'd be dividing by 9 and not by 9+1.
This occurs due to the issue of the
�null� option.� A common misconception is
that a 9-block only has nine options (from 1 � 9 green squares);
however, there is also the option of 0 squares with the target-color, the �null�
option.
The confusion may occur because the user relates the �Sample
Stalagmite� situation to one involving dice, where there is no �null� option.
Thus, once one realizes that there are (k + 1) options, the
range of n / (k + 1) to (n + 1) / (k + 1) makes more sense.
Approximating the most prevalent n-z
Now, suppose that one, armed with knowledge of the fact that the range of 1-z starts at the
probability of 10%, attempts to guess the most probable n-z from a given probability-to-be-green.
Given the probability of 9%, one might say that the 1-z is most
prevalent, since 9% is closer to 10% than to 0% (the beggining of the range of
prevalence of the 0-z). Right? Wrong!!!
Such rounding does not work in this situation.
If one sets the side to 3 and the probability to 9%, it is closer to
10% (where the 1-z becomes more dominant) than to 0% (where 0-z is more
dominant).� However, it turns out that
the 0-z is the most dominant at the probability of 9%.
This may be surprising. However, if one thinks in terms of range-partitions,
the world starts making sense again:
What one may do is divide the range of probabilities
into (k + 1) portions, each with its own dominant n-z.
So, the most dominant n-z of a 9-block's first range-partition is the 0-z.
This first range-partition extends over the range of 1/10 = 10%; thus, the 0-z
range is 0% - 10%.
Therefore, it does not matter if the probability-to-be-green is 9% or 1%; it is
still within the 10% range of the 0-z, making the 0-z the mode.