In the General Case approach, we considered situations where the target-total (hence variable k) is one more, two more, and three more than the number for which we are looking (the addend, hence variable n). Using these cases, we attempted to prove that the number of target addends (n) is expected to be 1/n of the number of addends of 1. Recall that showing that the frequency of n is 1/n as compared to the frequency of '1' is equivalent to showing that the frequency of n+1 is n/(n+1) as compared to the frequency of n.
The general cases
We will look at the case of k = 7, and, within that case, n will begin as being equivalent to k and then decrease by 1. For each of these sub-cases, we will give the combinations and express the probabilities and expectation values both in terms of the specific integers and in terms of the general case (see table, below).
(We chose the target-total of 7, instead of 5, which was used in other examples, because it gives greater flexibility in writing this proof.)For target-total of k = n (the case in which n = 7 and k = 7): ( for n > 0)
Combination
Probability
Expectation
7
(1/7)
1/7
Total
1/7
General case: (k = n proof)
Combination
Probability
Expectation
n
(1/n)
1/(n)
Total
1/n
For target-total of k = n + 1 (the case in which n = 6 and k = 7): ( for n > 1)
Combination
Probability
Expectation
61
(1/7)
1/7
16
(1/7)(1/6)
1/42
Total
7/42 = 1/6
General case: (k = n + 1 proof)
Combination
Probability
Expectation
n1
1/(n+1)
1/(n+1)
1n
[1/(n+1)](1/n)
1/[(n)(n+1)]
Total
(n+1)/[(n)(n+1)] = 1/n
For target-total of k = n + 2 (the case in which n = 5 and k = 7): (for n > 2)
Combination
Probability
Expectation
511
(1/7)(1/2)
1/14
52
(1/7)(1/2)
1/14
151
(1/7)(1/6)
1/42
115
(1/7)(1/6)(1/5)
1/210
25
(1/7)(1/5)
1/35
Total
42 / 210 = 1/5
General case: (k = n + 2 proof)
Combination
Probability
Expectation
n11
[1/(n+2)](1/2)
1/[(2)(n+2)]
n2
[1/(n+2)](1/2)
1/[(2)(n+2)]
1n1
[1/(n+2)][1/(n+1)]
1/[(n+2)(n+1)]
11n
[1/(n+2)](1/(n+1)](1/n)
1/[(n)(n+1)(n+2)]
2n
[1/(n+2)](1/n)
1/[(n)(n+2)]
Total
(n+1)(n+2)/[(n)(n+1)(n+2)]=1/n
For target-total of k = n + 3 (the case in which n = 4 and k = 7): (for n > 3)
Combination
Probability
Expectation
4111
(1/7)(1/3)(1/2)
1/42
421
(1/7)(1/3)
1/21
412
(1/7)(1/3)(1/2)
1/42
43
(1/7)(1/3)
1/21
1411
(1/7)(1/6)(1/2)
1/84
142
(1/7)(1/6)(1/2)
1/84
1141
(1/7)(1/6)(1/5)
1/210
1114
(1/7)(1/6)(1/5)(1/4)
1/840
124
(1/7)(1/6)(1/4)
1/168
241
(1/7)(1/5)
1/35
214
(1/7)(1/5)(1/4)
1/140
34
(1/7)(1/4)
1/28
Total
210 / 840 = 1/4
General case: (k = n + 3 proof)
Combination
Probability
Expectation
n111
[1/(n+3)](1/3)(1/2)
1/[(6)(n+3)]
n21
[1/(n+3)](1/3)
1/[(3)(n+3)]
n12
[1/(n+3)](1/3)(1/2)
1/[(6)(n+3)]
n3
[1/(n+3)](1/3)
1/[(3)(n+3)]
1n11
[1/(n+3)][1/(n+2)](1/2)
1/[(2)(n+3)(n+2)]
1n2
[1/(n+3)][1/(n+2)](1/2)
1/[(2)(n+3)(n+2)]
11n1
[1/(n+3)][1/(n+2)](1/(n+1)]
1/[(n+3)(n+2)(n+1)]
111n
[1/(n+3)][1/(n+2)][1/(n+1)](1/n)
1/[(n+3)(n+2)(n+1)(n)]
12n
[1/(n+3)][1/(n+2)](1/n)
1/[(n+3)(n+2)(n)]
2n1
[1/(n+3)][1/(n+1)]
1/[(n+3)(n+1)]
21n
[1/(n+3)][1/(n+1)](1/n)
1/[(n+3)(n+1)(n)]
3n
[1/(n+3)](1/n)
1/[(n+3)(n)]
Total
(n+1)(n+2)(n+3) / [(n+3)(n+2)(n+1)(n)] = 1/n
The above examples show a case for which the expectation value of n in the partition of a target total of n+1, n+2, n+3 is 1/n. We will use this to conjecture about the general case.
If k = n + 1 , the ratio of n's to (n+1)'s is n : n+1 (since the probability of (n+1)'s is 1/(n+1), by the k = n proof, and the probability of n's is 1/n, by the k = n+1 proof)
If k = n + 2 , the ratio of (n+1)'s to (n+2)'s is n+1 : n+2 (since the probability of (n+2)'s is 1/(n+2), by the k = n proof, and the probability of (n+1)'s is 1/(n+1), by the k = n+1 proof).
Also, the ratio of n's to (n+1)'s is n : n+1 (since the probability of (n+1)'s is 1/(n+1), by the k = n+1 proof, and the probability of n's is 1/n by the k = n+2 proof).
Thus, utilizing the two ratios, one may say that the ratio of n's to (n+2)'s is n : n+2 .If k = n + 3 , the ratio of (n+2)'s to (n+3)'s is n+2 : n+3 (since the probability of (n+3)'s is 1/(n+3), by the k = n proof, and the probability of (n+2)'s is 1/(n+2), by the k = n+1 proof).
Also, the ratio of (n+1)'s to (n+2)'s is n+1 : n+2 (since the probability of (n+2)'s is 1/(n+2), by the k = n+1 proof, and the probability of (n+1)'s is 1/(n+1), by the k = n+2 proof).
Also, the ratio of n's to (n+1)'s is n : n+1 (since the probability of (n+1)'s is 1/(n+1), by the k = n+2 proof, and the probability of n's is 1/n, by the k = n+3 proof).
Thus, utilizing the ratios above, one may say that the ratio of n's to (n+1)'s is n : n+1.
Also, utilizing the ratios above, one may say that the ratio of (n+1)'s to (n+2)'s is n+1 : n+2.
Also, utilizing the ratios above, one may say that the ratio of (n+2)'s to (n+3)'s is n+2 : n+3 .
What is prevalent in all three scenarios is that the ratio of n's to (n+1)'s stays constant at n : n+1 .
Furthermore, if we were to go on with the next scenario of k = n + 4, we would have found that the ratio of (n+1)'s to (n+2)'s is n+1 : n+2 , just like in the previous scenarios.Though we have not presented the k = n + 4 scenario, we can assume that there, too, the ratio of n : n+1 would bear.
Thus, though this proof is incomplete, one may see the trend of the probability of getting a certain number, n, in the partitions is 1/n, given any k and any n.