sloonz comments on A very strange probability paradox

[sloonz]

Note that a slightly different worded problem gives the intuitive result :

A_k is the event “I roll a dice k times, and it end up with 66, with no earlier 66 sequence”.

B_k be the event “I roll a dice k times, and it end up with a 6, and one and only one 6 before that (but not necessarily the roll just before the end : 16236 works)”.

C_k is the event “I roll a dice k times, and I only get even numbers”.

In this case we do have the intuitive result (that I think most mathematicians intuitively translate this problem into) :

Σ[k * P(A_k|C_k)] > Σ[k * P(B_k|C_k)]

Now the question is : why are not the two formulations equivalent ? How would you write “expected number of runs” more formally, in a way that would not yield the above formula, and would reproduce the numbers of your Python program ?

(this is what I hate in probability theory, where slightly different worded problems, seemingly equivalent, yields completely different results for no obvious reason).

Also, the difference between the two processes is not small :

Expected rolls until two 6s in a row (given all even): 2.725588
Expected rolls until second 6 (given all even): 2.999517

vs (n = 10 millions)

k	P(A_k|C_k)	P(B_k|C_k)
-------------------------
1	0.000000	0.000000
2	0.111505	0.111505
3	0.074206	0.148227
4	0.074719	0.148097
5	0.066254	0.130536
6	0.060060	0.108174
7	0.053807	0.086706
8	0.049360	0.067133
9	0.046698	0.050944
10	0.040364	0.038915
11	0.038683	0.029835
12	0.030691	0.024297
13	0.034653	0.011551
14	0.036450	0.014263
15	0.024138	0.006897
16	0.007092	0.007092
17	0.012658	0.000000
18	0.043478	0.000000
19	0.000000	0.000000
20	0.000000	0.000000

Expected values:
E[k * P(A_k|C_k)] = 6.259532
E[k * P(B_k|C_k)] = 5.739979