I roll a fair die and write the accrued sums,(e.g. (1,3,6,2,2...)=>(1,4,10,12,14...))  after a while I let the software to continue generation of the series' members.
What is the probability that my 7digit phone number makes the list?
2/7
but not totally exactly.
I'm sure the 7digit proviso was given so as to make the number very large. Also, it leaves the solver without a knowledge of what the actual number is, so the probability would be based on knowing only that it is indeed a 7digit number.
The expected solution would be that on average 6 throws would advance the count 21 positions, so that 6/21 of sums would be achieved, for a probability of 6/21 = 2/7. Simulation shows in fact that even for numbers as small as 100, the average hovers around 2/7 statistically speaking. A simulation I just ran had 285366 hits out of 1000000 trials, trying to hit 100 exactly.
However, there is an initiation effect that makes different numbers have different probabilities. The following table shows the effect for small numbers:
goal probability numerator and denominator decimal
1 1 6 .1666666666666667
2 7 36 .1944444444444445
3 49 216 .226851851851852
4 343 1296 .2646604938271605
5 2401 7776 .3087705761316872
6 16807 46656 .3602323388203018
7 70993 279936 .2536043952903521
8 450295 1679616 .268094016727633
9 2825473 10077696 .2803689454414978
10 17492167 60466176 .2892884610397721
11 106442161 362797056 .293393122241874
12 633074071 2176782336 .2908302132602385
13 3647371105 13060694016 .2792631923335612
14 22219348327 78364164096 .2835396585074294
15 134526474769 470184984576 .2861139321373954
16 809860055095 2821109907456 .2870714299200451
17 4852905842113 16926659444736 .286701924733424
18 29004175431175 101559956668416 .2855867251486823
19 173492524161649 609359740010496 .284712810463423
20 1044275922856663 3656158440062976 .2856210801517332
21 6273265544452129 2.193695064037786D+16 .2859679837591171
Note that the decimal probability is converging on that for 2/7 ~= .2857142857142857. But this leads us to feel that 2/7 is only asymptotically approached but never reached. But for practical purposes the answer would be 2/7. Taking the average over all 7digit numbers would be even closer than for one individual such number, but still wouldn't be exact.
The program producing the above table:
DECLARE SUB roll (wh#)
DEFDBL AZ
CLEAR , , 25000
DIM SHARED tot, goodct, goal
OPEN "in or out.txt" FOR OUTPUT AS #2
FOR goal = 1 TO 21
tot = 0: goodct = 0
roll 1
PRINT goal, goodct; 6 ^ goal, goodct / 6 ^ goal
PRINT #2, goal, goodct; 6 ^ goal, goodct / 6 ^ goal
NEXT
CLOSE
SUB roll (wh)
FOR r = 1 TO 6
tot = tot + r
IF tot = goal THEN goodct = goodct + 6 ^ (goal  wh)
IF tot < goal THEN
roll wh + 1
END IF
tot = tot  r
NEXT
END SUB

Posted by Charlie
on 20140328 10:58:30 