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In or out? (Posted on 2014-03-28) Difficulty: 2 of 5
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 7-digit phone number makes the list?

No Solution Yet Submitted by Ady TZIDON    
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Solution very, very, very close to ... | Comment 1 of 2

2/7

but not totally exactly.

I'm sure the 7-digit 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 7-digit 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 7-digit 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 A-Z
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 2014-03-28 10:58:30
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