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'Snake-Eyes ' Joe (Posted on 2008-07-28) Difficulty: 3 of 5
"Snake-Eyes" Joe introduced a die of his own into a game of chance.

He was subsequently challenged that the die was biased.

Very rigorously test to see if there are grounds to substantiate this claim; don't accept just two or three trial runs. Are you able to offer a theoretical model consistent with your findings?

Test "Snake-Eyes" Joe's Die with this simulator which has a run of 60,000 at a time:

No:123456Total
Scores 0 0 0 0 0 0 0

Note: the data changes with each subsequent mouse-over visitation to the link.

See The Solution Submitted by brianjn    
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re: error in solution - fallacy? | Comment 7 of 13 |
(In reply to error in solution by Eigenray)

Your proposal aligns with the proposal put forward by Paul  but I am struggling with the concept of "11".

I roll 9 1's and increment that total.  On the 10th 1 the counter slot is overridden but the substitute value may be a 1.

In all fairness of discussion I copied 26*  data tables from the problem's data generator to form the following tableau:

  1    9273    10033    10240    10266    10092    10096    60000
  2    9242    10229    10072    10048    10231    10178    60000
  3    9158    10083    10274    10223    10161    10101    60000
  4    9231    10137    10046    10217    10228    10141    60000
  5    9344    10020    10187      9918    10271    10260    60000
  6    9176    10063    10182    10183    10158    10238    60000
  7    9061    10239    10126    10169    10256    10149    60000
  8    9270    10208    10347      9934    10097    10144    60000
  9    9322    10098    10338    10049    10189    10004    60000
10    9090    10333    10209    10154    10069    10145    60000
11    9167    10120    10148      9969    10352    10244    60000
12    9270    10090    10172    10169    10267    10032    60000
13    9317    10051    10203    10105    10181    10143    60000
14    9257    10135    10157    10045    10155    10251    60000
15    9257    10111    10147    10193    10009    10283    60000
16    9157    10116    10022    10256    10121    10328    60000
17    9172    10109    10245    10189    10115    10170    60000
18    9122    10096    10258    10096    10329    10099    60000
19    9200    10202    10226    10210    10032    10130    60000
20    9142    10319    10276    10175    10075    10013    60000
21    9271    10151    10127    10179    10079    10193    60000
22    9354    10204    10106    10062    10142    10132    60000
23    9131    10036    10169    10033    10363    10268    60000
24    9254    10354    10123    10079    10084    10106    60000
25    9260    10211    10023    10165    10223    10118    60000
26    9200    10073    10230    10178    10180    10139    60000
    239698   263821  264653  263264  264459  264105    
    9219.1  10146.9 10178.9 10125.5 10171.5 10157.8   10156.2
   
In the last line of the tableau the average value for 1 falls short of that my suggestion of Paul's and the average value for the other 5 is marginally higher than that calculated value.

That I have a value 60 less than Paul for "1" and 13 for the other values may seem insignificant but placed as Paul suggests, any discrepancy could advantage a casino.

Convince me that there is a fallacy in dealing with the program source and I will ensure an adjustment to the listed solution.

* This was arbitrary, where I stopped on the spreadsheet.

  Posted by brianjn on 2008-08-01 05:45:07

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