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 On Average (Posted on 2004-01-26)
What is the expected number of rolls of a fair, normal 6-sided die, one is required to make, so that each of the 6 numbers comes up at least once?

Hint: this is not necessarily an integer answer
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As an aside, it would be interesting to see the computer program simulation of this, but this would not be proof of the solution (merely evidence supporting the proof).

 See The Solution Submitted by SilverKnight Rating: 3.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: The other answers | Comment 8 of 11 |

Here's a result from a simulation verifying, in addition to the mean, also the mode and median previously posted:

```
mean= 14.69898

6  1490  1490

7  3874  5364

8  5929 11293

9  7494 18787

10  8174 26961

11  8632 35593

12  8296 43889

13  7533 51422

14  6834 58256

15  6187 64443

16  5386 69829

17  4643 74472

18  3947 78419

19  3420 81839

20  2996 84835

where the program is

DEFDBL A-Z

RANDOMIZE TIMER

DIM numT(50)

totTook = 0: totTrials = 0

FOR trial = 1 TO 100000

totTrials = totTrials + 1

numHad = 0: thisTook = 0

DO

r = INT(RND(1) * 6 + 1)

END IF

totTook = totTook + 1

thisTook = thisTook + 1

IF thisTook <= 50 THEN

numT(thisTook) = numT(thisTook) + 1

ELSE

numT(0) = numT(0) + 1

END IF

NEXT trial

PRINT \"mean=\"; totTook / totTrials

FOR i = 6 TO 20

cumP = cumP + numT(i)

PRINT USING \"## ##### #####\"; i; numT(i); cumP

NEXT

----

and for the pair-of-dice version:

mean= 61.05001

33  1702 20384

34  1680 22064

35  1803 23867

36  1791 25658

37  1759 27417

38  1667 29084

39  1783 30867

40  1810 32677

41  1716 34393

42  1623 36016

43  1609 37625

44  1644 39269

45  1590 40859

46  1578 42437

47  1582 44019

48  1502 45521

49  1497 47018

50  1462 48480

51  1396 49876

52  1400 51276

53  1313 52589

54  1343 53932

55  1280 55212

56  1309 56521

57  1237 57758

58  1204 58962

59  1152 60114

60  1144 61258

from

DEFDBL A-Z

RANDOMIZE TIMER

DIM numT(100)

totTook = 0: totTrials = 0

FOR trial = 1 TO 100000

totTrials = totTrials + 1

numHad = 0: thisTook = 0

DO

r = INT(RND(1) * 6 + 1)

r2 = INT(RND(1) * 6 + 1)

r = r + r2

END IF

totTook = totTook + 1

thisTook = thisTook + 1

IF thisTook <= 100 THEN

numT(thisTook) = numT(thisTook) + 1

ELSE

numT(0) = numT(0) + 1

END IF

NEXT trial

PRINT \"mean=\"; totTook / totTrials

FOR i = 11 TO 60

cumP = cumP + numT(i)

IF i > 32 THEN

PRINT USING "## ##### #####"; i; numT(i); cumP

END IF

NEXT

--------

--------

Posted by Charlie
on 2004-01-26 22:44:38

```

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