February 20th, 2012 at 10:46:00 PM
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Quote:WongBoIt doesn't really have an answer does it?

It just has a range of probabilities?

I'm sure the Wizard could use on of his iterative techniques to figure out the exact number of average rolls. That's well beyond my highschool math education, however.

February 21st, 2012 at 1:24:02 AM
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Quote:MathExtremistI don't think he should be able to turn that in.

Question: does a roll that puts you below zero not count, equivalent to Chutes and Ladders? Or is the ending criteria zero or lower, equivalent to Candy Land?

(It's a safe bet I have kids, huh?)

lets say is zero or below

February 21st, 2012 at 1:37:07 AM
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Quote:WongBoIt doesn't really have an answer does it?

It just has a range of probabilities?

actually, getting an "expected number of roll" is the easier part of the question. "a range of probabilities" would be more difficult as you would have to figure out the standard deviation (in order to find a 95% confidence interval etc).

February 21st, 2012 at 7:41:02 AM
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I calculate 33.37217893, which comes very close to another poster's simulation.

So, how many times did it take your son?

So, how many times did it take your son?

I heart Crystal Math.

February 21st, 2012 at 9:22:13 AM
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something tells me what the teacher thinks the answer is, is actually incorrect

the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder

February 21st, 2012 at 5:47:33 PM
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Quote:CrystalMathI calculate 33.37217893, which comes very close to another poster's simulation.

So, how many times did it take your son?

before i started i think it is going to take FOREVER, but luckily i reached 0 in about 20 rolls

February 21st, 2012 at 6:18:04 PM
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i tried calculating using excel assuming it takes 10 rolls to reach 0 (without hitting the 1), and in average you hit the 1 in every 6 rolls.

it shows you need about 40.2 rolls.

it shows you need about 40.2 rolls.

February 21st, 2012 at 8:14:24 PM
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Yeah, I'm having a hard time getting this calculated.

I know that if you just discarded 1s, the probabilities are:

The problem is what happens when a one is rolled?

I know that if you just discarded 1s, the probabilities are:

# of Rolls | Probability |
---|---|

7 | 0.0004608 |

8 | 0.0291072 |

9 | 0.1779712 |

10 | 0.336336384 |

11 | 0.28534636544 |

12 | 0.129685766144 |

13 | 0.034703417344 |

14 | 0.00574724915200001 |

15 | 0.000600678760448001 |

16 | 0.0000393485418496 |

17 | 0.00000155612348416 |

18 | 0.00000003414556672 |

19 | 0.000000000347602944 |

20 | 0.000000000001048576 |

The problem is what happens when a one is rolled?

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You want the truth! You can't handle the truth!

February 21st, 2012 at 8:14:25 PM
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Doubled

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You want the truth! You can't handle the truth!

February 22nd, 2012 at 11:21:04 AM
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This, I calculated through the use of a Markov chain. Yesterday, I used a stochastic matrix with an absorbing state to calculate only the average number of rolls to end. There is almost exactly a 40% chance of finishing in 20 rolls or fewer. About 1 in 20 kids would have to go to 85 or more rolls.

# of Rolls | Probability |
---|---|

7 | 0.000128601 |

8 | 0.006790838 |

9 | 0.035641678 |

10 | 0.06121854 |

11 | 0.054355896 |

12 | 0.036897524 |

13 | 0.028020105 |

14 | 0.025764809 |

15 | 0.025430765 |

16 | 0.025400113 |

17 | 0.025389245 |

18 | 0.025307461 |

19 | 0.024970031 |

20 | 0.024189534 |

21 | 0.023035933 |

22 | 0.021817628 |

23 | 0.020776793 |

24 | 0.019942031 |

25 | 0.019230398 |

26 | 0.018567709 |

27 | 0.017919202 |

28 | 0.017274663 |

29 | 0.016634267 |

30 | 0.016003349 |

31 | 0.015389895 |

32 | 0.01480094 |

33 | 0.01423942 |

34 | 0.013703981 |

35 | 0.013191251 |

36 | 0.012698135 |

37 | 0.012222728 |

38 | 0.011764137 |

39 | 0.011322 |

40 | 0.010896106 |

I heart Crystal Math.