November 3rd, 2011 at 6:55:34 AM
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First time at this site. I was hoping someone could help me derive a classical process to calculate the expected number of dice throws with two dice to obtain all of the eleven sums. I understand the same problem with one die and the six possible outcomes, but I'm having difficulty when the outcomes are not equally likely. Thanks for any help.

November 3rd, 2011 at 7:00:38 AM
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The faces are equally likely, they just add up differently.

November 3rd, 2011 at 10:29:44 AM
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Quote:FleaStiffThe faces are equally likely, they just add up differently.

Really? You think that is the answer?

November 3rd, 2011 at 11:17:06 AM
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Let X be a random variable (r.v.) that represents the sum of both dice.

Let X1 be r.v. for outcome of die 1

Let X2 be r.v. for outcome of die 2

P( X = 2 ) = P( X1 = 1 ) * P( X2 = 1 ) = 1/36

P( X = 3 ) = P( X1 = 1 ) * P( X2 = 2 ) + P( X1 = 2 ) * P( X2 = 1 ) = 2/36

and so forth.

E[X] = 7

Edit: I clearly misunderstood the OP

Let X1 be r.v. for outcome of die 1

Let X2 be r.v. for outcome of die 2

P( X = 2 ) = P( X1 = 1 ) * P( X2 = 1 ) = 1/36

P( X = 3 ) = P( X1 = 1 ) * P( X2 = 2 ) + P( X1 = 2 ) * P( X2 = 1 ) = 2/36

and so forth.

E[X] = 7

Edit: I clearly misunderstood the OP

Wisdom is the quality that keeps you out of situations where you would otherwise need it

November 3rd, 2011 at 1:45:28 PM
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removed

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November 6th, 2011 at 11:39:11 AM
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I tried to find BrunoZ and could not find the post. I did arrive at the same results in different simulation. Thanks for the info, but I would like to find BrunoZ.

November 6th, 2011 at 5:38:56 PM
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Quote:mustangsallyThis was solved this past summer over at 2+2 forum by the handsome math guru BruceZ.

I used the BruceZ method and agree with his answer.

For now, I don't think I can explain his answer any better than he did, but will try for my next Ask the Wizard column. Meanwhile, the following table shows the expected number of rolls to get a 2, 2-3, 2-4, 2-5, ...

Highest Number | Prob | Exp wait | Prob rolled before last | Prob rolled last | Exp rolls |
---|---|---|---|---|---|

2 | 0.027778 | 36.0 | 0.000000 | 1.000000 | 36.000000 |

3 | 0.055556 | 18.0 | 0.666667 | 0.333333 | 42.000000 |

4 | 0.083333 | 12.0 | 0.850000 | 0.150000 | 43.800000 |

5 | 0.111111 | 9.0 | 0.922222 | 0.077778 | 44.500000 |

6 | 0.138889 | 7.2 | 0.956044 | 0.043956 | 44.816484 |

7 | 0.166667 | 6.0 | 0.973646 | 0.026354 | 44.974607 |

8 | 0.138889 | 7.2 | 0.962994 | 0.037006 | 45.241049 |

9 | 0.111111 | 9.0 | 0.944827 | 0.055173 | 45.737607 |

10 | 0.083333 | 12.0 | 0.911570 | 0.088430 | 46.798765 |

11 | 0.055556 | 18.0 | 0.843824 | 0.156176 | 49.609939 |

12 | 0.027778 | 36.0 | 0.677571 | 0.322429 | 61.217385 |

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan

November 6th, 2011 at 8:50:59 PM
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Thanks heaps Sally - I knew it was something simple. SixHorse

November 6th, 2011 at 10:40:56 PM
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Quote:mustangsallyYou want to know the average wait time for all 11 numbers to roll from 2 dice. Excellent math question.

This is not even a normal distribution.

The mean of 61.2 is not even close to a mode of around 35-38 and they are not close to the median of 52.

I would like to see someone here at WoV create a formula that would solve for the distribution.

It's not normal, it's Poisson. See this post for the formula (from a textbook by Sheldon Ross).

"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563

November 8th, 2011 at 4:52:07 PM
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removed

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