## Poll

16 votes (51.61%) | |||

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**31 members have voted**

Quote:ThatDonGuyMedian? I don't know - all I know is the mean.

Update - run 75,776 is still in progress, and is approaching 200 million rolls.

Oh. I thought you were getting output of how long each trial took. And would be simple enough to take the median over the 75,000 trials.

Quote:unJonOh. I thought you were getting output of how long each trial took. And would be simple enough to take the median over the 75,000 trials.

I changed my code to count how many times each length occurs rather than trying to save each length individually, and now I pretty much constantly get a median of 10 rolls.

Quote:ThatDonGuyI changed my code to count how many times each length occurs rather than trying to save each length individually, and now I pretty much constantly get a median of 10 rolls.

Actually, a median of 10 can be calculated without simulation.

Assume one die is red and the other is blue, and let "the difference" be the red sum minus the blue sum.

After one roll, the difference is {-5, -4, ..., 4, 5} with probability {1/36, 2/36, ..., 2/36, 1/36}

For the second roll, if the difference after one roll is -5, the new difference is {-10, -9, ..., 1, 0) with probability {1/36 x 1/36, 1/36 x 2/36, ..., 1/36 x 2/36, 1/36 x 1/36};

if it is -4, the new difference is {-9, -8, ..., 0, 1} with probability with probability {2/36 x 1/36, 2/36 x 2/36, ..., 2/36 x 2/36, 2/36 x 1/36};

and so on through +5 becoming {0, 1, ..., 10}, except that you skip where the previous difference is zero as you would have stopped rolling at that point.

Repeat for the third, fourth, etc. rolls until the sum of all of the probabilities that the sum is zero > 1/2.

Here are the probabilities that you will get matching sums in N rolls or fewer for various values of N:

1: 1 / 6

2: 163 / 648

3: 1,211 / 3,888

4: 300,139 / 839,808

5: 993,427 / 2,519,424

6: 231,266,905 / 544,195,584

7: 1,472,751,395 / 3,265,173,504

8: 222,694,111,193 / 470,184,984,576

9: 4,176,367,195,507 / 8,463,329,722,368

10: 116,789,309,836,109 / 228,509,902,503,936

9 rolls has a probability < 1/2, and 10 has one > 1/2, so the median is 10.

Quote:ThatDonGuyActually, a median of 10 can be calculated without simulation.

Assume one die is red and the other is blue, and let "the difference" be the red sum minus the blue sum.

After one roll, the difference is {-5, -4, ..., 4, 5} with probability {1/36, 2/36, ..., 2/36, 1/36}

For the second roll, if the difference after one roll is -5, the new difference is {-10, -9, ..., 1, 0) with probability {1/36 x 1/36, 1/36 x 2/36, ..., 1/36 x 2/36, 1/36 x 1/36};

if it is -4, the new difference is {-9, -8, ..., 0, 1} with probability with probability {2/36 x 1/36, 2/36 x 2/36, ..., 2/36 x 2/36, 2/36 x 1/36};

and so on through +5 becoming {0, 1, ..., 10}

Repeat for the third, fourth, etc. rolls until the sum of all of the probabilities that the sum is zero > 1/2.

Here are the probabilities that you will get matching sums in N rolls or fewer for various values of N:

1: 1 / 6

2: 163 / 648

3: 1,211 / 3,888

4: 300,139 / 839,808

5: 993,427 / 2,519,424

6: 231,266,905 / 544,195,584

7: 1,472,751,395 / 3,265,173,504

8: 222,694,111,193 / 470,184,984,576

9: 4,176,367,195,507 / 8,463,329,722,368

10: 116,789,309,836,109 / 228,509,902,503,936

9 rolls has a probability < 1/2, and 10 has one > 1/2, so the median is 10.

And there’s our gambling game. The house offers an even money bet that the red and blue dice are equal. Gambler can take the yes within nine rolls or the no within 10 rolls.

Rolls | Probability |
---|---|

1 | 0.166667 |

2 | 0.112654 |

3 | 0.092850 |

4 | 0.080944 |

5 | 0.072693 |

6 | 0.066539 |

7 | 0.061722 |

8 | 0.057819 |

9 | 0.054573 |

10 | 0.051819 |

11 | 0.049443 |

12 | 0.047367 |

13 | 0.045532 |

14 | 0.043895 |

15 | 0.042423 |

16 | 0.041089 |

Excel shows a very close fit to this curve is y = 0.1784*x

^{-1.011}, where x = number of rolls and y = probability.

And the "average" waiting time for equality is the dot product (or sumproduct in Excel) of those two columns, if extended to the infinite Excel spreadsheet in the sky..Quote:WizardThis table shows the probability of equal totals after exactly a given number of rolls for the first time.

Rolls Probability 1 0.166667 2 0.112654 3 0.092850 4 0.080944 5 0.072693 6 0.066539 7 0.061722 8 0.057819 9 0.054573 10 0.051819 11 0.049443 12 0.047367 13 0.045532 14 0.043895 15 0.042423 16 0.041089

Excel shows a very close fit to this curve is y = 0.1784*x^{-1.011}, where x = number of rolls and y = probability.

If you think about this integral for a few minutes and what function the sin(x) function approximates near x = 0, the answer to this integral is obvious. Getting there is the hard part.

Quote:teliotI saw this integral on Twitter yesterday. The method the person gave to get the answer seemed too advanced, and I thought there must be an easier way. I believe I found an easier way.

If you think about this integral for a few minutes and what function the sin(x) function approximates near x = 0, the answer to this integral is obvious. Getting there is the hard part.

I can't come up with an "exact" solution, but I think I have found a reasonable estimate:

Integrate by parts:

u = sin (x^n); dv = x^(-n) dx

du = n x^(n-1) cos (x^n) dx; v = x^(1-n) / (1-n)

INTEGRAL{sin (x^n) / x^n dx) = x^(1-n) sin (x^n) / (1-n) - INTEGRAL{n / (1-n) * x^(n-1) x^(1-n) cos (x^n) dx)

= x^(1-n) sin (x^n) / (1-n) - n / (1-n) * INTEGRAL{cos (x^n) dx)

= x^(1-n) sin (x^n) / (1-n) + n / (n-1) * INTEGRAL{cos (x^n) dx)

On the left half of the plus sign, x = 1 has a value of sin 1 / (1-n), and x = 0 has a value of 0, so, as n approaches +INF, the value approaches 0.

On the right side, note that, for 0 < x < 1, as n approaches +INF, x^n approaches 0, so the graph approaches a horizontal line from (0,1) to (1,1) and then a vertical line from (1,1) to (1,0), whose integral from 0 to 1 = 1.

The solution is 0 + 1 = 1.

I hate to just toss a sudoku on the table but, if you haven't played in a long time or only dabble, you might enjoy the cutting edge of these puzzles.

The one above--by the Dutch master Aad van der Wetering--at first glance seems impossible. However, in addition to the normal sudoku rules (which I assume you're familiar with), this puzzle has three special rules to help you solve it.

1: Both marked diagonals must contain the digits 1-9. (Just like a standard row or column.)

2: Cells that are a chess knight's move apart cannot contain the same digit. (Thus if you put a 6 in a cell and there's another 6 a knight's move away, something is wrong.)

3: The central 3x3 box (shaded gray) must form a magic square. (That is, each row, column and diagonal must add up to the same amount.)

A challenging puzzle. Good luck.