brifri238

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About brifri238

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  1. I am thinking the average length of a random chord is pi/4 (0.7854) as a first thought.
  2. I have run two simulations, each over 1,000,000 runs. The first simulates the scenario where one stick is randomly broken and then one of the resulting two sticks is randomly choosen and then randomly broken. The second scenario simulates where the stick is randomly broken and each of the two resulting sticks are then randomly broken and then one of the resulting four sticks is randomly choosen and thrown away: Scenario 1: Probability of Sucessful Triangle: 0.19378 Scenario 2: Probability of Sucessful Triangle: 0.19371 Assuming my simulations are correct, it practically makes no difference which scenario is choosen
  3. FYI: If you extend the sequence from 2 remainder 1 .... to bigger numbers possible solutions are (I am not absoutely sure they are the smallest number that works for each but these work!) 20 remainder 19 ......................232,792,559 22 remainder 21.......................232,792,559 23 remainder 22....................5,354,228,879 24 remainder 23....................5,354,228,879 30 remainder 29.............2,329,089,562,799 31 remainder 30...........72,201,776,446,799 36 remainder 35.........144,403,552,893,599 40 remainder 39......5,342,931,457,063,199
  4. For those of those familiar with SQL, these types of problems are easily solved by perfoming a query on a Numbers (or Tally) table. For example I have run the following selection on a numbers table (Numbers) containing 100,000,000 Rows (1,2,3....100,000,000) for the first 17 rows of the sequence (2 remainer 1 up to 18 remaining 17) The answer is 12,252,239 SELECT Min(Number) FroM Numbers WHERE Number%2 = 1 AND Number%3 = 2 AND Number%4 = 3 AND Number%5 = 4 AND Number%6 = 5 AND Number%7 = 6 AND Number%8 = 7 AND Number%9 = 8 AND Number%10 =9 AND Number%11 = 10 AND Number%12 = 11 AND Number%13 = 12 AND Number%14 = 13 AND Number%15 = 14 AND Number%16 = 15 AND Number%17 = 16 AND Number%18 = 17 (To get the solutions to next rows in the sequence: 19 remainder 18, 20 remainder 19...) requires a bigger Number table
  5. I re-ran both my simulations: scenario 1: with an infinite number of card decks, and the more real world scenario 2: with one deck of cards where cards are dealt without replacement. Both simulations were run over 1000,000,000 hands and the previous results and rankings were confirmed: Scenario 1: (infinite number of decks): Hand 1 (34.2521%), Hand3 (33.8808%) and Hand 2 (31.8671%) Scenario 2: (one deck without replacement): Hand 3 (34.0031%), Hand 1(33.9674%) and Hand 2 (32.0294%)
  6. Similar to Superprismatic I too ran a simulation (over only a mere 100,000,000 hands) and got similar results with Hand1 winning and Hand 3 in second.(Hand1: 34.257% / Hand3: 33.854% / Hand2: 31.889%). I also ran the simulation for a 'REAL WORLD' execution using one deck of cards without replacement. (Other assumptions the same, and in addition: if a hand goes over 21 no further cards are dealt to that person during that round-[this assumption is irrelevant when using an infinite number of decks]). In this case Hand 3 is the most likely to win! - (Hand 3: 34.0045 %, Hand 1 in a close second with 33.9777 % and Hand 2 with 32.0178%)