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Dice Probabilties
| January 12th, 2012 at 6:12:16 AM permalink | |
| Swlong1981 Member since: Jan 12, 2012 Threads: 1 Posts: 2 | First time posting, I have 2 questions. In 15 rolls of a a 6 sided die, what is the probability that any 1 number does not appear? In 3 rolls of 5 six sided dice, what is the probability that any 1 number does not appear? I understand that the probability of 1 number not appearing is 15.38/1. What I am more interested in is if you didn't have to choose which number will not appear, what is the overall chance of any specific number not showing up. Thanks in advanced for your answers. |
| January 12th, 2012 at 9:30:44 AM permalink | |
| Ayecarumba Member since: Nov 17, 2009 Threads: 113 Posts: 2047 | Homework? The probability any one number not appearing in a single roll of a six sided die is 5/6. The odds of it not appearing in 15 rolls is (5/6)^15. This is the same whether a single die is rolled 15 times in sucession, or five dice are thrown three times, as each roll is independent. If it doesn't matter which number, I think you multiply the above by 6 to get the probability of any of the six numbers not appearing in 15 rolls. But that is just a guess on my part. |
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| guido111 Member since: Sep 16, 2010 Threads: 5 Posts: 480 | I will answer the first question then look at the second one when I have more time. For "any" number instead of a "specific" number this can be solved with inclusion-exclusion principle. In Roulette (0,00) board, what is the probability that any number will not have hit by the 200th spin? 5th question from the bottom shows how to do the math. He also covers the principle here What is the expected number of rolls of two dice for every total from 2 to 12 to occur at least once? My answer using his formula is: For at least 1 number not appearing. 0.389432829 - 0.034255919 + 0.000610352 = ~0.3557872614049 simulation results below 64.41% showed all 6 numbers. p = 1/6added And a table for rolls 6 to 40. 1 - 6 sided die at least 1 number not showing in n rolls
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| January 12th, 2012 at 9:51:00 AM permalink | |
| guido111 Member since: Sep 16, 2010 Threads: 5 Posts: 480 | For any "specific" number the above formula is correct. That is to say before any trials begin, a number is chosen. |
| January 12th, 2012 at 10:19:39 AM permalink | |
| Swlong1981 Member since: Jan 12, 2012 Threads: 1 Posts: 2 | Thanks for the quick response. I assure you it's not homework, just a question that came up around the office. We have been playing shake a day in the office, which no one has one yet, and have been trying to come up with other "games" with the 5 dice. So my understanding is that ~35% of the time at least one number will not show up when you roll 5 dice 3 times? We are trying to figure out if the "player" chooses a specific set of numbers (i.e 1, 3, 5 or 2,3,4,6) or all of the numbers 1-6. What would be the likely hood of those events happening Again, thanks for the help! |
| January 12th, 2012 at 10:33:06 AM permalink | |
| guido111 Member since: Sep 16, 2010 Threads: 5 Posts: 480 | No. I only answered your first question above. "In 15 rolls of a a 6 sided die, what is the probability that any 1 number does not appear?" answer at least 1 number ~35% exactly 1 number ~32% |
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