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Random # Generation Error?
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Transitional_Mil wrote
at 1:13 PM, Wednesday March 7, 2007 EST
Over the past week or so, I have noticed that a very high percentage of 8v8 rolls contain adjacent doublings of dice scores. For example:
gunsnbombs's turn Kaio defeated 8v8: 29 to 25 (4,4,5,4,2,6,1,3 to 5,1,4,1,2,1,6,5) Kaio defended 8v8: 20 to 24 (1,5,3,2,1,4,2,2 to 2,2,1,6,2,2,4,5) 2fast's turn jacobsladder defeated 8v7: 23 to 17 (5,5,2,1,2,5,2,1 to 1,1,6,1,3,4,1) jacobsladder defeated 7v5: 26 to 18 (5,5,2,1,3,6,4 to 1,3,2,6,6) jacobsladder's turn 2fast defeated 8v8: 37 to 28 (6,5,5,6,4,4,3,4 to 6,6,1,4,3,4,1,3) Kaio's turn 2fast defeated 8v8: 34 to 31 (6,4,3,3,3,4,6,5 to 6,6,2,6,1,4,3,3) 2fast's turn jacobsladder defeated 8v8: 32 to 26 (1,6,1,5,6,6,6,1 to 6,4,5,1,5,2,2,1) Of these these 14 rolls (1 each for offense and defence), all but 1 contained at least one instance of adjacent doubling. Is this something that should happen statistically, or is it a possible bug in the random number generator? I have seen myriad examples of this tendency toward doubling. Any thoughts? |
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Transitional_Mil wrote
at 3:51 PM, Wednesday March 7, 2007 EST In response to no_wolf:
The probability of rolling the same number twice in a row in a two dice roll = 1/36 There are seven adjacent pairs of dice in an 8 die roll. The chances are equal that any of these pairs could be doubles, and you can sum the probability for one pair times the number of pairs. This is how I reached the 7/36 figure for an 8-die roll. Is this not correct? |
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fuzzycat wrote
at 3:54 PM, Wednesday March 7, 2007 EST Well, Your statistics *are* tainted.
Just do it as simple as i did, make on a sheet on one side strokes for every number that is NOT doubled (first digit does not count, as it is not a "dice after a dice" event) make on the side strokes for every number that IS doubled (same number as the digit before) Calculate doubles / (doubles + not doubles), and you will be pretty near the 16,7%, a little higher is not a surprise when your sample is not taken by random, but taken what you consider to be significant evidence for doubles. |
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Semagon wrote
at 3:59 PM, Wednesday March 7, 2007 EST No it's not :)
The chances of no pairs is (5/6)^7. The first dice is a probability of 1. Just because you don't care about the first number. The probability of the next dice not matching is 5/6. Then given whatever that number is, the probability of the next dice not being the same is again 5/6. Multiply that 5/6 7 times for 7 potential pairs, and you get approx. .28. So, there's a 28% chance you _won't_ get a pair, and (1-.28)= .72 = = 72% chance you will get at least one. |
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Transitional_Mil wrote
at 4:00 PM, Wednesday March 7, 2007 EST Ah! I was wrong...the two dice pair probability is 16.7%. You were right!
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Semagon wrote
at 4:01 PM, Wednesday March 7, 2007 EST And yeah, the singular of "dice" is "die", let's face it, we all do it, and I didn't mean to type "= =" at the end there, just "="
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no_Wolf wrote
at 4:01 PM, Wednesday March 7, 2007 EST TM:
The probability of two dice being: 1 and 1; 1/36 2 and 2; 1/36 3 and 3; 1/36 4 and 4; 1/36 5 and 5; 1/36 6 and 6; 1/36 any pair; 6/36 = 1/6 |
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Ryan wrote
at 4:01 PM, Wednesday March 7, 2007 EST I see one mistake, (i think), in your equation. Out of 36 combinations of 2 dice there are 6 doubles not 1, (1-1,2-2,3-3 etc) which is 6/36 or 1/6.
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Ryan wrote
at 4:03 PM, Wednesday March 7, 2007 EST damn you no wolf!
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fuzzycat wrote
at 4:03 PM, Wednesday March 7, 2007 EST "
The probability of rolling the same number twice in a row in a two dice roll = 1/36 " This is wrong. Your probability of getting at least one double is: 7 events of a dice after a dice. every event has a 1/6 chance to have the same number as the digit before. or have a 5/6 chance having NOT the same digit as before. You can multiply the chances of getting NO double. (they do *not* add, take look e.g. http://www.edcollins.com/backgammon/diceprob.htm) So your total probability of NOT getting at least one double is: 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 = 0,28% Your probality to get ONE double in 8 dice is 1-0.28 = 72% So 3 out of 4 rolls do have a double! I hope to have clarified this now for good. |
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Transitional_Mil wrote
at 4:03 PM, Wednesday March 7, 2007 EST Thank you Semagon. Your stats are correct. I am feeling more comfortable now. Thanks for the input everyone!
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