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 Posted: Thu Jul 12, 2007 7:45 am
I was watching a Lotto on my guild that rolled 6 20-sided dice. If you got 2 out of the 6 right or 6 of 6 right, you won something. I was wondering this: What is the probability that 2 of your 6 dice will be the same as 2 others.
At first I figured 1/400 chance of getting two right but that would only be true if you were rolling only 2 dice. Since you're rolling 6, what would the probability be then?
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Posted: Thu Jul 12, 2007 7:50 am
Hmm... I'm new here and I didn't see the Mathematics subforum. I was wondering a mod could move this to their, since this is obviously misplaced...
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Posted: Thu Jul 12, 2007 9:12 am
Robertp3001 Hmm... I'm new here and I didn't see the Mathematics subforum. I was wondering a mod could move this to their, since this is obviously misplaced... done
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Posted: Thu Jul 12, 2007 12:51 pm
Thank you but this question is still on my mind.
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Posted: Thu Jul 12, 2007 1:10 pm
Robertp3001 I was watching a Lotto on my guild that rolled 6 20-sided dice. If you got 2 out of the 6 right or 6 of 6 right, you won something. I was wondering this: What is the probability that 2 of your 6 dice will be the same as 2 others.
At first I figured 1/400 chance of getting two right but that would only be true if you were rolling only 2 dice. Since you're rolling 6, what would the probability be then? What is the probability that you get two pairs when rolling six 20-sided dice? As in: 1, 18, 5, 18, 1, 7 ?
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Posted: Thu Jul 12, 2007 8:43 pm
What is the probability that if you roll g 20-sided dice that another person will roll at least 2 of the same numbers as you.
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Posted: Thu Jul 12, 2007 9:38 pm
I just read across a similar problem in a book I was reading a few days ago. tomorrow, I'll post up an answer for you. Just to clarify things, you're rolling 6 dice 20-siders, and you want to know what the odds are that two of them will be the same?
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Posted: Fri Jul 13, 2007 11:48 am
The odds that 2 of the dice you roll were the same as the new roll. I forgot to add the fact that the must be in the same position too. Like if you roll a 6 on your first roll or the six rolls. The next six rolls must also have a 6 in the first roll or it doesn't count.
Ex.
1st roll: 1,2,3,4,5,6
2nd roll:1,5,1,2,5,4
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Posted: Fri Jul 13, 2007 12:45 pm
Oh, thats a bit different than that problem I saw, but I'll still get on it. Okay, I've got an answer. Now in your first post it said 2 out of 6 or 6 out of 6, but I wasn't sure whether you meant only 2 out of 6 (i.e. if you got 3 out of 6, you got nothing) or at least 2 out of 6, so I calculated both. Luckily, the second case just builds on the first, so I didn't have to do much more.
In the first case, only 2 out of 6, we want to figure out the odds of the first positions and second positions being the same or the first and third, or the second and sixth, etc. By listing these out, we have 15 differant ways of grouping 1 through six in non repeating pairs. All of them also have equal odds, so once we figure one out, we have them all. That simplifies the problem down to 15 * P(1 and 2). Since the probability of 1s and 2s being the same are independent events, it simplifies again into 15 * P(1) * P(2). Since they are twenty sided dice, P(1) = P(2) = (1/20), because once you roll the first roll, the odds of it appearing in the second roll are 1/20, and the same with each of the 6 die you are using. Now that we have the numbers we need, the odds of having only 2 be the same are 15 / 400 -> 3 / 80 -> 3.75%
For the second case, you do the same thing as the first case and try to find the odds of only 3, only 4, only 5, and all 6 being the same, then you add them to the odds of only 2 being the same to get the odds of at least 2 being the same. Of course, that isn't much more than only 2. If we do the grouping thing with 3, we get 20 differant ways to arrange 6 numbers non-repeating. The Odds are 20 * P(1) * P(2) * P(3), and since we know what P(1) though P(6) are already (1/20), we can tab it up. In the end, all of them added together becomes:
(15 * P^2) + (20 * P^3) + (10 * P^4) + (6 * P^5) + (1 * P^6) where P = 1/20.
Adding that up--hopefully by calculator and not by hand--gives 4.006439%.
I hope that helps.
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Posted: Sat Jul 14, 2007 6:05 am
Sorry but I'm not good with putting problems into words or vice-versa. It looks like what I was looking for but I'm going to look over it later today. Thanks for helping.
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Posted: Sat Jul 14, 2007 3:54 pm
I actually worked on writing up equations for dice rolling, and theres a minature calculator too: http://bugaremote.kwikphp.com/dice.phpI end up getting: 2097525/64000000 In the chances that you roll the same number AT LEAST twice on a 20 sided-die with 6 rolls. I hope this is what you mean. That comes up to be about: 3.28%
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Posted: Sat Jul 14, 2007 4:57 pm
Another way to look at it is to ask how likely is it that on the second six rolls, none of them match the corresponding roll from the first six rolls, and then what's the chance that only one will match up?
So we say: the chance that none of them will match is (19/20)^6, since each roll can only be one of 19 choices since one number is already taken due to the first set of rolls, and the chance that only one of them will match is 6*1/20*(19/20)^5 (i.e. there are six choices for the matching roll, that roll has a 1/20 chance of actually matching, and then for each successive roll there are only 19 choices for each)
So we end up with (19/20+6/20)*(19/20)^5 = (25/20)*(19/20)^5 = 96.7226171875%.
Now, we know that one of the following must be true: none of them match, one pair matches, or at least two pairs match. Since we want the last case, and we know the first two cases, we can subtract the probability for the first two cases from 1 to get the probability of the last case.
So the probability that at least two match is 100%-96.7226171875% = 3.2773828125%, as Buga calculated.
Codelabmaster got the wrong result, because, for instance, in his two-matches case, he neglected to take into account what the other dice are doing. There are a bunch of (19/20)^n factors that he's missing.
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