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5:16 PM Z<Zajt> If I throw a dice 3 times, how can I calculate the probability that two of the dices(not all three) will give the same result?
5:17 PM I<Indivicivet> Zajt: is it still okay if all three have the same result, or do you want a 2/1 split?
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5:17 PM Z<Zajt> it should only be two of the dices, not all three Indivicivet
5:17 PM F<FilipinosRich> Zajt how many stats can you do
5:17 PM I<Indivicivet> Zajt: hint - you can assume the first dice lands on 1
5:18 PM I<Indivicivet> (roll one dice first; if it's not a 1, relabel the dice such that it is a 1 :P)
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5:18 PM Z<Zajt> the probability is 1/6 to get 1
5:18 PM Z<Zajt> but don't know how to calculate the probability so two dices get the same
5:19 PM I<Indivicivet> Zajt use the hint :P then consider the next two die rolls
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5:19 PM Z<Zajt> why relabel it?
5:19 PM I<Indivicivet> it makes it easier to think about the next two rolls; you don't *have* to relabel it
5:20 PM I<Indivicivet> but relabelling means I can work it out in my head :P
5:20 PM I<Indivicivet> (I hope...)
5:20 PM I<Indivicivet> Zajt oh, alternative way to consider it is to think about the chance of getting 3 of the same, and the chance of getting all different, then subtract these from 1
5:21 PM Z<Zajt> hmm alright but I don't know how to calculate it still :/
5:23 PM T<thoolihan> prob of getting all diff should be 5/6 * 4/6 (no prob on first dice, 5 unique numbers left on second dice, 4 left on third...)
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5:24 PM Z<Zajt> why 5/6 * 4/6 there thoolihan ?
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5:24 PM J<joel135> [precisely two equal] = [1st=2nd!=3rd] + [1st=3rd!=2nd] + [2nd=3rd!=1st] = (1/6)*(5/6) + (1/6)*(5/6) + (1/6)*(5/6) = 15/36 = 5/12
5:24 PM Z<Zajt> 5/6 is the probability to not get one number
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5:25 PM Z<Zajt> joel135: the first one there, 1/6 is to get the probability on the first dice, but why multiply it with 5/6?
5:26 PM J<joel135> [2nd!=3rd] = 5/6
5:26 PM S<stefan27> or combinatorically, since each outcome 111 , 112 , ... , 666 are equally likely. you just need to count strings of type aab, aba or baa with a!=b and then divide by 6^3  and there's 6*5 of type aab, 6*5 of type aba, 6*5 of type baa... so 30*3/6^3=5/12
5:27 PM S<stefan27> where 112 would be interpreted as dice1 landed on 1 dice2 landed on 1 dice3 landed on 2
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5:28 PM Z<Zajt> why is the probability 5/6 when the second dice and the third dice shouldn't be the same joel135 ?
5:28 PM J<joel135> (by symmetry, 1st=2nd and 2nd=3rd are independent events so [1st=2nd!=3rd] = [1st=2nd][2nd!=3rd])
5:29 PM J<joel135> Zajt: [2nd=3rd] = 1/6, so [2nd!=3rd] = 5/6
5:29 PM T<thoolihan> "lternative way to consider it is to think about the chance of getting 3 of the same, and the chance of getting all different, then subtract these from 1". That comes out to 36/36 - 1/36 - 20/36 = 15/36.   1/36 comes from 1/6 * 1/6.  20/36 comes from 5/6 * 4/6.
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5:32 PM Z<Zajt> joel135: so to get the same number in the second and the third, it is 5/6 because 1/6 is in the second and 1/6 in the third and then you take 1-1/6 to fix so they are not the same?
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5:32 PM J<joel135> I don't know what "1/6 is in the second" means.
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5:33 PM T<Trashlord> a sixth of the value is in the second result
5:33 PM T<Trashlord> or something like that
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5:35 PM S<Sterile> Zajt, what's the original question
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5:36 PM Z<Zajt> joel135: the probability when you throw the dice
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5:41 PM J<joel135> I don't see in your explanation how you combine the two 1/6 to get a single 1/6 for use in calculating 1-1/6=5/6. Otherwise it sounds good.
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