Will the rolling of Six fair dice, one after another, give me $\left(\frac{1}{6}\right)^6$? I could not figure out the answer of second one.
Six fair dice are rolled one after the other. What is the probability of getting $111111$? and what is the probability of getting $111222$?
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probability-distributions
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0What is the chance of getting a 2 with a fair dice? – 2017-01-29
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2The second one is also $\frac{1}{6^6}$. What's so special about $1$ or $2$? They both have the same probability of coming up on a fair roll of the dice. – 2017-01-29
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0@Student it is 1/6? – 2017-01-29
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0So, you could find the answer for 111111... How did you do that? And could you apply it to 111222? – 2017-01-29
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1@астонвіллаолофмэллбэрг Thanks for reply. So you mean the probability of getting 111111 and 111222 when we roll six dice, ONE AFTER ANOTHER is same. That is (1/6)^6 – 2017-01-29
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1Just for clarification: I hope you don't mean to say the outcome has three $1'$s and three$2'$s because then more possibilities will occur. – 2017-01-29
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0@Student thanks for reply. This is what I am thinking. Chance of getting 1 in each roll is 1/6. For six dice, rolled one after another it should be (1/6)^6? Am I right? – 2017-01-29
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0Yes. However, this is only the case if we want that specific order! If the order does not matter, there are more ways in having three 2's and 1's with 6 throws – 2017-01-29
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0@AnuragA. Exactly, you are right. The order is 111222. No other possibilities. Thanks for Reply. – 2017-01-29
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0@Student. Exactly. The required order is only 111222. – 2017-01-29
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0@Acrobat Exactly. – 2017-01-29