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Here's the question: A casino game rolls three 6-sided dice
a) How many possible outcomes are there?
b) Suppose a player wins if at least two of the three dice end up with the same number. How many possible winning outcomes are there?

I am still bit stuck on how to do this, and what formula to use.
Would anyone help me by explaining in a way that I am able to understand??
It would be much appreciated

Thanks

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  • $\begingroup$ For a) do you sum the outcomes? $\endgroup$ – Alex Mar 2 '15 at 22:10
  • $\begingroup$ Yes you do. But I am unsure what to use, either the Combinations or the Permutations formula??!! $\endgroup$ – Alexei Murphy Mar 2 '15 at 22:13
  • $\begingroup$ I find it easiest to think about problems like this assuming that the dice are distinguishable (i.e. a red one, a blue one, and a yellow one). Then, see what you have to do to change it for dice that are indistinguishable, i.e. all white dice. $\endgroup$ – TravisJ Mar 2 '15 at 22:15
  • $\begingroup$ Yeah true, this is actually the first time I've really got a question like this. But thanks for the tip $\endgroup$ – Alexei Murphy Mar 2 '15 at 22:17
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For a):

Since there are $6$ possible outcomes for each roll, the number of different (ordered) outcomes on 3 rolls is $6\cdot{6}\cdot{6} = 216$.

For b):

Hint: The easiest way might be to first compute the number of outcomes when the player does NOT win i.e. where all dice end up with different numbers.

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  • $\begingroup$ Cool thanks. But, did you have to use any formulas?? $\endgroup$ – Alexei Murphy Mar 2 '15 at 22:16
  • $\begingroup$ This assumes the dice are distinguishable... i.e. a red/yellow/and blue die, or that the 3 dice are rolled in sequence. $\endgroup$ – TravisJ Mar 2 '15 at 22:16
  • $\begingroup$ So, just wondering, how do you know if it is ordered or not? Because can it also be not ordered?? $\endgroup$ – Alexei Murphy Mar 2 '15 at 22:34
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    $\begingroup$ You can use as sample space the three numbers showing, say listed in non-decreasing order. So for example an outcome could be $2,2,5$ or $3,4,5$ or $6,6,6$. Unfortunately, these outcomes are not equally likely, which makes computing probabilities more cumbersome. If instead we imagine the dice to be blue, white, and red, and list the result of a roll in that order, then we get $6^3$ equally likely outcomes, and calculating probability comes down to counting. $\endgroup$ – André Nicolas Mar 2 '15 at 22:58
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For this you would want to use combinations not permutations, since the dice are indistinguishable. $6^3$ gives the number of possible rolls under the assumption that $(3,3,2)$ is different than $(3,2,3)$ but since this is a casino and casinos don't have numbered or different colored die, $(3,3,2)$ should be the same as $(3,2,3)$ since there is no way to tell these two results are distinct. Aka, order doesn't matter, so this is a combinatorics problem.

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  • $\begingroup$ The question isn't clear whether it wants to count these as different outcomes or not. However, I suspect that the intention is to consider the dice as distinguishable, since in that case dividing the number of successful outcomes (b) by the total number of outcomes (a) gives the probability. $\endgroup$ – Especially Lime Nov 12 at 12:50

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