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What is the probability of getting two numbers the same having three dices? I had this on my exam, it sounded super easy but isn't so.

Solution: I know that there are $$6^3$$ combinations for all three dices. Then I just wrote the following:

enter image description here

And the combinations that I want are: $$first-second:(1,1,x),(2,2,x),(3,3,x),(4,4,x),(5,5,x),(6,6,x)$$

$$second-third:(x,1,1),(x,2,2),(x,3,3),(x,4,4),(x,5,5),(x,6,6)$$ $$first-third:(1,x,1),(2,x,2),(3,x,3),(4,x,4),(5,x,5),(6,x,6)$$ wich gives me the combination of $6+6+6=18$ and final result $18/216$, which is incorrect. What am I doing wrong?

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    $\begingroup$ Exactly two the same, or at least two the same? $\endgroup$ – tilper Sep 7 '16 at 14:27
  • $\begingroup$ Exactly the same $\endgroup$ – eugene_sunic Sep 7 '16 at 14:28
  • $\begingroup$ Refer to Lovsovs' answer-hint. What you've done so far doesn't exclude, for example, (1,1,1). $\endgroup$ – tilper Sep 7 '16 at 14:29
  • $\begingroup$ Still unclear. Do you mean exactly 2 the same? $\endgroup$ – drhab Sep 7 '16 at 14:29
  • $\begingroup$ Yes exactly to the same... $\endgroup$ – eugene_sunic Sep 7 '16 at 14:29
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Hint: For each of your eighteen cases, what can $x$ be?

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  • $\begingroup$ depends, for (1,1x) x can be [2-6] but still confused..., and that 5/6 but what do I do with it now? $\endgroup$ – eugene_sunic Sep 7 '16 at 14:28
  • $\begingroup$ @wesewx The point is that for each of the cases you listed, $x$ can be $5$ different things. Can you see the solution now? $\endgroup$ – Bobson Dugnutt Sep 7 '16 at 14:36
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    $\begingroup$ @wesewx For each of the eighteen cases, you have $5$ different possibilities for $x$. That gives $18\times 5$ possibilities in all. $\endgroup$ – Bobson Dugnutt Sep 7 '16 at 14:39
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    $\begingroup$ @wesewx And your reputation has grown uptil now with $15$. Hold on!! $\endgroup$ – drhab Sep 7 '16 at 14:43
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    $\begingroup$ @drhab you mean +17, answer accpeted-->+2 xd $\endgroup$ – eugene_sunic Sep 7 '16 at 14:48
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Alternative solution:

$$\Pr(\text{exactly }2\text{ the same})=1-\Pr(3\text{ distinct})-\Pr(3\text{ the same})=1-\frac56\frac46-\frac16\frac16$$

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