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Each of the six sides of a dice is marked with one of the numbers $1,2,3,4,5,6$. The numbers are chosen at random and independently. The dice is thrown twice. What is the probability that the two throws show the same number?

Choosing one out of six, is done with the probability $1/6$. A dice shows a certain side upp also with probability $1/6$. So for example, the probability of choosing the number $1$ and then throwing a number $1$ is $1/6\cdot 1/6=1/36.$

I'm trying to solve this by drawing a tree diagram. For the first throw, the probability to get a certain number is $6\cdot 1/36=1/6$, and to get the same number on the second throw is then $1/6\cdot 1/6 = 1/36.$

The correct answer is $11/36.$ What am I missing?

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    $\begingroup$ The term "dice" (for which the singular is "die") is potentially misleading. What's going on here is that each side of an initially blank cube is being marked, independently and at random, with a number between $1$ and $6$. $\endgroup$ – Barry Cipra May 20 '18 at 17:46
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When you throw the die the second time, it either lands with the same face showing as on the first throw, or with one of the other five faces showing. If it lands with the same face showing (which happens with probability $1/6$), the numbers certainly match. If it lands with a different face showing (which happens with probability $5/6$), the numbers match with probability $1/6$. The total probability is thus

$${1\over6}\cdot1+{5\over6}\cdot{1\over6}={11\over36}$$

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  • $\begingroup$ Why would the numbers ever match if it landed on a different face? $\endgroup$ – Rhys Hughes May 20 '18 at 17:38
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    $\begingroup$ @RhysHughes, because of what the OP says: The numbers are chosen at random and independently. You (and fgrieu) are picturing a standard die; that's not what the OP describes. I've added a comment below the OP to that effect. $\endgroup$ – Barry Cipra May 20 '18 at 17:47

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