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So I know that rolling a fair six-sided die twice would expect the complete feasible outcomes would be 36, and also rolling the very same number twice would certainly be 2/36 or 1/18, but I feel choose that"s wrong. What am I doing that isn"t right?


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I think this diagram could help. It"s lifted from one more webwebsite, so just imagine that the "white die" denotes your initially roll, and the "red die" denotes your second roll.

You"ll notification that there are exactly $6$ pairs of rolls (out of all $36$ feasible outcomes) in which the initially roll equates to the second roll.

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The theoretical factor for this is well-defined in amWhy"s answer.


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Tright here are 6 possible numbers that can be rolled twice.

For a specific number, say $1$, the probability of rolling it twice is equal to $$ underbracedfrac 16_1 extst roll; imes ;underbracedfrac 16_2 extnd roll=dfrac 136$$ by the preeminence of the product.

Because this have the right to occur for $2, 3, 4, 5, 6,$ and also for $1$,

the probcapacity of rolling the very same number twice is $$6cdot dfrac 16cdot dfrac 16 = dfrac 16$$


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The initially roll doesn"t matter. It just sets a tarobtain. The possibilities of hitting a tarobtain in one roll are one in SIX, not one in 3. You can have actually someone contact out a number for the taracquire. That doesn"t issue.

See more: Analysis Of The Penalty Of Death By Hl Mencken Summary, The Penalty Of Death Mencken Analysis


Hint: as you proclaimed, there is a complete of $36$ outcomes. The outcomes you"re interested in are $(1,1),....$ , a full of _ outcomes out of $36$.


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