# If Grace and her friends were to roll the two dice 100 times, how many of the 100 times could they expect to roll the same number on each die? [closed]

need some help on this question I tried the question out but still don't understand

## closed as off-topic by Saad, Peter Foreman, Jean-Claude Arbaut, Lee David Chung Lin, darij grinbergApr 23 at 15:39

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The probability of both dice having the same number is $$\frac{6}{36}$$, as there are $$36$$ different outcomes, $$6$$ of which have two of the same number, i.e. $$(1,1), (2,2),...$$.

The expected number of rolls of this type in $$100$$ pairs of dice rolls is $$100*\frac{6}{36}$$

It is like,

($$1$$ of grace and $$1$$ of his friend) or ($$2$$ of grace and $$2$$ of his friend) or ($$3$$ of grace and $$3$$ of his friend)...........

$$\implies (\frac{1}{6} \times \frac{1}{6}) + (\frac{1}{6} \times \frac{1}{6})+(\frac{1}{6} \times \frac{1}{6})+(\frac{1}{6} \times \frac{1}{6})+(\frac{1}{6} \times \frac{1}{6})+(\frac{1}{6} \times \frac{1}{6})$$

$$\implies 6 \times \frac{1}{36}$$

$$\implies \frac{1}{6}$$

This means

If they roll $$6$$ times they get $$1$$ time the same

So,

If they roll $$100$$ times they get $$\frac{100}{6}$$ the same.