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The expected number of times you would roll a 3 in 100 rolls is equal to the probability of rolling a 3 on one roll times 100. So - do you know the probability of rolling a 3? Hint: there are 6 sides on a die, and there is one 3.

BTW- when you do the math you'll find that you get a non-integer answer. Which means if you run this experiment you can never actually get the expected value. So despite its name you can't "expect" to get the "expected value." What you can expect is that if you do this experiment over many trials 1/2 the time you should get less than the expected value and 1/2 the time you'll get more than the expected value.

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Originally Posted by ebaines

What you can expect is that if you do this experiment over many trials 1/2 the time you should get less than the expected value and 1/2 the time you'll get more than the expected value.

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Strictly speaking, not exactly so. This is a discrete outcome experiment, and in general, the most probable outcome is not the same as the average outcome after (infinitely) many experiments. 100/6 = 16.(6), i.e. closer to 17 than to 16. No matter how many times one tries this experiment, the probability of getting 16 threes will always be slightly lower then the probability of getting 17 threes and so on. I guess the problem is really confusing: one never gets (or expects) in this single experiment anything but some whole number of threes. I would say the answer is: the number of threes to expect most is 16 or 17, but one can expect anything from 0 to 100.