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Help me about expected value

I read the article about 'Expected Value' but didn't really understand the concept of expected value. Please someone explain in a bit more details. And if you can, then specifically explain what does it mean " If we roll a standard 6-sided die, there are 6 possibilities, each occurring with probability \(\frac{1}{6}\), so the expected value is \(\frac{1}{6}(1)+\frac{1}{6}(2)+\frac{1}{6}(3)+\frac{1}{6}(4)+\frac{1}{6}(5)+\frac{1}{6}(6)=3.5\). Thanks in advance.

Note by Aiman Rafeed
4 years, 4 months ago

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The expected value is sometimes called a weighted average. In your example, all the outcomes have the same probability (which you can think of as "weight"), so the weighted average is equivalent to the average of 1,2,3,4,5,6. But let's say there was a game with the 6-sided die in which if you rolled a perfect square, you won 5 points; otherwise you lose 2 points. The probability of rolling a perfect square = 1/3 (there are only two perfect squares, 1 or 4). The probability of not rolling a perfect square is therefore 2/3. If you play the game multiple times, your expected value (the number of points you will earn) = (1/3)(5) + (2/3)(-2) = 1/3. So If points were money, this would be a winning game since the expected value is positive.

Kathleen Kasper - 4 years, 4 months ago

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Thanks a lot. I understand the concept now.

Aiman Rafeed - 4 years, 4 months ago

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