**" 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.

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TopNewestThe 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.Log in to reply

Thanks a lot. I understand the concept now.

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