Mean/Expected Value for Discrete Quantities

[Agent Smith]

Say we get bulbs in boxes, 10 bulbs per box, and the mean/expected value of broken bulbs (in any given box) = 2.5. This I was told to interpret as an event "in the long run" like so: If you get 10 boxes, you should see around 25 broken bulbs (10 boxes = 100 bulbs).

What meaning does 2.5 have for ONE box? Is there going to be 2 and a HALF broken bulbs in ONE box? Is 2.5 meaningless for ONE box?

 

[khansaheb]

What meaning does 2.5 have for ONE box?

The expected breakage will be 2 to 3.

Expected value is a useful parameter for "many" boxes

 

[khansaheb]

The expected breakage will be 2 to 3.

Expected value is a useful parameter for "many" boxes

Continued from above -

Expected value is a useful parameter applied to "many" boxes. It tells you that "next" box you open, you can "expect" to have 2 to 3 breakage. You "may" find 10 breakage in the next box- but that would be "rare". If you open 10000 boxes, you should not be surprised to find 2500 breakage.

 

[Agent Smith]

@khansaheb that made a whole lot of sense. Now what about (say) a game of chance. Price of entry 10 dollars. Prize 30 dollars. Chance of winning 1/10 and chance of losing 9/10. Where random variable X is expected net gain ...
\(\displaystyle \text{E(X)} = \frac{1}{10} \times 20 - \frac{9}{10} \times 10 = 2 - 9 = -7\) dollars.

It doesn't look like this value \(\displaystyle \text{E(X)} = -7\) dollars has meaning for 1 attempt at playing this game.

 

[Dr.Peterson]

@khansaheb that made a whole lot of sense. Now what about (say) a game of chance. Price of entry 10 dollars. Prize 30 dollars. Chance of winning 1/10 and chance of losing 9/10. Where random variable X is expected net gain ...
\(\displaystyle \text{E(X)} = \frac{1}{10} \times 20 - \frac{9}{10} \times 10 = 2 - 9 = -7\) dollars.

It doesn't look like this value \(\displaystyle \text{E(X)} = -7\) dollars has meaning for 1 attempt at playing this game.

Of course not. Everything in probability theory is based on the Law of Large Numbers; it says nothing about any individual trial, but a lot about long-term averages (though even that is not certain).

The probability of getting heads on one toss of a coin is 1/2, but that doesn't mean you get half a head on the next toss; in fact, it tells you absolutely nothing about the next toss. But it tells you a lot about what can be expected to happen overall on the next hundred, or million, tosses.