Translating mathematical problem statement into natural language

A lottery is implied.

You pay $\$1$.

You win $\$350$ with probability $\dfrac1{1000}$ (your chance to have drawn the winning ticket).

So "on average", you get $\$0.35$ worth of television and your expected loss is indeed $\$0.65$.


In the usual context, they expect you to buy a ticket but not the television. If your ticket is drawn, you win the television incurring no further cost.

So the gamble is $-1$ (cost of ticket and no gain) with probability $999/1000$ and $350-1=349$ (value of prize net of ticket purchase) with probability $1/1000$.


I can understand the results of such calculations if I can imagine that the experiment described can be repeated many times under the same circumstances and independently. Say, somebody organizes the television experiment $1\ 000\ 000$ times and you take part always and there are always $999$ other gamblers present. The question is your average gain.

For sure, you lose $1$ dollar every time but you win the television set with a chance of $1$ to $1\ 000$. That is, you will win (about) $1\ 000$ times out of the $1 \ 000\ 000$ cases. So your total loss is $ \$1 \ 000 \ 000$ and your total gain is $\$350\ 000$. The bottom line is $\$350\ 000-\$1\ 000\ 000=-\$650\ 000.$

As far as the average gain: You simply divide your total gain by $1\ 000\ 000$, the number of experiments.

$$\frac{-\$650\ 000}{1\ 000\ 000}=-\$0.65.$$

Probability theory does not work such a naïve way. We say that the probability of winning is $\frac1{1\ 000}=$ and the probability of not wining is $1-\frac1{1\ 000}=0.999.$ The expected gain is the sum of the products of the gains and the probabilities:

$$(-1)\cdot0.999+349\cdot0.001=-0.999+0.349=-0.65.$$

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Probability