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Let us consider following problem :

One thousand tickets are sold at $\$1$ each for a color television valued at $\$350$. What is the expected value of the gain if you purchase one ticket?

I would like to describe my basic problem related to this task. English is not my native language therefore I did not understand logic of statement. What I know is that we had $1000$ tickets that are sold at $\$1$ each. This means that the total revenue from $1000$ tickets is $\$1000$. Is this correct? We have a color television that costs $\$350$.

My question is:

What is the expected value of the gain if you purchase one ticket, really I don't understand the connection of purchase=buy. If I bought one ticket that cost $\$1$, and if I bought a television which costs $\$350$, then my gain will be $\$349$ correct? If I lost, then my gain will be $\$-1$ because I paid this $\$1$. What about their probability? If the probability that out of $\$1000$ I will win is $0.001$, then the probability that I will loose is $0.999$, so expected value will be

$$349\times 0.001+(-1)\times 0.999=-0.65$$

But still I do not understand logically this statement, could you describe please in a simple manner this problem?

Mutantoe
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dato datuashvili
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3 Answers3

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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 $\$0.65$.

4

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 $1/1000$.

mlc
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  • what about original $1000$ tickets that are sold at $1$ – dato datuashvili Mar 13 '17 at 15:18
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    @datodatuashvili The 1000 tickets comes into this calculation as the chance of winning is 1/1000. Your expected value would be the same 65 cent loss if you paid a dollar for your ticket and all the others were given away free. Of course it would not be the same for the organization running the raffle. – Ethan Bolker Mar 13 '17 at 15:23
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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:

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

zoli
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  • You say it doesn't work in such a naïve way, but if you assign the USD 1 cost to all events, and the $350 benefit only to the win, you do get the right answer using the "naïve way". You just have to keep straight that there are in fact 1000/1000 events in which you spend the USD 1 and only 1/1000 in which you get the TV. Then the two formulae are mathematically equivalent. – Monty Harder Mar 13 '17 at 21:36