Payoff of a dice game

Question

I came across this question today:

"A fair die is tossed. If 2,3 or 5 occurs, the player wins that number of rupees, but if 1, 4 or 6 occurs, the player loses that number if rupees. Then find the possible payoffs for the player".

My textbook has then proceeded to solve it like this:

What is the logic behind directly adding all the values? Secondly, how is it that the game is unfavorable to the player? There is an equal chance that the player will get 2, 3 or 5 and 1,4 or 6!

Please help. (Also note that this doubt is not specific to this problem alone, since this concept is crucial to understanding this part of the chapter "Probability".) Much thanks in advance :) Regards.

Edit: Thanks ever so much for these answers :) I read up on "Expected values" but encountered another important doubt. I have posted it here and hope that you will clear my doubt :)

Answer 1

What is the logic behind directly adding all the values?

This is due to the Law of total probability; essentially, each possible outcome is weighted with the probability of that outcome, which gives the expected (average) value over an infinite number of trials.

How is it that the game is unfavorable to the player? There is an equal chance that the player will get 2, 3 or 5 and 1,4 or 6!

Yes, but the results of getting 2, 3 or 5 isn't the same as for 1, 4 or 6, i.e., \begin{align} 2+3+5&<1+4+6. \end{align}

Answer 2

Perhaps instead of "Then find the possible payoffs for the player" it meant "would someone expect to make/lose money playing this game?" They are testing a concept called "Expected Value".

Let's run through an easier example, then go back to the one your book mentions. If I have a fair 50-50 coin, and I say, "Heads, you get one dollar, tails you give me two dollars", you wouldn't want to play that because you'd expect to lose money. How much money would you expect to lose? It would be $ \$1\cdot\frac{1}{2} + \$(-2)\frac{1}{2} = \$ (-0.50) $. In other words, you might expect to lose fifty cents each game.

Apply that to this situation: the calculation your book gives in the solution shows that, if you were to play that game, you would expect lose about a sixth of a rupee each round.

Answer 3

For your second question, let's exaggerate the values: suppose that with probability 1/2, you lose 10000 and with probability 1/2 you win 1. Then even though the probabilities are equal, you'd admit that the game is unfavorable.

The expected value in the above example is $1/2 \times 1 - 1/2 \times 100000$.

The conclusion is however not appropriate. Consider another game/bet where you win $10^{12}$ with probability $10^{-6}$ and lose 10^5 with the remaining probability. The expected value is positive and large (close to $10^6-10^5$), but if you play this bet just once, it would be foolish to take it on (depending on how valuable/large $10^5$ is).

In the absence of complete information about the probability distribution, the expected value alone does not tell you whether the outcome is likely to be close to that value. For this, you will need to look into concepts like variance and standard deviation.

Answer 4

For such simple excercise you can imagine to throw the die a big number of times, obtaining perfect statistics.

In your case let's throw the die 600 times, obtaining 100 ones, 100 twos and exactly 100 polls for all the figures.

Our earnings would have been $$100*(-1)+100*2+100*(3)+100*(-4)+100*(5)+100*(-6)=-100$$ so, for 600 throws we are "expected" to lose 100, dividing these values we have the game payoff of $-1/6$.

If you do the division before the sum you have exactly the solution of your book: $${100*(-1)+100*2+100*(3)+100*(-4)+100*(5)+100*(-6) \over 600}={-100 \over 600}$$.

The game have a negative payoff because, although you have equal probabilities to win or lose one match, not all matches have the same earnings so you can't assume the game is fair, but you have to evaluate your earnings as I did before.