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I am given a table of possible $X_1$ and $X_2$ values that can be generated in a casino. In the game, both are generated with each turn.

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The questions asks me to determine the minimum fee that should be charged per turn so that that casino doesn't lose money, if the payouts are:

$a) \ 8X_1$

$b) \ 4X_1 + 8(X_2)^2-\frac{51}{128}$

$c) \ 8X_1X_2$

For a), I added up the total possible odds for each value $X_1$ could take, $$ 0\cdot\frac{2}{16} + 1\cdot\frac{4}{16}+2\cdot\frac{6}{16} + 3\cdot\frac{4}{16} = \frac{7}{4}$$ and determined that the casino would need to charge at least $\$1.75$ per turn.

For b) I tried the pretty much the same thing, only breaking it down into individual situations, for example; $$P\{X_1=0, X_2 =1\} = \frac{1}{16}$$ $$P\{X_1=0, X_2 =2\} = \frac{1}{16}$$ $$etc...$$

My question is, is there a faster, more efficient way to solve these types of problems? What if there were many more possible outcomes?

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CodyBugstein
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3 Answers3

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What you are trying to do is to find the value of

$$\mathbb{E}(Y)=\sum_yy\mathbb{P}(Y=y)$$

for various random variables $Y$ depending on $X_1,X_2$. Notice that this is exactly what you did in the first question: you worked out the possible values for $Y=8X_1$, and then multiplied each value by the probability of it occurring, and then added up the results.

If $Y=4X_1+8(X_2)^2-\frac{51}{128}$, you could work out the possible values of that rather nasty expression, find the probability of each one occurring, and then do the same process to find $\mathbb{E}(Y)$, but it is actually easier to use the following identity:

$$\mathbb{E}(aU+bV)=\sum_ux\mathbb{P}(aU+bV=x)$$ $$=a\sum_uu\mathbb{P}(U=u)+b\sum_v\mathbb{P}(V=v)=a\mathbb{E}(U)+b\mathbb{E}(V)$$

Getting from the first line to the second is non-trivial, but I'll leave that to you as an exercise, if you're not allowed just to quote it. Using that identity should make the problem a lot easier.

John Gowers
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  • I should point out that the identity I have given is valid even if $X$ and $Y$ are not independent (they aren't, as Ross has pointed out). For part (c) you have no choice but to use the longer method. – John Gowers Oct 12 '12 at 15:42
  • Can you explain the formula more? What is $Y$, and what is $y$? – CodyBugstein Oct 14 '12 at 00:11
  • You say "Notice that this is exactly what you did in the first question: you worked out the possible values for Y=8X1" Thats not what I did - I worked out possible values for $X_1$. – CodyBugstein Oct 14 '12 at 02:00
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    In that case, you got the wrong answer, as the question states that the payoff is $8X_1$. If the casino is only charging \frac{7}{4} per game then they stand to lose a lot of money. – John Gowers Oct 14 '12 at 09:19
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For a) you need to multiply by $8$, as you are paying $8X_1$

For b) you are doing the right thing. The only way to simplify it is to find some pattern. For example, in c) if $X_1$ and $X_2$ were independent (they are not) you could just multiply the expectations.

Ross Millikan
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After playing with it, I found a simpler answer:

Each potential payoff can simply be calculated using $E[X]$. For example,

$a) \ 8X_1 = 8E[X_1]$

$b) \ 4X_1 + 8(X_2)^2-\frac{51}{128} \ = \ 4E[X_1] + 8(E[X_2])^2-\frac{51}{128} $

$c) \ 8X_1X_2 \ = \ c) \ 8X_1E[X_1]E[X_2]$

Then, this whole thing can be calculated easily by simple finding only two things: $E[X_1]$ and $E[X_2]$

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CodyBugstein
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    That doesn't look right to me: in general, $E[X_1X_2] \ne E[X_1]E[X_2]$ if $X_1$ and $X_2$ are correlated. In particular, generally $E[X^2] \ne E[X]^2$. Donkey_2009's answer seems like the correct one here. – Ilmari Karonen Oct 15 '12 at 17:12