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We consider the following dice game:

You bet 1 euro against the bank and choose one number from $\{1, 2, 3, 4, 5, 6\}$ and roll three dice.

If at least one of the dice shows your number, you get the stake back and also for each dice that shows your number you get an additional 1 euro.

If your number does not appear, the stake expires.

Let X be your winnings (i.e. payout minus stake) on such a game.

(a) Determine the probability function $p_X$.

(b) What is the profit that you expect on average?

(c) If the staked number falls three times, in addition to the reimbursement of the stake, no longer 3 euros, but c euros should be given. How must c be chosen, to make the game fair?

$$$$

I have done the following :

(a) When three dice are thrown, the number of total events is $6^3 = 216$.

Is the number of desired events equal to $3$ ?

So would the probability function be $p_X(x)=\frac{3}{216}$ ?

(b) We give 1 euro.

If none of the dice shows the number that we have chosen then the profit is $\text{Payout} - \text{stake} =0-1=-1$ euro.

If one of the dice shows the number that we have chosen then the profit is $\text{Payout} - \text{stake} =(1+1)-1=1$ euro.

If two of the dice show the number that we have chosen then the profit is $\text{Payout} - \text{stake} =(1+2)-1=2$ euro.

If all three of the dice show the number that we have chosen then the profit is $\text{Payout} - \text{stake} =(1+3)-1=3$ euro.

Do we multiply each of these outcomes with the corresponding probabilities and sum them up?

(c) We get the formula of (b) and instead of $3$ we write $c$ and the result must be $0$ ?

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    $\begingroup$ Die is singular; dice is plural; dices is the third person singular form of the verb to dice, which means to cut into small cubes. $\endgroup$ Nov 27, 2021 at 10:07
  • $\begingroup$ I corrected it! :-) @N.F.Taussig $\endgroup$
    – Mary Star
    Nov 27, 2021 at 14:30

1 Answer 1

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Is the number of desired events equal to 3 ?

No, its not. You are counting the number of dice that shows a given number, after throwing three dice. The number of events is four: zero, one, two or three dice shows the chosen number. And the choosing of the number is not a probability event at all, what is probabilistic is the number of dice with the chosen number.

Do we multiply each of these outcomes with the corresponding probabilities and sum them up?

Yes, its exactly that. There is a theorem than says that

$$ \mathrm{E}[h(Y)]=\int_{\mathbb{R}}h(y)P(dy) $$

In the case of $Y$ discrete then the above integral can be written as the sum

$$ \mathrm{E}[h(Y)]=\sum_{k\in \mathbb{N}}h(k)\Pr [Y=k] $$

Now, if $Y$ is the probability distribution that count the number of "good numbers" after throwing three dice, then its easily seen that $X=h(Y)$ for some function $h$.

(c) We get the formula of (b) and instead of 3 we write c and the result must be 0 ?

Yes, that is. Now you must choose $c$ such that $\mathrm{E}[X]=0$.

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  • $\begingroup$ (a) So you mean that the probability function is $p_X(x)=\frac{1}{4}$ ? $$$$ (b) Are the outcomes that I calculated correct? So do we have $$E[X]=-1\cdot p_X(-1)+1\cdot p_X(1)+2\cdot p_X(2)+3\cdot p_X(3)=-1\cdot \frac{1}{4}+1\cdot \frac{1}{4}+2\cdot \frac{1}{4}+3\cdot \frac{1}{4}=\frac{5}{4}$$ ? $$$$ (c) Do we have : $$E[X]=0 \Rightarrow -1\cdot p_X(-1)+1\cdot p_X(1)+2\cdot p_X(2)+c\cdot p_X(3)=0\\ \Rightarrow -1\cdot \frac{1}{4}+1\cdot \frac{1}{4}+2\cdot \frac{1}{4}+c\cdot \frac{1}{4}=0 \\ \Rightarrow c=-2$$ Is everything correct and complete? $\endgroup$
    – Mary Star
    Nov 27, 2021 at 9:05
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    $\begingroup$ your probabilities are wrong, they are not $1/4$ all, what you have is a binomial diitribution with parameters $n=4$ and $p=1/6$. That is: the number of dice with the chosen number follow this binomial distribution $\endgroup$
    – Masacroso
    Nov 27, 2021 at 21:32
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    $\begingroup$ @Mary yes, $n=3$, sorry. And $p=1/6$, that is, the probability that a dice show up the desired value. $\endgroup$
    – Masacroso
    Nov 27, 2021 at 21:49
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    $\begingroup$ No, $p_X$ is not a constant function, its the mass function of a binomial distribution, with parameters $n=3$ and $p=1/6$ $\endgroup$
    – Masacroso
    Nov 27, 2021 at 21:59
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    $\begingroup$ yes, that is. -- $\endgroup$
    – Masacroso
    Nov 27, 2021 at 22:03

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