In the game of Two-Finger Morra, $2$players show $1$or $2$fingers and simultaneously guess the number of fingers their opponent will show. If only one of the players guesses correctly, he wins an amount (in dollars) equal to the sum of the fingers shown by him and his opponent. If both players guess correctly or if neither guesses correctly, then no money is exchanged. Consider a specified player, and denote by X the amount of money he wins in a single game of Two-Finger Morra.

(a) If each player acts independently of the other, and if each player makes his choice of the number of fingers he will hold up and the number he will guess that his opponent will hold up in such a way that each of the $4$possibilities is equally likely, what are the possible values of $X$and what are their associated probabilities?

(b) Suppose that each player acts independently of the other. If each player decides to hold up the same number of fingers that he guesses his opponent will hold up, and if each player is equally likely to hold up $1$or $2$ fingers, what are the possible values of$X$ and their associated probabilities?

In the game of Two-Finger Morra, $2$players show $1$or $2$ fingers and simultaneously guess the number of fingers their opponent will show. If only one of the players guesses correctly, he wins an amount (in dollars) equal to the sum of the fingers shown by him and his opponent. If both players guess correctly or if neither guesses correctly, then no money is exchanged. Consider a specified player, and denote by$X$ the amount of money he wins in a single game of Two-Finger Morra.

The possible values of $X$are $0$if both or neither player guesses correctly, $2,3$, or $4$if he guesses correctly and the other player doesn't, and $-2,-3$, or $-4$if his opponent guesses correctly and he doesn't. We compute

$P\{X=2\}$

$=P($he guesses $1$$)P($he holds up $1$) $P($opponent guesses $2$) $P($opponent holds up $1$)

$=\frac{1}{2^4}$

$=\frac{1}{16}$

By similar computations we compute that

$P\{X=-2\}=P\{X=4\}=P\{X=-4\}=\frac{1}{16}$

and

$P\{X=3\}=P\{X=-3\}=\frac{1}{8}$

Hence

$P\{X=0\}$

$=1-P\{X=2\}-P\{X=-2\}-P\{X=4\}-P\{X=-4\}-P\{X=3\}-P\{X=-3\}$

$=1-\frac{1}{16}-\frac{1}{16}-\frac{1}{16}-\frac{1}{16}-\frac{1}{8}-\frac{1}{8}$

$=\frac{1}{2}$

$P\{X=2\}$

$=\frac{1}{16}$

$P\{X=-2\}=$$\frac{1}{16}$

$P\{X=3\}=$$\frac{1}{8}$

$P\{X=0\}$$=\frac{1}{2}$

In the game of Two-Finger Morra, $2$players show $1$or $2$ fingers and simultaneously guess the number of fingers their opponent will show. If only one of the players guesses correctly, he wins an amount (in dollars) equal to the sum of the fingers shown by him and his opponent. If both players guess correctly or if neither guesses correctly, then no money is exchanged. Consider a specified player, and denote by $X$ the amount of money he wins in a single game of Two-Finger Morra.

If both players guess the same, then both players guess correctly and $X=0$.

If both players guess differently, then both players guess wrong and $X=0$.

Thus the only possible value for $X$is $0$and $P(X=0)=1.$

The only possible value for $X$is $0$and $P(X=0)=1.$