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

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


A
Asked by 2 years ago
45 points

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
Adrian

1 Answer

Answered by 2 years ago
8.7k points

For this type of problem, it is helpful to construct a table,

Player 1 Player 2 Random Variable
Guess Show Guess Show X
1 1 1 1 0
1 1 1 2 -3
1 1 2 1 2
1 1 2 2 0
1 2 1 1 3
1 2 1 2 0
1 2 2 1 0
1 2 2 2 -4
2 1 1 1 -2
2 1 1 2 0
2 1 2 1 0
2 1 2 2 3
2 2 1 1 0
2 2 1 2 4
2 2 2 1 -3
2 2 2 2 0

part (a)

There are 16 possibilities in the table above and so we have,

$$P(X=0)=\frac{8}{16}=\frac{1}{2}$$

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

$$P(X=3)=P(X=-3)=\frac{2}{16}=\frac{1}{8}$$

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

part (b)

from the table above the following set can only occur {1111,1122,2211,2222} all with output X=0, so, $P(X=0)=1$

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Asked: 2 years ago

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