Question

You and a friend play a game where you each toss a balanced coin. If the upper faces on the coins are both tails, you win $1; if the faces are both heads, you win $6; if the coins do not match (one shows a head, the other a tail), you lose $3 (win (−$3)). Calculate the mean and variance of Y, your winnings on a single play of the game. Note that E(Y) > 0.

Answer 1

Answer:  The mean and variance of Y is $0.25 and $6.19 respectively.

Step-by-step explanation:

Given : You and a friend play a game where you each toss a balanced coin.

sample space for tossing two coins : {TT, HT, TH, HH}

Let Y denotes the  winnings on a single play of the game.

You win $1; if the faces are both heads

then P(Y=1)=P(TT)=$\dfrac{1}{4}=0.25$

You win $6; if the faces are both heads

then P(Y=6)=P(HH)=$\dfrac{1}{4}=0.25$

You loose $3; if the faces do not match.

then P(Y=1)=P(TH, HT)=$\dfrac{2}{4}=0.50$

The expected value to win : E(Y)=$\sum_{i=1}^{i=3} y_ip(y_1)$

$=1\times0.25+6\times0.25+(-3)\times0.50=0.25$

Hence, the mean of Y : E(Y)= $0.25

$E(Y^2)=\sum_{i=1}^{i=3} y_i^2p(y_i)\\\\=1^2\times0.25+6^1\times0.25+(-3)^2\times0.5\\\\=0.25+1.5+4.5=6.25$

Variance = $E[Y^2]-E(Y)^2$

$=6.25-(0.25)^2=6.25-0.0625=6.1875\approx6.19$

Hence, variance of Y = $ 6.19