Question

Consider the following strategy playing the roulette. Bet $1 on red. If red appears (which happens with probability 18/38), then take $1 and stop playing for the day. If red does not appear, then bet additional $1 on red each of the following two rounds, and then stop playing for the day no matter the outcome. Let X be the net gain (a negative gain means a loss). (a) Find the probability mass function of X; (b) Compute P.X > 0/: Is this a good strategy

Answer 1

Answer:

a.$\int(x_i)=p(X=x_i)$

b.$P(X>0)=0.5918$

C.No

Explanation:

Let X be the probability of a net gain.

a.Given that $X$ is a discrete random variable,$\int(x_i)$ is its probability mass function defined as:

$\int(x_i)=p(X=x_i)$    #for all $i$.

b. Possible outcomes of X are $1,-1,-3$. A win of $1 occurs when the gambler wins/loses the first bet but is successful in the next two. A win of -$1 occurs in the, if the gambler loses the first and second but wins $3^r^d$ or loses first and third but wins $2^n^d$. The gambler wins -$3 if he loses all three bets.

$P(X>0)=P(X=1)=\frac{18}{38}+\frac{20}{38}(\frac{18}{38})^2=\frac{4059}{6859}\\=0.5918$

c.No. The gambler has a more than 50% probability of winning. The winnings are however high risk and less than $zero.$