(a)

Observe that possible outcomes of X are 1,-1,-3.

The gambler will win 1$ if he wins in the first bet or he loses the first bet, but he have success in the second and the third.

He will win -1$ if he loses the first and the second bet, but wins the third or loses the first, but wins the second.

Finally, he will win -3$ if he loses all the bets. Hence \(\displaystyle{P}{\left({X}{>}{0}\right)}={P}{\left({X}={1}\right)}=\frac{{18}}{{38}}+\frac{{20}}{{38}}\cdot{\left(\frac{{18}}{{38}}\right)}^{{2}}=\frac{{4059}}{{6859}}\approx{0.5918}\)

(b)

No, the strategy is not the winning one.

Even though that the probability that the gambler wins in this system is greater that 1/2, we will show in (c) that the expected winnings is less than zero since there is a big risk of losing all the bucks.

(c)

See that

\(\displaystyle{P}{\left({X}=-{1}\right)}={2}{\left(\frac{{20}}{{38}}\right)}^{{2}}\cdot\frac{{18}}{{38}}=\frac{{1800}}{{6859}}\)

\(\displaystyle{P}{\left({X}=-{3}\right)}={\left(\frac{{20}}{{38}}\right)}^{{3}}=\frac{{1000}}{{6859}}\)

so, using definition of expectation, we nave that

\(\displaystyle{E}{\left({X}\right)}={1}\cdot\frac{{4059}}{{6859}}-{1}\cdot\frac{{1800}}{{6859}}-{3}\cdot\frac{{1000}}{{6859}}\approx-{0.11}\)

which says that the gambler is expected to lose 11 cents.

Observe that possible outcomes of X are 1,-1,-3.

The gambler will win 1$ if he wins in the first bet or he loses the first bet, but he have success in the second and the third.

He will win -1$ if he loses the first and the second bet, but wins the third or loses the first, but wins the second.

Finally, he will win -3$ if he loses all the bets. Hence \(\displaystyle{P}{\left({X}{>}{0}\right)}={P}{\left({X}={1}\right)}=\frac{{18}}{{38}}+\frac{{20}}{{38}}\cdot{\left(\frac{{18}}{{38}}\right)}^{{2}}=\frac{{4059}}{{6859}}\approx{0.5918}\)

(b)

No, the strategy is not the winning one.

Even though that the probability that the gambler wins in this system is greater that 1/2, we will show in (c) that the expected winnings is less than zero since there is a big risk of losing all the bucks.

(c)

See that

\(\displaystyle{P}{\left({X}=-{1}\right)}={2}{\left(\frac{{20}}{{38}}\right)}^{{2}}\cdot\frac{{18}}{{38}}=\frac{{1800}}{{6859}}\)

\(\displaystyle{P}{\left({X}=-{3}\right)}={\left(\frac{{20}}{{38}}\right)}^{{3}}=\frac{{1000}}{{6859}}\)

so, using definition of expectation, we nave that

\(\displaystyle{E}{\left({X}\right)}={1}\cdot\frac{{4059}}{{6859}}-{1}\cdot\frac{{1800}}{{6859}}-{3}\cdot\frac{{1000}}{{6859}}\approx-{0.11}\)

which says that the gambler is expected to lose 11 cents.