Question

A roulette wheel has 38 numbers. Eighteen of the numbers are black, eighteen are red, and...

A roulette wheel has 38 numbers. Eighteen of the numbers are black, eighteen are red, and two are green. When the wheel is spun, the ball is equally likely to land on any of the 38 numbers. Each spin of the wheel is independent of all other spins of the wheel. One roulette bet is a bet on black—that the ball will stop on one of the black numbers. The payoff for winning a bet on black is \$2 for every \$1 one bets; that is, if you win, you get the dollar ante back and an additional dollar, for a net gain of \$1, while if you lose, you gets nothing back, for a net loss of \$1. Each \$1 bet thus results in the gain or loss of \$1. Suppose one repeatedly places \$1 bets on black, and plays until either winning \$7 more than he has lost, or loses \$7 more than he has won. Equivalently, one plays until the first time that | net winnings | = | \$won − \$bet | = |\$2×(#bets won) − \$1×#bets | = \$7, where |x| is the absolute value of x.

The chance that one places exactly 6 bets before stopping is __0__

The chance that one places exactly 7 bets before stopping is __0.01653813___

The chance that one places exactly 8 bets before stopping is __0___

The chance that one places exactly 9 bets before stopping is (Q12) ?

from the given data,

1)You cannot win 7\$ by just playing 6 bets as the maximum win amount for each bet is 1\$.

The chance that one places exactly 6 bets before stopping is 0.

2)Hence Probability = P(one places 7 bets before stopping)

= P(one loses first 7 bets or one wins first 7 bets)

= (20/38)7 + (18/38)7

= 0.016538132 =0.0165

3)You can't get a net loss or net gain of \$7.00 on 8 bets.

The chance that one places exactly 8 bets before stopping is 0.

4)

If you stop after placing 9 bets, then you must have won 8 bets and lost 1, or lost 8 bets and won 1. But if you win 8 and lose 1, at least one of the first 7 bets must be a loss (or else you would have stopped after 7 bets). Similarly, if you lose 8 and win 1, at least one of the first 7 bets must be a win)

P(one places 9 bets before stopping)
= P(lose 8 bets and win 1 bet, with the winning bet = 1st to 7th bet)
+ P(win 8 bets and lose 1 bet, with the losing bet = 1st to 7th bet)

= 7 (20/38)8(18/38) + 7 (20/38)(18/38)8
= 0.02886156 =0.0289

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