a)
expected value = P(winning)*net profit +P(losing) * net loss = 5/38*25+33/38*(-5) = -$1.05
b)
a$ 5 bet on number 16 is better because net loss is less in this as compared to other one.
33 When playing roulette at a casino, a gambler is trying to decide whether to bet...
When playing roulette at a casino, a gambler is trying to decide whether to bet $1515 on the number 2727 or to bet $1515 that the outcome is any one of the threethree possibilities 00 comma 0 comma or 100, 0, or 1. The gambler knows that the expected value of the $1515 bet for a single number is negative 0.79 cents−0.79¢. For the $1515 bet that the outcome is 00 comma 0 comma or 100, 0, or 1, there...
When playing roulette at a casino, a gambler is trying to decide whether to bet $10 on the number 15 or to bet $10 that the outcome is any one of the three possibilities 00 comma 0 comma or 100, 0, or 1. The gambler knows that the expected value of the $1010 bet for a single number is −$1.06. For the $10 bet that the outcome is 00 .0 or , or 1, there is a probability of Fraction3/38...
When playing roulette at a casino, a gambler is trying to decide whether to bet $20 on the number 13 or to bet $20 that the outcome is any one of the three possibilities 00 comma 0 comma or 1. The gambler knows that the expected value of the $20 bet for a single number is negative $ 1.05. For the $20 bet that the outcome is 00 comma 0 comma or 1, there is a probability of StartFraction 3...
Score: 0 of 1 pt 11 of 12 (10 complete) Hw Score: 70.83%, 8.5 of 12 pts 5.1.28 Question Help When playing roulette at a casino, a gambler is trying to decide whether to bet $20 on the number 25 or to bet $20 that the outcome is any one of the five possibilities 00, 0, 1,2, or 3. The gambler knows that the expected value of the $20 bet for a single number is $1.05. For the $20 bet...
In the game of roulette, when a player gives the casino $1 for a bet on the number 24, the player has a 37/38 probability of losing $1 and a 1/38 probability of making a net gain of $35. (The prize is $36, but the player's $1 bet is not returned, so the net gain is $35.) If a player bets $1 that the outcome is an odd number, the probability of losing $1 is 20/38 and the probability of...
24. A gambler decides to bet on the first 12 numbers on a roulette table. There are 12 chances in 38 to win. This bet pays 2 to 1. The gambler bets $3 on each play and plays 190 times. (a) (4 points) Construct a box model for one bet. (b) (2 points) Compute the expected value for the sum of all 190 plays (c) (3 points) Compute the standard error for the sum of all 190 plays. (d) (5...
If a gambler places a bet on the number 7 in roulette, he or she has a 1/38 probability of winning. Find the mean and standard deviation for the number of wins of gamblers who bet on the number 7 one hundred and seventy times The value of the mean is μ=___?
11. If a gambler places a bet on the number 7 in roulette, he or she has a 1/38 probability of winning. a. Find the mean and standard deviation for the number of wins of gamblers who bet on the number 7 two hundred and fifty times. b. Would 0 wins in two hundred and fifty bets be an unusually low number of wins?
Casino games of pure chance (e.g., craps, roulette, baccarat, and keno) always yield a "house advantage." For example, in the game of double-zero roulette, the expected casino win percentage is5.37% on bets made on whether the outcome will be either black or red. (This implies that for every $5 bet on black or red, the casino will earn a net of about 37 cents.) It can be shown that in 100 roulette plays on black/red, the average casino win percentage...
Casino games of pure chance (e.g., craps, roulette, baccarat, and keno) always yield a "house advantage." For example, in the game of double-zero roulette, the expected casino win percentage is 5.37% on bets made on whether the outcome will be either black or red. (This implies that for every $5 bet on black or red, the casino will earn a net of about 37cents.) It can be shown that in 100 roulette plays on black/red, the average casino win percentage...