Net Gain in Roulette Betting - Expectation & Standard Deviation

[habman_6]

30) A special roulette wheel has seven equally likely outcomes: 0, 1, 2, 3, 4, 5, 6. If you bet that an odd number comes up, you win or lose $10 according to whether or not that event occurs. If X denotes your ‘net’ gain, X=10 if we see 1, 3, or 5, and X = -10 otherwise. Suppose that you play this game 400 times. Let Y be your net gain after these 400 plays. The mean (expectation) and standard deviation of Y are, respectively (accurate to the number of figures shown):
A) -270; 198
B) -571; 198
C) -270; 988
D) -571; 988
E) none of the other answers displayed




I can get the mean (-571), but for the standard deviation, I only get the answer (apparently 198) when I use the formula:

sigma² _ new = b * sigma²_old

When the formula SHOULD use b², instead of just b.

(Im using sigma² = Sum((x_i-u)²*p) to get the original variance.)

 

[HallsofIvy]

30) A special roulette wheel has seven equally likely outcomes: 0, 1, 2, 3, 4, 5, 6. If you bet that an odd number comes up, you win or lose $10 according to whether or not that event occurs. If X denotes your ‘net’ gain, X=10 if we see 1, 3, or 5, and X = -10 otherwise. Suppose that you play this game 400 times. Let Y be your net gain after these 400 plays. The mean (expectation) and standard deviation of Y are, respectively (accurate to the number of figures shown):
A) -270; 198
B) -571; 198
C) -270; 988
D) -571; 988
E) none of the other answers displayed




I can get the mean (-571), but for the standard deviation, I only get the answer (apparently 198) when I use the formula:

sigma² _ new = b * sigma²_old

When the formula SHOULD use b², instead of just b.

(Im using sigma² = Sum((x_i-u)²*p) to get the original variance.)

On anyone spin your probability of winning is 3/7 and of losing is 4/7. This is a binomial distribution with p= 3/7, q= 4/7 and N= 400. The mean value, and standard deviation for a binomial distribution with p, q, n are np and $\sqrt{npq}$ respectively. With your values, yes, -571 is the expectation and 988, not 198, is the is the standard deviation. I can't say anything about your formula, sigma² _ new = b * sigma²_old, since you haven said what "b" and "sigma_old" are.

 

[habman_6]

What I did was the mean is:

$$400*[(3/7)(10)+(4/7)(-10)]$$
= -571

Now, for standard deviation, it should be:

(For a single turn) variation =

$$sigma^2=[10-(-1.428)]^2*(3/7)+[-10-(-1.428)]^2*(4/7)$$

So, sigma_new² (after 400 turns) SHOULD =

b^2*sigma^2
=400²*([10-(-1.428)]²*(3/7)+[-10-(-1.428)]²*(4/7))²

However, I only get the right answer (which is 198), when I simply use 400 instead of 400².

 

[habman_6]

Nevermind, I got it using binomial distribution. The formula I was using before was for when you linearly transform by multiplying by 400, not when you repeat 400 times.

Thanks