In order to help pay your tuition bills you have decided to go to Las Vegas and

Question

In order to help pay your tuition bills you have decided to go to Las Vegas and try your luck at Blackjack using the following methodology for card counting.

Cards ..Count ...Cards .....Count ...Cards ...Count ...Cards ..Count
Ace..................3.................... 7 ..................Jack
1 ........-2......... 4 ..........-1 ......8........ +1 ......Queen ...+2
2 ....................5 ....................9.................. King
......................6 ....................10

What is the mean and variance of your card counting distribution?

If we make the assumption of normality, what is the probability that your count will be less than -3 or greater than +3?

Explanation / Answer

Every card has (at the start) the same chance of showing up. Thus to find the mean you can multiply the number of cards times there value and add them all up: e.g. 2 cards with value -2 3 cards with value -1 3 cards with value 1 2 cards with value 2 would give average=mean= 2*-2+3*-1+3*1+2*2=0 But the weird thing about your picture thing is there is a 1 which is by my knowing not in a regular card deck. The variance can be calculated by -^2 is average of the square of the values minus the average of the numbers squared In this example 2*(-2)^2 =8 3*(-1)^2 =3 3*(1)^2 =3 2*(2)^2 =8 =22 22/(2+3+3+2)= 22/8 The average though was 0 thus the variance would be 22/8 -0=22/8 (standard deviation is the square root of the variance) for the last question the answer for -3 and 3 are the same if the mean is 0 you can use normdist() in excel using the calculated standard dev and the average and the x as value -3 or 3 and non cummulative

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