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strojnjashka [21]
3 years ago
12

A roulette wheel has the numbers 1 through 36, 0, and 00. A bet on two numbers pays 17 to 1 (that is, if you bet $1 and one of t

he two numbers you bet comes up, you get back your $1 plus another $17). How much do you expect to win with a $1 bet on two numbers
Mathematics
1 answer:
damaskus [11]3 years ago
8 0

Answer:

-$0.05

Step-by-step explanation:

The computation is shown below:

The loss case = -1

The win case = 11 + 17 + 1 = 17

Now the number of pairs could be formed from (1 to 356, 0, 00) i.e.

= \frac{38!}{2!36!}

= 703

Now

Pr (x = 17) is

= \frac{1\times37}{703}\\\\ = \frac{37}{703}

And, Pr (x = -1) is

= 1 - \frac{37}{703}\\\\ = \frac{666}{703}

Now

E(x) = (-1) Pr (x = -1) + (17) Pr (x = 17)

= -1 \times \frac{666}{703} + 17 \times \frac{37}{703}  \\\\ = \frac{-37}{703}

= -$0.05

hence, the -$0.05 would be expected to win that associated with a $1 bet on two numbers

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