# textbook question

#### aptiva

##### New Member
im having trouble with this question - i dont know how the textbook arrived at the answer of losing 1.3 cents per game on average

A gambler stakes \$1.00 on a red number on a roulette wheel that contains 18 red numbers, 18 black numbers and a 0. If a red number turns up, he gets back twice his stake. if a black number turns up, he loses his stake. If the 0 turns up, he gets a second spin (for free) and either losses his stake if a black number or another number 0 turns up, or wins back his stake if a red number turns up.

If the gambler plays this game a large number of times, on average what would his net winnings (or losings) be on each game?

#### JohnM

##### TS Contributor
Using the formula for expected value (same as a weighted average), here are the possible outcomes:

red --> wins --> +1.00 --> probability = (18/37)
black --> loses --> -1.00 --> probability = (18/37)
zero and black --> loses --> -1.00 --> probability = (18/37^2)
zero and zero --> loses --> -1.00 --> probability = (1/37^2)
zero and red --> breaks even --> 0.00 --> probability = (18/37^2)

Now, sum the products of the winnings and their respective probabilities, and you get (-19 / 37^2) = -0.013

#### aptiva

##### New Member
Hi John
Do you mind explaing how you obtained the bottom three probabilities (and also why is it 37^2?)

zero and black --> loses --> -1.00 --> probability = (18/37^2)
zero and zero --> loses --> -1.00 --> probability = (1/37^2)
zero and red --> breaks even --> 0.00 --> probability = (18/37^2)

thanks

#### JohnM

##### TS Contributor
For example:

zero and black --> loses --> -1.00 --> probability = (18/37^2)

P(zero and black) = P(zero) * P(black) = (1/37) * (18/37) = (18/37^2)