Question

1. A game is said to pay “3 to 1” if for every $1 bet you win, you get paid $3. In roulette, there are 40 spaces on the wheel: 19 red, 19 black, and 2 green. Red and black pay 2 to 1 and green pays 20 to 1.
if I bet 10$ on red for 6 games, what can I expect to win (or lose)?
I know I need to use the expected value formula just not sure how to set it up

Answer 1

not exactly sure how roulette is played, but do you have to land on red every time (all 6 times) to win if you bet on red?

Answer 2

would it be (-10)(1)+(2)(19/40)+(3)(19/40)+(4)(19/40)+(5)(19/40)+(6)(19/40)=

Answer 3

no I don't think so

Answer 4

that's wrong because it say for every 1$ bet you win 3$

Answer 5

well if you're just focusing on one game, and you bet $10 on red, the expected winnings are

E[X] = P(winning)*V(Winning) + P(losing)*V(losing)

E[X] = (19/40)*(20) + (21/40)*(-10)

E[X] = 9.5 + (-5.25)

E[X] = 4.25

Answer 6

not 100% sure, but it seems like it's as simple as saying

6*(E[X]) = 6*(4.25) = 25.5

would be your expected winnings after 6 games...but then again, that might be too high

Answer 7

I'll let you know

Answer 8

alright thanks

Answer 9

(-$10)(21/40) + ($30)(19/40)
-5.25 + 14.25 =$9.00
$9.00 *6 = $54.00

Answer 10

I thought red paid 2 to 1 (not 3 to 1)?

Answer 11

it say 3 to 1 odds

Answer 12

hmm at the very top it says "Red and black pay 2 to 1"

Answer 13

oh wait I see what you are talking about

Answer 14

there might be a typo? idk

Answer 15

(-10)(21/40) + (20)(19/40)
-5.25 + 9.50 = $3.75 *6 = 22.50
that sounds better?

Answer 16

(-10)(21/40) + (20)(19/40) = 4.25 though

Answer 17

yep you are right thank you I would of gotten it wrong twice

Answer 18

you're welcome