Question

If you bet on “red” at roulette, you have chance 18/38 of winning. (There will be more on roulette later in the course; for now, just treat it as a generic gambling game.) Suppose you make a sequence of independent bets on “red” at roulette, with the decision that you will stop playing once you’ve won 5 times. What is the chance that after 15 bets you are still playing?

Answer 1

since you are still plaing after 15 bets, the first 15 bets you can win 0,1,2,3,or 4 times.

Answer 2

so you have played roullette 15 times, the probability of success (by betting at red) is equal to 18/38.
Let X = number of times you win betting on red
X is a binomial random variable, X can equal 0,1,2,3,4...15
we are interested in P( X <= 4) = P( X = 0 or X = 1 or X=2 or X=3 or X=4)

Answer 3

= P(X=0) + P(X=1) + ... P(X=4)
= 15 C0 * (18/38)^0 *(20/38)^15 + 15C1 *(18/38)^1 * (20/38)^14+...

Answer 4

15 C0 * (18/38)^0 *(20/38)^15 + 15C1 *(18/38)^1 * (20/38)^14+ 15C2 *(18/38)^2 * (20/38)^13 +15C3 *(18/38)^3 * (20/38)^12 + 15C4 *(18/38)^4 * (20/38)^11
Please check over

Answer 5

ok, I follow you

Answer 6

P( X <= 4)
= P( X = 0 or X = 1 or X=2 or X=3 or X=4)
= P(X=0) + P(X=1) + ... P(X=4)
=15 C0 * (18/38)^0 *(20/38)^15 + 15C1 *(18/38)^1 * (20/38)^14+ 15C2 *(18/38)^2 * (20/38)^13 +15C3 *(18/38)^3 * (20/38)^12 + 15C4 *(18/38)^4 * (20/38)^11
Now on calculator this is fast
binomcdf(15, 18/38, 4) = .087399

Answer 7

unless i read the problem incorrectly

Answer 8

my question is why you just have P(4) ?

Answer 9

P(4) is one possibility, you can also have P(3) and P(2), etc

Answer 10

you only stop playing if you get 5 wins, so you keep playing if you get less than 5 wins

Answer 11

no worries. you answered tough linear algebra :)

Answer 12

thank you i didnt found it :)