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 Post subject: Probability HelpPosted: Sun, 10 Jun 2012 02:29:05 UTC
 S.O.S. Newbie

Joined: Sun, 10 Jun 2012 02:27:27 UTC
Posts: 3
In a betting game a player gets to 3 chances to play after betting 6 dollars. The player must roll a 4 on a die. If he doesn't he loses. If he does then he can advance to the next round where he can then draw 3 cards from a deck of 8 consisting of 2 kings, two queens, two jacks and 2 jokers. If he draws the 2 jokers then he wins the game (does't matter what the third card is). The prize for winning the game is 30 dollars plus the 6 he originally used to play the game. So in total 36 dollars.

a) what's the probability of rolling a 4?

is it 1/6?

b) what's the probability of drawing 2 jokers?

is it (2C2)(6C1)/(8C2)

which is, 6/28

c) what is the probability of winning?

is it
(1/6)(6/28)
which is 1/28 ?

d) What is the expected value of one bet?

is it

E(x)= ( ∑ xi)(P(x))=((-6.00)(27/28))((30.00)(1/28))
= -1215/196 dollars?

e) Calculate his expected number of wins in those 3 trails

is it (3)(1/28) ?

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 Post subject: Re: Probability HelpPosted: Sun, 10 Jun 2012 06:39:14 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Mon, 19 May 2003 19:55:19 UTC
Posts: 7961
Location: Lexington, MA
Hello, Chee!

A few corrections . . .

Quote:
In a betting game a player gets 3 chances to play after betting 6 dollars.
The player must roll a 4 on a die. If he doesn't he loses.
If he does, then he can advance to the next round where he can then draw 3 cards
from a deck of 8 consisting of 2 kings, two queens, two jacks and 2 jokers.
If he draws the 2 jokers, then he wins the game (does't matter what the third card is).
The prize for winning the game is 30 dollars plus the 6 he originally used to play the game.
So in total 36 dollars.

a) What's the probability of rolling a 4?
Is it 1/6? . Yes!

b) What's the probability of drawing 2 jokers?
Is it (2C2)(6C1)/(8C2)? . no

There are: outcomes.

Quote:
c) What is the probability of winning?
is it (1/6)(6/28)? . no, for two reasons.

First, the probability of winning a game is: .

Second, for his $6 bet, he gets to play three times. I assume that if he wins, he stops playing. There are three ways he can. Win on the 1st try: . Win on the 2nd try: . Win on the 3rd try: . Therefore: . Also: . Quote: d) What is the expected value of one bet? He can expect to win$30 with probability

He can expect to lose $6 with probability . . . . . . . . . . . . . He can expect to lose an average of about$4.11 per game.

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 Post subject: Re: Probability HelpPosted: Sun, 10 Jun 2012 13:25:38 UTC
 S.O.S. Newbie

Joined: Sun, 10 Jun 2012 02:27:27 UTC
Posts: 3
Soroban wrote:

First, the probability of winning a game is: .

Second, for his $6 bet, he gets to play three times. I assume that if he wins, he stops playing. There are three ways he can. Win on the 1st try: . Win on the 2nd try: . Win on the 3rd try: . Therefore: . Also: .[/size] isn't the probability on the second and third still (1/56)? why on the win on the second try you multiplied (1/56) by (55/56) and then on the third you squared (55/56) then multiplied (1/56)? is there a formula or something? Top  Post subject: Re: Probability HelpPosted: Sun, 10 Jun 2012 18:00:33 UTC  Member of the 'S.O.S. Math' Hall of Fame Joined: Sun, 24 Jul 2005 20:12:39 UTC Posts: 3724 Location: Ottawa Ontario Soroban wrote: a) What's the probability of rolling a 4? Is it 1/6? . Yes! Not if "for 3 rolls"...problem's wording seems to allow 3 rolls. _________________ Walked over to a beggar...he gave me a quarter... Top  Post subject: Re: Probability HelpPosted: Mon, 11 Jun 2012 00:57:40 UTC  S.O.S. Newbie Joined: Sun, 10 Jun 2012 02:27:27 UTC Posts: 3 Soroban wrote: First, the probability of winning a game is: . Second, for his$6 bet, he gets to play three times.

I assume that if he wins, he stops playing.
There are three ways he can.

Win on the 1st try: .
Win on the 2nd try: .
Win on the 3rd try: .

Therefore: .

Also: .[/size]

but wait, aren't there many possibilities on how to win on the second try? One possibility, roll the die & don't get 4 (you lost so second try) then roll the die and get 4, then draw 2 jokers. Second possibility roll die and get 4, then don't draw 2 jokers (you lost so second try), roll 4, then draw 2 jokers? which results in two different probabilities of winning on the second attempt. can you apply the geometric distribution equation? :s

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