Probability Help

Chee · **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) ?

Soroban · **Posted:** Sun, 10 Jun 2012 06:39:14 UTC

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: _8C_3

outcomes.

$$P(\text{2 Jokers}) \,=\,\frac{(_2C_2)(_6C_1)}{_8C_2} \;=\;\frac{6}{56} \:=\:\frac{3}{28}$

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: . $\frac{1}{6}\cdot\frac{3}{28} \:=\:\frac{1}{56}$

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: . $\frac{1}{56}$
Win on the 2nd try: . $\left(\frac{55}{56}\right)\left(\frac{1}{56}\right)$
Win on the 3rd try: . $\left(\frac{55}{56}\right)^2\left(\frac{1}{56}\right)$

Therefore: . $P(\text{win}) \:=\:\frac{1}{56} + \left(\frac{55}{56}\right)\!\left(\frac{1}{56}\right) + \left(\frac{55}{56}\right)^2\!\left(\frac{1}{56}\right) \:=\:\dfrac{9,\!241}{175,\!616}$

Also: . $P(\text{lose}) \:=\:1 - \frac{9,241}{175,616} \:=\:\frac{166,375}{175,616}$

Quote:

d) What is the expected value of one bet?

He can expect to win $30 with probability $\frac{9,241}{175,616}$

He can expect to lose $6 with probability $\frac{166,375}{175,616}$

$\text{Expected Value} \:=\:(30)\left(\frac{9,241}{175,616}\right) + (\text{-}6)\left(\frac{166,375}{175,616}\right) \:=\:\frac{\text{-}721,020}{175,616}$

. . . . . . . . . . . . . $=\;\text{-}4.105662354 \;\approx\;\text{-}4.11$

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

Chee · **Joined:** Sun, 10 Jun 2012 02:27:27 UTC **Posts:** 3

Soroban wrote:

First, the probability of winning a game is: . $\frac{1}{6}\cdot\frac{3}{28} \:=\:\frac{1}{56}$

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: . $\frac{1}{56}$
Win on the 2nd try: . $\left(\frac{55}{56}\right)\left(\frac{1}{56}\right)$
Win on the 3rd try: . $\left(\frac{55}{56}\right)^2\left(\frac{1}{56}\right)$

Therefore: . $P(\text{win}) \:=\:\frac{1}{56} + \left(\frac{55}{56}\right)\!\left(\frac{1}{56}\right) + \left(\frac{55}{56}\right)^2\!\left(\frac{1}{56}\right) \:=\:\dfrac{9,\!241}{175,\!616}$

Also: . $P(\text{lose}) \:=\:1 - \frac{9,241}{175,616} \:=\:\frac{166,375}{175,616}$ [/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?

Denis · **Posted:** Sun, 10 Jun 2012 18:00:33 UTC

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.

Chee · **Joined:** Sun, 10 Jun 2012 02:27:27 UTC **Posts:** 3

Soroban wrote:

First, the probability of winning a game is: . $\frac{1}{6}\cdot\frac{3}{28} \:=\:\frac{1}{56}$

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: . $\frac{1}{56}$
Win on the 2nd try: . $\left(\frac{55}{56}\right)\left(\frac{1}{56}\right)$
Win on the 3rd try: . $\left(\frac{55}{56}\right)^2\left(\frac{1}{56}\right)$

Therefore: . $P(\text{win}) \:=\:\frac{1}{56} + \left(\frac{55}{56}\right)\!\left(\frac{1}{56}\right) + \left(\frac{55}{56}\right)^2\!\left(\frac{1}{56}\right) \:=\:\dfrac{9,\!241}{175,\!616}$

Also: . $P(\text{lose}) \:=\:1 - \frac{9,241}{175,616} \:=\:\frac{166,375}{175,616}$ [/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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