Interesting questions on probability Kevin.Qi UTA Seminar 2012 Spring

April 17,2012

Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

1 / 12

Probablity game A popular dice gambling game during 17 centery

Problem There are two people A and B, they play a gambling game. Each of them pay 30 dollar to start the game(60 total in the jackpot). A and B have equal chance to win, which is 0.5 . They make a agreement, who first win three games, who can take all the 60 dollar. Now, A win twice and B win once, but something happened the game is interrupted. But they want find a way to devide the money from jackpot.

Question How would you divide the 60 dollars between these two people?

Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

2 / 12

Possible solution

Some people say since A won twice and B only won once. They should get paid as a ratio 2:1. In an other word. A should get 40 dollar and B should get 20 dollar.

Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

3 / 12

Possible solution

Some people say since A won twice and B only won once. They should get paid as a ratio 2:1. In an other word. A should get 40 dollar and B should get 20 dollar. What do you think?

Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

3 / 12

The correct solution

Suppose person A and person B keep playing this game. Four equally likely outcomes: < A, A >, < A, B >, < B, A >, < B, B >

A has a triple chance to win this game compared to B. So they should get paid in the ratio 3:1. In other words, A should get 45$ and B should get 15$. This is a very famous question called “chip dividing problem”.

Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

4 / 12

History of probablity pheory

The foundation of the classic probablity theory: 1

Huygens: On the calculation of gambling

2

Bernoulli fould“Bernoulli large number theory”

3

1812, Laplace published the Analysis of probablity theory: it defined the classic probability.

4

Through Gauss and Poission’s hard work, the probablity theory was accepted by the academia.

Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

5 / 12

Definitions

Definition P(A) is probability measure of the event A.

Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

6 / 12

Definitions Definition P(A) is probability measure of the event A.

Example Rolling a fair dice. Let An be the event “obtain n”, and B the event “obtain a score bigger than 3”. Then P(An ) = 1/6 and P(B) = P(A4 ∪ A5 ∪ A6 ) = P(get4) + P(get5) + P(get6) = 1/6 + 1/6 + 1/6 = 1/2 Here we can do that since it is disjoint union. Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

6 / 12

Definitions

Definition P(A) is probability measure of the event A.

Definition (Conditional probability) Define p(A|B) as the the probabilty of A given the B happened.

Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

6 / 12

Interesting question 1

Question 1 Tomorrow there will be either rain or snow but not both. The probablity of rain is 0.4, and the probablity of snow is 0.6. If it rains, then the probablity that I will be late for my lecture is 0.2, while the corresponding probablity in the event of snow is 0.6. What is the probablity that I will be late?

Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

7 / 12

Solution for question 1:

solution P(late) = P(late|rain) · P(rain) + p(late|snow ) · p(snow ) P(late) = 0.2 · 0.4 + 0.6 · 0.6 = 11/25

Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

8 / 12

More definitions

Definition (Random variable) Define Random variable: Random Variable is a mapping X on probablity space: ω → R . For the previous example, define X =1 if outcome is 1,2,3 X=0 if outcome is 4,5,6 Then P(X=0)=P(4,5,6)=P(B) from the previous example.

Definition (Expectation) Expect value P is the the mean or the expectation about the outcome. We define E(x)= xP(X = x) For this example E(X)=0.5

Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

9 / 12

Interesting question 2 You friend want play a game with you. You don’t need pay anything to start the game. There is a box contain 20 balls. 10 are red and 10 are white. You need grab 10 balls out of these 20. The rule as following: 1 If 10 balls are in same color either all red or all white. He will give you 300 2 If 9 balls are in same color. Which means either 9 red ball 1 white ball or 9 white ball 1 red ball. He will give you 30 3 If 8 balls are in same color. Which means either 8 red ball 2 white ball or 8 white ball 2 red ball. He will give you 3 4 If 7 balls are in same color. Which means either 7 red ball 3 white ball or 3 white ball 7 red ball. He will give you 2 5 If 6 balls are in same color. Which means either 6 red ball 4 white ball or 4 white ball 6 red ball. He will give you 1 6 If 5 balls are in same color. Which means either 5 red 5 while or 5 white 5 red. You have to pay him 5 Will you play this game with Probablity him? Theory Kevin.Qi (UA Math Department) April 17,2012 10 / 12

Solution for question 2:

solution To analyze this P game. The best method is to calculate the expect value. Remeber E(x)= xP(X = x) When X=300 P(300)=0.00001 When X=30 P(30)=0.00108 When X=3 P(3)=0.02192 When X=2 P(2)=0.15588 When X=1 P(1)=0.47739 When X=-5 P(-5)=0.34 E(X)=300 · 0.00001 + 30 · 0.00108 + 3 · 0.02192 + 2 · 0.15588 + 1 · 0.47739 + (−5) · 0.34 =-0.809 This expect velue means you are losing 0.809 dollar each game

Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

11 / 12

The End

Kevin.Qi (UA Math Department)

Probablity Theory

April 17,2012

12 / 12