Lottery odds when order matters?

Clash Royale CLAN TAG#URR8PPP

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Let's say there is a lottery where one can play 10 numbers chosing from a total of 20. The lottery will pull out 10 numbers randomly. What's the chance that the first 2 number played exactly match (order included) the first 2 of the lottery?

probability

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asked Jul 18 at 20:51

Dumberino

31

add a comment | 

up vote
0
down vote

favorite

Let's say there is a lottery where one can play 10 numbers chosing from a total of 20. The lottery will pull out 10 numbers randomly. What's the chance that the first 2 number played exactly match (order included) the first 2 of the lottery?

probability

share|cite|improve this question

asked Jul 18 at 20:51

Dumberino

31

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let's say there is a lottery where one can play 10 numbers chosing from a total of 20. The lottery will pull out 10 numbers randomly. What's the chance that the first 2 number played exactly match (order included) the first 2 of the lottery?

probability

share|cite|improve this question

asked Jul 18 at 20:51

Dumberino

31

Let's say there is a lottery where one can play 10 numbers chosing from a total of 20. The lottery will pull out 10 numbers randomly. What's the chance that the first 2 number played exactly match (order included) the first 2 of the lottery?

probability

share|cite|improve this question

asked Jul 18 at 20:51

Dumberino

31

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 20:51

Dumberino

31

asked Jul 18 at 20:51

Dumberino

31

asked Jul 18 at 20:51

Dumberino

31

31

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
0
down vote



accepted

Hence of symmetry we can assume the first two chosen numbers are $(1,2)$. This because your ambiguity when choosing $(1,2)$ is as much as choosing $(2,3)$ or any other couple, therefore $$Pr(textbeing matched|(x,y)text have been chosen)=Pr(textbeing matched|(w,z)text have been chosen)$$let's define$$P_x,y=Pr(textbeing matched|(x,y)text have been chosen)$$then we can obtain$$Pr(textbeing matched)=sum_(x,y)Pr(textchoosing (x,y))P_x,y\=P_x,ysum_(x,y)Pr(textchoosing (x,y))\=P_x,y\=P_1,2$$ then the probability is $$P=Pr((x,y)=(1,2))=dfrac12binom202=dfrac1380$$

share|cite|improve this answer

edited Jul 18 at 22:10

answered Jul 18 at 21:44

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  what's the basic reasoning?
  – Dumberino
  Jul 18 at 22:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This because your ambiguity when choosing $(1,2)$ is as much as choosing $(2,3)$ or any other couple. Therefore $$Pr(textbeing matched|(x,y)text have been chosen)=Pr(textbeing matched|(w,z)text have been chosen)$$
  – Mostafa Ayaz
  Jul 18 at 22:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  the guy picks 10 numbers(out of 20) where only the first two matches the house number. How doh you comoute all the favorable permutations?
  – Dumberino
  Jul 18 at 22:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  'cause once the first two numbers matched pairwise we don't care the others e.g. (1,2,5,6,4,6,7,8,6,4) and $(1,2,4,5,6,5,4,3,4,4)$ are equal for us and for the lottery
  – Mostafa Ayaz
  Jul 18 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  kk I was overcomplicating things
  – Dumberino
  Jul 18 at 22:35
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

Hint:

1. What is the chance that the first number matches?


2. What is the chance that the second number matches? (It's not the same as above. Why not?)

share|cite|improve this answer

answered Jul 18 at 20:56

John

22k32346

• 
  
  
  

  
  

  
  
  
  
  
  

  
  45/1847560? maybe
  – Dumberino
  Jul 18 at 21:07
  
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Hence of symmetry we can assume the first two chosen numbers are $(1,2)$. This because your ambiguity when choosing $(1,2)$ is as much as choosing $(2,3)$ or any other couple, therefore $$Pr(textbeing matched|(x,y)text have been chosen)=Pr(textbeing matched|(w,z)text have been chosen)$$let's define$$P_x,y=Pr(textbeing matched|(x,y)text have been chosen)$$then we can obtain$$Pr(textbeing matched)=sum_(x,y)Pr(textchoosing (x,y))P_x,y\=P_x,ysum_(x,y)Pr(textchoosing (x,y))\=P_x,y\=P_1,2$$ then the probability is $$P=Pr((x,y)=(1,2))=dfrac12binom202=dfrac1380$$

share|cite|improve this answer

edited Jul 18 at 22:10

answered Jul 18 at 21:44

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  what's the basic reasoning?
  – Dumberino
  Jul 18 at 22:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This because your ambiguity when choosing $(1,2)$ is as much as choosing $(2,3)$ or any other couple. Therefore $$Pr(textbeing matched|(x,y)text have been chosen)=Pr(textbeing matched|(w,z)text have been chosen)$$
  – Mostafa Ayaz
  Jul 18 at 22:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  the guy picks 10 numbers(out of 20) where only the first two matches the house number. How doh you comoute all the favorable permutations?
  – Dumberino
  Jul 18 at 22:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  'cause once the first two numbers matched pairwise we don't care the others e.g. (1,2,5,6,4,6,7,8,6,4) and $(1,2,4,5,6,5,4,3,4,4)$ are equal for us and for the lottery
  – Mostafa Ayaz
  Jul 18 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  kk I was overcomplicating things
  – Dumberino
  Jul 18 at 22:35
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote



accepted

Hence of symmetry we can assume the first two chosen numbers are $(1,2)$. This because your ambiguity when choosing $(1,2)$ is as much as choosing $(2,3)$ or any other couple, therefore $$Pr(textbeing matched|(x,y)text have been chosen)=Pr(textbeing matched|(w,z)text have been chosen)$$let's define$$P_x,y=Pr(textbeing matched|(x,y)text have been chosen)$$then we can obtain$$Pr(textbeing matched)=sum_(x,y)Pr(textchoosing (x,y))P_x,y\=P_x,ysum_(x,y)Pr(textchoosing (x,y))\=P_x,y\=P_1,2$$ then the probability is $$P=Pr((x,y)=(1,2))=dfrac12binom202=dfrac1380$$

share|cite|improve this answer

edited Jul 18 at 22:10

answered Jul 18 at 21:44

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  what's the basic reasoning?
  – Dumberino
  Jul 18 at 22:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This because your ambiguity when choosing $(1,2)$ is as much as choosing $(2,3)$ or any other couple. Therefore $$Pr(textbeing matched|(x,y)text have been chosen)=Pr(textbeing matched|(w,z)text have been chosen)$$
  – Mostafa Ayaz
  Jul 18 at 22:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  the guy picks 10 numbers(out of 20) where only the first two matches the house number. How doh you comoute all the favorable permutations?
  – Dumberino
  Jul 18 at 22:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  'cause once the first two numbers matched pairwise we don't care the others e.g. (1,2,5,6,4,6,7,8,6,4) and $(1,2,4,5,6,5,4,3,4,4)$ are equal for us and for the lottery
  – Mostafa Ayaz
  Jul 18 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  kk I was overcomplicating things
  – Dumberino
  Jul 18 at 22:35
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Hence of symmetry we can assume the first two chosen numbers are $(1,2)$. This because your ambiguity when choosing $(1,2)$ is as much as choosing $(2,3)$ or any other couple, therefore $$Pr(textbeing matched|(x,y)text have been chosen)=Pr(textbeing matched|(w,z)text have been chosen)$$let's define$$P_x,y=Pr(textbeing matched|(x,y)text have been chosen)$$then we can obtain$$Pr(textbeing matched)=sum_(x,y)Pr(textchoosing (x,y))P_x,y\=P_x,ysum_(x,y)Pr(textchoosing (x,y))\=P_x,y\=P_1,2$$ then the probability is $$P=Pr((x,y)=(1,2))=dfrac12binom202=dfrac1380$$

share|cite|improve this answer

edited Jul 18 at 22:10

answered Jul 18 at 21:44

Mostafa Ayaz

8,6023630

Hence of symmetry we can assume the first two chosen numbers are $(1,2)$. This because your ambiguity when choosing $(1,2)$ is as much as choosing $(2,3)$ or any other couple, therefore $$Pr(textbeing matched|(x,y)text have been chosen)=Pr(textbeing matched|(w,z)text have been chosen)$$let's define$$P_x,y=Pr(textbeing matched|(x,y)text have been chosen)$$then we can obtain$$Pr(textbeing matched)=sum_(x,y)Pr(textchoosing (x,y))P_x,y\=P_x,ysum_(x,y)Pr(textchoosing (x,y))\=P_x,y\=P_1,2$$ then the probability is $$P=Pr((x,y)=(1,2))=dfrac12binom202=dfrac1380$$

share|cite|improve this answer

edited Jul 18 at 22:10

answered Jul 18 at 21:44

Mostafa Ayaz

8,6023630

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 22:10

edited Jul 18 at 22:10

edited Jul 18 at 22:10

answered Jul 18 at 21:44

Mostafa Ayaz

8,6023630

answered Jul 18 at 21:44

Mostafa Ayaz

8,6023630

answered Jul 18 at 21:44

Mostafa Ayaz

8,6023630

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  what's the basic reasoning?
  – Dumberino
  Jul 18 at 22:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This because your ambiguity when choosing $(1,2)$ is as much as choosing $(2,3)$ or any other couple. Therefore $$Pr(textbeing matched|(x,y)text have been chosen)=Pr(textbeing matched|(w,z)text have been chosen)$$
  – Mostafa Ayaz
  Jul 18 at 22:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  the guy picks 10 numbers(out of 20) where only the first two matches the house number. How doh you comoute all the favorable permutations?
  – Dumberino
  Jul 18 at 22:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  'cause once the first two numbers matched pairwise we don't care the others e.g. (1,2,5,6,4,6,7,8,6,4) and $(1,2,4,5,6,5,4,3,4,4)$ are equal for us and for the lottery
  – Mostafa Ayaz
  Jul 18 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  kk I was overcomplicating things
  – Dumberino
  Jul 18 at 22:35
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  what's the basic reasoning?
  – Dumberino
  Jul 18 at 22:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This because your ambiguity when choosing $(1,2)$ is as much as choosing $(2,3)$ or any other couple. Therefore $$Pr(textbeing matched|(x,y)text have been chosen)=Pr(textbeing matched|(w,z)text have been chosen)$$
  – Mostafa Ayaz
  Jul 18 at 22:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  the guy picks 10 numbers(out of 20) where only the first two matches the house number. How doh you comoute all the favorable permutations?
  – Dumberino
  Jul 18 at 22:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  'cause once the first two numbers matched pairwise we don't care the others e.g. (1,2,5,6,4,6,7,8,6,4) and $(1,2,4,5,6,5,4,3,4,4)$ are equal for us and for the lottery
  – Mostafa Ayaz
  Jul 18 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  kk I was overcomplicating things
  – Dumberino
  Jul 18 at 22:35
  

  
  
  
  

what's the basic reasoning?
– Dumberino
Jul 18 at 22:01

what's the basic reasoning?
– Dumberino
Jul 18 at 22:01

This because your ambiguity when choosing $(1,2)$ is as much as choosing $(2,3)$ or any other couple. Therefore $$Pr(textbeing matched|(x,y)text have been chosen)=Pr(textbeing matched|(w,z)text have been chosen)$$
– Mostafa Ayaz
Jul 18 at 22:08

This because your ambiguity when choosing $(1,2)$ is as much as choosing $(2,3)$ or any other couple. Therefore $$Pr(textbeing matched|(x,y)text have been chosen)=Pr(textbeing matched|(w,z)text have been chosen)$$
– Mostafa Ayaz
Jul 18 at 22:08

the guy picks 10 numbers(out of 20) where only the first two matches the house number. How doh you comoute all the favorable permutations?
– Dumberino
Jul 18 at 22:28

the guy picks 10 numbers(out of 20) where only the first two matches the house number. How doh you comoute all the favorable permutations?
– Dumberino
Jul 18 at 22:28

'cause once the first two numbers matched pairwise we don't care the others e.g. (1,2,5,6,4,6,7,8,6,4) and $(1,2,4,5,6,5,4,3,4,4)$ are equal for us and for the lottery
– Mostafa Ayaz
Jul 18 at 22:30

'cause once the first two numbers matched pairwise we don't care the others e.g. (1,2,5,6,4,6,7,8,6,4) and $(1,2,4,5,6,5,4,3,4,4)$ are equal for us and for the lottery
– Mostafa Ayaz
Jul 18 at 22:30

kk I was overcomplicating things
– Dumberino
Jul 18 at 22:35

kk I was overcomplicating things
– Dumberino
Jul 18 at 22:35

 | 
show 1 more comment

up vote
1
down vote

Hint:

1. What is the chance that the first number matches?


2. What is the chance that the second number matches? (It's not the same as above. Why not?)

share|cite|improve this answer

answered Jul 18 at 20:56

John

22k32346

• 
  
  
  

  
  

  
  
  
  
  
  

  
  45/1847560? maybe
  – Dumberino
  Jul 18 at 21:07
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

Hint:

1. What is the chance that the first number matches?


2. What is the chance that the second number matches? (It's not the same as above. Why not?)

share|cite|improve this answer

answered Jul 18 at 20:56

John

22k32346

• 
  
  
  

  
  

  
  
  
  
  
  

  
  45/1847560? maybe
  – Dumberino
  Jul 18 at 21:07
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Hint:

1. What is the chance that the first number matches?


2. What is the chance that the second number matches? (It's not the same as above. Why not?)

share|cite|improve this answer

answered Jul 18 at 20:56

John

22k32346

Hint:

1. What is the chance that the first number matches?


2. What is the chance that the second number matches? (It's not the same as above. Why not?)

share|cite|improve this answer

answered Jul 18 at 20:56

John

22k32346

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 20:56

John

22k32346

answered Jul 18 at 20:56

John

22k32346

answered Jul 18 at 20:56

John

22k32346

22k32346

• 
  
  
  

  
  

  
  
  
  
  
  

  
  45/1847560? maybe
  – Dumberino
  Jul 18 at 21:07
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  45/1847560? maybe
  – Dumberino
  Jul 18 at 21:07
  
  

  
  
  
  

45/1847560? maybe
– Dumberino
Jul 18 at 21:07

45/1847560? maybe
– Dumberino
Jul 18 at 21:07

add a comment | 

 

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