Probability that the second number picked is greater than the first [closed]

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Given the set 1,2,3,4,5 we randomly pick one element. Without replacing(returning) the number we picked, we pick another random element. Find the probability that the second number picked is greater than the first.

probability

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edited Aug 30 at 15:00

asked Aug 30 at 14:35

user3140379

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closed as off-topic by John Ma, Gibbs, Chris Godsil, Xander Henderson, HK Lee Sep 1 at 2:42

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Given the set 1,2,3,4,5 we randomly pick one element. Without replacing(returning) the number we picked, we pick another random element. Find the probability that the second number picked is greater than the first.

probability

share|cite|improve this question

edited Aug 30 at 15:00

asked Aug 30 at 14:35

user3140379

213

closed as off-topic by John Ma, Gibbs, Chris Godsil, Xander Henderson, HK Lee Sep 1 at 2:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Gibbs, Chris Godsil, Xander Henderson, HK Lee

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
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down vote

favorite

1

up vote
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favorite

1

1

Given the set 1,2,3,4,5 we randomly pick one element. Without replacing(returning) the number we picked, we pick another random element. Find the probability that the second number picked is greater than the first.

probability

share|cite|improve this question

edited Aug 30 at 15:00

asked Aug 30 at 14:35

user3140379

213

Given the set 1,2,3,4,5 we randomly pick one element. Without replacing(returning) the number we picked, we pick another random element. Find the probability that the second number picked is greater than the first.

probability

share|cite|improve this question

edited Aug 30 at 15:00

asked Aug 30 at 14:35

user3140379

213

share|cite|improve this question

share|cite|improve this question

edited Aug 30 at 15:00

edited Aug 30 at 15:00

edited Aug 30 at 15:00

asked Aug 30 at 14:35

user3140379

213

asked Aug 30 at 14:35

user3140379

213

asked Aug 30 at 14:35

user3140379

213

213

closed as off-topic by John Ma, Gibbs, Chris Godsil, Xander Henderson, HK Lee Sep 1 at 2:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Gibbs, Chris Godsil, Xander Henderson, HK Lee

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by John Ma, Gibbs, Chris Godsil, Xander Henderson, HK Lee Sep 1 at 2:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Gibbs, Chris Godsil, Xander Henderson, HK Lee

If this question can be reworded to fit the rules in the help center, please edit the question.

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5 Answers
5

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oldest

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up vote
3
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accepted

We have that

• $E_1=1 implies P_1(E_2>E_1)=1$


• $E_1=2 implies P_2(E_2>E_1)=frac34$


• $E_1=3 implies P_3(E_2>E_1)=frac12$


• $E_1=4 implies P_4(E_2>E_1)=frac14$


• $E_1=5 implies P_5(E_2>E_1)=0$

therefore

$$P(E_2>E_1)=frac15 cdot left(1+frac34+frac12+frac14+0right)=frac12$$

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answered Aug 30 at 14:39

gimusi

70.8k73786

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up vote
4
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Yes, we can find the probability and it is $frac 12$. The pedestrian way is to list all the pairs, of which there are $20$, and count that $10$ of them have the second number greater. You can also argue by symmetry that the first number is just a likely to be the greater as the second to show that the answer is $frac 12$.

You can certainly compute the probability that the second is greater conditioned on what the first is. The probability that the second is greater given that the first is $2$ is $frac 34$ because $3$ of the $4$ remaining numbers are greater than $2$. This is another route to come up with the overall $frac 12$ answer. You can average this conditional probabilty over all the possible first numbers.

share|cite|improve this answer

answered Aug 30 at 14:39

Ross Millikan

279k22188355

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The argument by symmetry is especially interesting, since it won't depend on the distribution/set that the two values are chosen from (right?).
  – Justin
  Aug 30 at 19:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Justin: that is correct. It does demand that the two values not be the same, which here is accomplished by drawing without replacement. It could also be accomplished by drawing from a continuous distribution so the chance of repeating a value is zero. It comes up here a lot-a typical one would be drawing from a deck of cards and asking what the chance the third card drawn is a heart. People get excited about the fact that you may or may not have drawn hearts in the first two cards, but it doesn't matter. Think of drawing three cards and then looking at the first, not the third.
  – Ross Millikan
  Aug 30 at 20:19
  

  
  
  
  

add a comment | 

up vote
1
down vote

Let call $X$ the first number and $Y$ the second number you pick.

If $X = 1$, there are 4 possibilities for $Y$ that all work.

If $X = 2$, there are still 4 possibilities. But if $Y=1$, then it doesn't work, so only 3 work.

If $X = 3$, only 2 work.

If $X = 4$, only 1 work.

If $X = 5$ only 0 work.

The number of possibilities that work is $sum_k=0^4 k = 10$.
The number of all possibilities is $4*5 = 20$.
The probability is $10/20 = 50%$.

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answered Aug 30 at 14:40

PackSciences

38914

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up vote
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The easiest way to see that the answer is .5 is to realize that we are picking two non-equal numbers A and B and there is no reason to suppose that we pick A first and B second rather than B first and A second. Thus A > B and A < B are both equally likely.

This observation also has the advantage of extending to a set of numbers of any size. Provided none of the numbers repeat, it doesn't matter if your set contains 5 numbers or 50 numbers or 500 numbers -- the probability that the second number is greater than the first is always 50%.

share|cite|improve this answer

answered Aug 30 at 19:58

Galendo

1111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your logic seems good, but this type of explanation, with phrases like there is no reason to suppose leads to mistakes in cases where the actual probability differs greatly from our intuition, such as the Monty Hall Problem or the Boy or Girl paradox. Your answer is not bad, but more rigor is required for good mathematics.
  – Eric Wilson
  Aug 30 at 21:01
  

  
  
  
  

add a comment | 

up vote
0
down vote

Illustrated above are all the possible ways you could select two numbers from 1,2,3,4,5 without replacement. (The diagonal is excluded because it requires picking the same number twice).

Of the 20 possible selections, 10 have the second choice as higher than the first, and 10 are the other way around. Hence, the probability is 10/20 = 50%

It is also trivial to see that this percentage will hold true for any ordered set of any size (assuming there are at least two elements), due to the symmetry of the diagram.

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answered Aug 31 at 1:04

Arcanist Lupus

1812

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

We have that

• $E_1=1 implies P_1(E_2>E_1)=1$


• $E_1=2 implies P_2(E_2>E_1)=frac34$


• $E_1=3 implies P_3(E_2>E_1)=frac12$


• $E_1=4 implies P_4(E_2>E_1)=frac14$


• $E_1=5 implies P_5(E_2>E_1)=0$

therefore

$$P(E_2>E_1)=frac15 cdot left(1+frac34+frac12+frac14+0right)=frac12$$

share|cite|improve this answer

answered Aug 30 at 14:39

gimusi

70.8k73786

add a comment | 

up vote
3
down vote



accepted

We have that

• $E_1=1 implies P_1(E_2>E_1)=1$


• $E_1=2 implies P_2(E_2>E_1)=frac34$


• $E_1=3 implies P_3(E_2>E_1)=frac12$


• $E_1=4 implies P_4(E_2>E_1)=frac14$


• $E_1=5 implies P_5(E_2>E_1)=0$

therefore

$$P(E_2>E_1)=frac15 cdot left(1+frac34+frac12+frac14+0right)=frac12$$

share|cite|improve this answer

answered Aug 30 at 14:39

gimusi

70.8k73786

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

We have that

• $E_1=1 implies P_1(E_2>E_1)=1$


• $E_1=2 implies P_2(E_2>E_1)=frac34$


• $E_1=3 implies P_3(E_2>E_1)=frac12$


• $E_1=4 implies P_4(E_2>E_1)=frac14$


• $E_1=5 implies P_5(E_2>E_1)=0$

therefore

$$P(E_2>E_1)=frac15 cdot left(1+frac34+frac12+frac14+0right)=frac12$$

share|cite|improve this answer

answered Aug 30 at 14:39

gimusi

70.8k73786

We have that

• $E_1=1 implies P_1(E_2>E_1)=1$


• $E_1=2 implies P_2(E_2>E_1)=frac34$


• $E_1=3 implies P_3(E_2>E_1)=frac12$


• $E_1=4 implies P_4(E_2>E_1)=frac14$


• $E_1=5 implies P_5(E_2>E_1)=0$

therefore

$$P(E_2>E_1)=frac15 cdot left(1+frac34+frac12+frac14+0right)=frac12$$

share|cite|improve this answer

answered Aug 30 at 14:39

gimusi

70.8k73786

share|cite|improve this answer

share|cite|improve this answer

answered Aug 30 at 14:39

gimusi

70.8k73786

answered Aug 30 at 14:39

gimusi

70.8k73786

answered Aug 30 at 14:39

gimusi

70.8k73786

70.8k73786

add a comment | 

add a comment | 

up vote
4
down vote

Yes, we can find the probability and it is $frac 12$. The pedestrian way is to list all the pairs, of which there are $20$, and count that $10$ of them have the second number greater. You can also argue by symmetry that the first number is just a likely to be the greater as the second to show that the answer is $frac 12$.

You can certainly compute the probability that the second is greater conditioned on what the first is. The probability that the second is greater given that the first is $2$ is $frac 34$ because $3$ of the $4$ remaining numbers are greater than $2$. This is another route to come up with the overall $frac 12$ answer. You can average this conditional probabilty over all the possible first numbers.

share|cite|improve this answer

answered Aug 30 at 14:39

Ross Millikan

279k22188355

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The argument by symmetry is especially interesting, since it won't depend on the distribution/set that the two values are chosen from (right?).
  – Justin
  Aug 30 at 19:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Justin: that is correct. It does demand that the two values not be the same, which here is accomplished by drawing without replacement. It could also be accomplished by drawing from a continuous distribution so the chance of repeating a value is zero. It comes up here a lot-a typical one would be drawing from a deck of cards and asking what the chance the third card drawn is a heart. People get excited about the fact that you may or may not have drawn hearts in the first two cards, but it doesn't matter. Think of drawing three cards and then looking at the first, not the third.
  – Ross Millikan
  Aug 30 at 20:19
  

  
  
  
  

add a comment | 

up vote
4
down vote

Yes, we can find the probability and it is $frac 12$. The pedestrian way is to list all the pairs, of which there are $20$, and count that $10$ of them have the second number greater. You can also argue by symmetry that the first number is just a likely to be the greater as the second to show that the answer is $frac 12$.

You can certainly compute the probability that the second is greater conditioned on what the first is. The probability that the second is greater given that the first is $2$ is $frac 34$ because $3$ of the $4$ remaining numbers are greater than $2$. This is another route to come up with the overall $frac 12$ answer. You can average this conditional probabilty over all the possible first numbers.

share|cite|improve this answer

answered Aug 30 at 14:39

Ross Millikan

279k22188355

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The argument by symmetry is especially interesting, since it won't depend on the distribution/set that the two values are chosen from (right?).
  – Justin
  Aug 30 at 19:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Justin: that is correct. It does demand that the two values not be the same, which here is accomplished by drawing without replacement. It could also be accomplished by drawing from a continuous distribution so the chance of repeating a value is zero. It comes up here a lot-a typical one would be drawing from a deck of cards and asking what the chance the third card drawn is a heart. People get excited about the fact that you may or may not have drawn hearts in the first two cards, but it doesn't matter. Think of drawing three cards and then looking at the first, not the third.
  – Ross Millikan
  Aug 30 at 20:19
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

Yes, we can find the probability and it is $frac 12$. The pedestrian way is to list all the pairs, of which there are $20$, and count that $10$ of them have the second number greater. You can also argue by symmetry that the first number is just a likely to be the greater as the second to show that the answer is $frac 12$.

You can certainly compute the probability that the second is greater conditioned on what the first is. The probability that the second is greater given that the first is $2$ is $frac 34$ because $3$ of the $4$ remaining numbers are greater than $2$. This is another route to come up with the overall $frac 12$ answer. You can average this conditional probabilty over all the possible first numbers.

share|cite|improve this answer

answered Aug 30 at 14:39

Ross Millikan

279k22188355

Yes, we can find the probability and it is $frac 12$. The pedestrian way is to list all the pairs, of which there are $20$, and count that $10$ of them have the second number greater. You can also argue by symmetry that the first number is just a likely to be the greater as the second to show that the answer is $frac 12$.

You can certainly compute the probability that the second is greater conditioned on what the first is. The probability that the second is greater given that the first is $2$ is $frac 34$ because $3$ of the $4$ remaining numbers are greater than $2$. This is another route to come up with the overall $frac 12$ answer. You can average this conditional probabilty over all the possible first numbers.

share|cite|improve this answer

answered Aug 30 at 14:39

Ross Millikan

279k22188355

share|cite|improve this answer

share|cite|improve this answer

answered Aug 30 at 14:39

Ross Millikan

279k22188355

answered Aug 30 at 14:39

Ross Millikan

279k22188355

answered Aug 30 at 14:39

Ross Millikan

279k22188355

279k22188355

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The argument by symmetry is especially interesting, since it won't depend on the distribution/set that the two values are chosen from (right?).
  – Justin
  Aug 30 at 19:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Justin: that is correct. It does demand that the two values not be the same, which here is accomplished by drawing without replacement. It could also be accomplished by drawing from a continuous distribution so the chance of repeating a value is zero. It comes up here a lot-a typical one would be drawing from a deck of cards and asking what the chance the third card drawn is a heart. People get excited about the fact that you may or may not have drawn hearts in the first two cards, but it doesn't matter. Think of drawing three cards and then looking at the first, not the third.
  – Ross Millikan
  Aug 30 at 20:19
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The argument by symmetry is especially interesting, since it won't depend on the distribution/set that the two values are chosen from (right?).
  – Justin
  Aug 30 at 19:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Justin: that is correct. It does demand that the two values not be the same, which here is accomplished by drawing without replacement. It could also be accomplished by drawing from a continuous distribution so the chance of repeating a value is zero. It comes up here a lot-a typical one would be drawing from a deck of cards and asking what the chance the third card drawn is a heart. People get excited about the fact that you may or may not have drawn hearts in the first two cards, but it doesn't matter. Think of drawing three cards and then looking at the first, not the third.
  – Ross Millikan
  Aug 30 at 20:19
  

  
  
  
  

The argument by symmetry is especially interesting, since it won't depend on the distribution/set that the two values are chosen from (right?).
– Justin
Aug 30 at 19:33

The argument by symmetry is especially interesting, since it won't depend on the distribution/set that the two values are chosen from (right?).
– Justin
Aug 30 at 19:33

@Justin: that is correct. It does demand that the two values not be the same, which here is accomplished by drawing without replacement. It could also be accomplished by drawing from a continuous distribution so the chance of repeating a value is zero. It comes up here a lot-a typical one would be drawing from a deck of cards and asking what the chance the third card drawn is a heart. People get excited about the fact that you may or may not have drawn hearts in the first two cards, but it doesn't matter. Think of drawing three cards and then looking at the first, not the third.
– Ross Millikan
Aug 30 at 20:19

@Justin: that is correct. It does demand that the two values not be the same, which here is accomplished by drawing without replacement. It could also be accomplished by drawing from a continuous distribution so the chance of repeating a value is zero. It comes up here a lot-a typical one would be drawing from a deck of cards and asking what the chance the third card drawn is a heart. People get excited about the fact that you may or may not have drawn hearts in the first two cards, but it doesn't matter. Think of drawing three cards and then looking at the first, not the third.
– Ross Millikan
Aug 30 at 20:19

add a comment | 

up vote
1
down vote

Let call $X$ the first number and $Y$ the second number you pick.

If $X = 1$, there are 4 possibilities for $Y$ that all work.

If $X = 2$, there are still 4 possibilities. But if $Y=1$, then it doesn't work, so only 3 work.

If $X = 3$, only 2 work.

If $X = 4$, only 1 work.

If $X = 5$ only 0 work.

The number of possibilities that work is $sum_k=0^4 k = 10$.
The number of all possibilities is $4*5 = 20$.
The probability is $10/20 = 50%$.

share|cite|improve this answer

answered Aug 30 at 14:40

PackSciences

38914

add a comment | 

up vote
1
down vote

Let call $X$ the first number and $Y$ the second number you pick.

If $X = 1$, there are 4 possibilities for $Y$ that all work.

If $X = 2$, there are still 4 possibilities. But if $Y=1$, then it doesn't work, so only 3 work.

If $X = 3$, only 2 work.

If $X = 4$, only 1 work.

If $X = 5$ only 0 work.

The number of possibilities that work is $sum_k=0^4 k = 10$.
The number of all possibilities is $4*5 = 20$.
The probability is $10/20 = 50%$.

share|cite|improve this answer

answered Aug 30 at 14:40

PackSciences

38914

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let call $X$ the first number and $Y$ the second number you pick.

If $X = 1$, there are 4 possibilities for $Y$ that all work.

If $X = 2$, there are still 4 possibilities. But if $Y=1$, then it doesn't work, so only 3 work.

If $X = 3$, only 2 work.

If $X = 4$, only 1 work.

If $X = 5$ only 0 work.

The number of possibilities that work is $sum_k=0^4 k = 10$.
The number of all possibilities is $4*5 = 20$.
The probability is $10/20 = 50%$.

share|cite|improve this answer

answered Aug 30 at 14:40

PackSciences

38914

Let call $X$ the first number and $Y$ the second number you pick.

If $X = 1$, there are 4 possibilities for $Y$ that all work.

If $X = 2$, there are still 4 possibilities. But if $Y=1$, then it doesn't work, so only 3 work.

If $X = 3$, only 2 work.

If $X = 4$, only 1 work.

If $X = 5$ only 0 work.

The number of possibilities that work is $sum_k=0^4 k = 10$.
The number of all possibilities is $4*5 = 20$.
The probability is $10/20 = 50%$.

share|cite|improve this answer

answered Aug 30 at 14:40

PackSciences

38914

share|cite|improve this answer

share|cite|improve this answer

answered Aug 30 at 14:40

PackSciences

38914

answered Aug 30 at 14:40

PackSciences

38914

answered Aug 30 at 14:40

PackSciences

38914

38914

add a comment | 

add a comment | 

up vote
1
down vote

The easiest way to see that the answer is .5 is to realize that we are picking two non-equal numbers A and B and there is no reason to suppose that we pick A first and B second rather than B first and A second. Thus A > B and A < B are both equally likely.

This observation also has the advantage of extending to a set of numbers of any size. Provided none of the numbers repeat, it doesn't matter if your set contains 5 numbers or 50 numbers or 500 numbers -- the probability that the second number is greater than the first is always 50%.

share|cite|improve this answer

answered Aug 30 at 19:58

Galendo

1111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your logic seems good, but this type of explanation, with phrases like there is no reason to suppose leads to mistakes in cases where the actual probability differs greatly from our intuition, such as the Monty Hall Problem or the Boy or Girl paradox. Your answer is not bad, but more rigor is required for good mathematics.
  – Eric Wilson
  Aug 30 at 21:01
  

  
  
  
  

add a comment | 

up vote
1
down vote

The easiest way to see that the answer is .5 is to realize that we are picking two non-equal numbers A and B and there is no reason to suppose that we pick A first and B second rather than B first and A second. Thus A > B and A < B are both equally likely.

This observation also has the advantage of extending to a set of numbers of any size. Provided none of the numbers repeat, it doesn't matter if your set contains 5 numbers or 50 numbers or 500 numbers -- the probability that the second number is greater than the first is always 50%.

share|cite|improve this answer

answered Aug 30 at 19:58

Galendo

1111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your logic seems good, but this type of explanation, with phrases like there is no reason to suppose leads to mistakes in cases where the actual probability differs greatly from our intuition, such as the Monty Hall Problem or the Boy or Girl paradox. Your answer is not bad, but more rigor is required for good mathematics.
  – Eric Wilson
  Aug 30 at 21:01
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

The easiest way to see that the answer is .5 is to realize that we are picking two non-equal numbers A and B and there is no reason to suppose that we pick A first and B second rather than B first and A second. Thus A > B and A < B are both equally likely.

This observation also has the advantage of extending to a set of numbers of any size. Provided none of the numbers repeat, it doesn't matter if your set contains 5 numbers or 50 numbers or 500 numbers -- the probability that the second number is greater than the first is always 50%.

share|cite|improve this answer

answered Aug 30 at 19:58

Galendo

1111

The easiest way to see that the answer is .5 is to realize that we are picking two non-equal numbers A and B and there is no reason to suppose that we pick A first and B second rather than B first and A second. Thus A > B and A < B are both equally likely.

This observation also has the advantage of extending to a set of numbers of any size. Provided none of the numbers repeat, it doesn't matter if your set contains 5 numbers or 50 numbers or 500 numbers -- the probability that the second number is greater than the first is always 50%.

share|cite|improve this answer

answered Aug 30 at 19:58

Galendo

1111

share|cite|improve this answer

share|cite|improve this answer

answered Aug 30 at 19:58

Galendo

1111

answered Aug 30 at 19:58

Galendo

1111

answered Aug 30 at 19:58

Galendo

1111

1111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your logic seems good, but this type of explanation, with phrases like there is no reason to suppose leads to mistakes in cases where the actual probability differs greatly from our intuition, such as the Monty Hall Problem or the Boy or Girl paradox. Your answer is not bad, but more rigor is required for good mathematics.
  – Eric Wilson
  Aug 30 at 21:01
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your logic seems good, but this type of explanation, with phrases like there is no reason to suppose leads to mistakes in cases where the actual probability differs greatly from our intuition, such as the Monty Hall Problem or the Boy or Girl paradox. Your answer is not bad, but more rigor is required for good mathematics.
  – Eric Wilson
  Aug 30 at 21:01
  

  
  
  
  

Your logic seems good, but this type of explanation, with phrases like there is no reason to suppose leads to mistakes in cases where the actual probability differs greatly from our intuition, such as the Monty Hall Problem or the Boy or Girl paradox. Your answer is not bad, but more rigor is required for good mathematics.
– Eric Wilson
Aug 30 at 21:01

Your logic seems good, but this type of explanation, with phrases like there is no reason to suppose leads to mistakes in cases where the actual probability differs greatly from our intuition, such as the Monty Hall Problem or the Boy or Girl paradox. Your answer is not bad, but more rigor is required for good mathematics.
– Eric Wilson
Aug 30 at 21:01

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Illustrated above are all the possible ways you could select two numbers from 1,2,3,4,5 without replacement. (The diagonal is excluded because it requires picking the same number twice).

Of the 20 possible selections, 10 have the second choice as higher than the first, and 10 are the other way around. Hence, the probability is 10/20 = 50%

It is also trivial to see that this percentage will hold true for any ordered set of any size (assuming there are at least two elements), due to the symmetry of the diagram.

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answered Aug 31 at 1:04

Arcanist Lupus

1812

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up vote
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down vote

Illustrated above are all the possible ways you could select two numbers from 1,2,3,4,5 without replacement. (The diagonal is excluded because it requires picking the same number twice).

Of the 20 possible selections, 10 have the second choice as higher than the first, and 10 are the other way around. Hence, the probability is 10/20 = 50%

It is also trivial to see that this percentage will hold true for any ordered set of any size (assuming there are at least two elements), due to the symmetry of the diagram.

share|cite|improve this answer

answered Aug 31 at 1:04

Arcanist Lupus

1812

add a comment | 

up vote
0
down vote

up vote
0
down vote

Illustrated above are all the possible ways you could select two numbers from 1,2,3,4,5 without replacement. (The diagonal is excluded because it requires picking the same number twice).

Of the 20 possible selections, 10 have the second choice as higher than the first, and 10 are the other way around. Hence, the probability is 10/20 = 50%

It is also trivial to see that this percentage will hold true for any ordered set of any size (assuming there are at least two elements), due to the symmetry of the diagram.

share|cite|improve this answer

answered Aug 31 at 1:04

Arcanist Lupus

1812

Illustrated above are all the possible ways you could select two numbers from 1,2,3,4,5 without replacement. (The diagonal is excluded because it requires picking the same number twice).

Of the 20 possible selections, 10 have the second choice as higher than the first, and 10 are the other way around. Hence, the probability is 10/20 = 50%

It is also trivial to see that this percentage will hold true for any ordered set of any size (assuming there are at least two elements), due to the symmetry of the diagram.

share|cite|improve this answer

answered Aug 31 at 1:04

Arcanist Lupus

1812

share|cite|improve this answer

share|cite|improve this answer

answered Aug 31 at 1:04

Arcanist Lupus

1812

answered Aug 31 at 1:04

Arcanist Lupus

1812

answered Aug 31 at 1:04

Arcanist Lupus

1812

1812

add a comment | 

add a comment |