## Poll

1 vote (12.5%) | |||

No votes (0%) | |||

1 vote (12.5%) | |||

5 votes (62.5%) | |||

1 vote (12.5%) | |||

No votes (0%) |

**8 members have voted**

May 28th, 2015 at 12:45:47 PM
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Suppose:

Player A picks a random number between 0 and 1. Call that number a.

Player B picks a random number between 0 and 2. Call that number b.

Player C picks a random number between 0 and 3. Call that number c.

What is the probability of:

Player A picks a random number between 0 and 1. Call that number a.

Player B picks a random number between 0 and 2. Call that number b.

Player C picks a random number between 0 and 3. Call that number c.

What is the probability of:

- a<b<c
- a<c<b
- b<a<c
- b<c<a
- c<a<b
- c<b<a

Please put answers in spoiler tags.

It's not whether you win or lose; it's whether or not you had a good bet.

May 28th, 2015 at 1:05:38 PM
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Where's the choice for "I shall solve it without calculus, geometry, or simulation"?

A's number is between 0 and 1

B's is between 0 and 1 1/2 of the time, and between 1 and 2 1/2 of the time

C's is between 0 and 1 1/3 of the time, between 1 and 2 1/3 of the time, and between 2 and 3 1/3 of the time.

Each of these six possibilities is equally likely:

{0-1, 0-1, 0-1} - each of the six results is also equally likely

{0-1, 0-1, 1-2} - a < b < c 1/2 of the time, and b < a < c 1/2 of the time

{0-1, 0-1, 2-3} - a < b < c 1/2 of the time, and b < a < c 1/2 of the time

{0-1, 1-2, 0-1} - a < c < b 1/2 of the time, and c < a < b 1/2 of the time

{0-1, 1-2, 1-2} - a < b < c 1/2 of the time, and a < c < b 1/2 of the time

{0-1. 1-2. 2-3} - a < b < c

The probabilities are:

a < b < c = 1/6 x (1/6 + 1/2 + 1/2 + 0 + 1/2 + 1) = 4/9

a < c < b = 1/6 x (1/6 + 0 + 0 + 1/2 + 1/2 + 0) = 7/36

b < a < c = 1/6 x (1/6 + 1/2 + 1/2 + 0 + 0 + 0) = 7/36

b < c < a = 1/6 x (1/6 + 0 + 0 + 0 + 0 + 0) = 1/36

c < a < b = 1/6 x (1/6 + 0 + 0 + 1/2 + 0 + 0) = 1/9

c < b < a = 1/6 x (1/6 + 0 + 0 + 0 + 0 + 0) = 1/36

A's number is between 0 and 1

B's is between 0 and 1 1/2 of the time, and between 1 and 2 1/2 of the time

C's is between 0 and 1 1/3 of the time, between 1 and 2 1/3 of the time, and between 2 and 3 1/3 of the time.

Each of these six possibilities is equally likely:

{0-1, 0-1, 0-1} - each of the six results is also equally likely

{0-1, 0-1, 1-2} - a < b < c 1/2 of the time, and b < a < c 1/2 of the time

{0-1, 0-1, 2-3} - a < b < c 1/2 of the time, and b < a < c 1/2 of the time

{0-1, 1-2, 0-1} - a < c < b 1/2 of the time, and c < a < b 1/2 of the time

{0-1, 1-2, 1-2} - a < b < c 1/2 of the time, and a < c < b 1/2 of the time

{0-1. 1-2. 2-3} - a < b < c

The probabilities are:

a < b < c = 1/6 x (1/6 + 1/2 + 1/2 + 0 + 1/2 + 1) = 4/9

a < c < b = 1/6 x (1/6 + 0 + 0 + 1/2 + 1/2 + 0) = 7/36

b < a < c = 1/6 x (1/6 + 1/2 + 1/2 + 0 + 0 + 0) = 7/36

b < c < a = 1/6 x (1/6 + 0 + 0 + 0 + 0 + 0) = 1/36

c < a < b = 1/6 x (1/6 + 0 + 0 + 1/2 + 0 + 0) = 1/9

c < b < a = 1/6 x (1/6 + 0 + 0 + 0 + 0 + 0) = 1/36

May 28th, 2015 at 1:13:05 PM
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Quote:ThatDonGuyWhere's the choice for "I shall solve it without calculus, geometry, or simulation"?

A's number is between 0 and 1

B's is between 0 and 1 1/2 of the time, and between 1 and 2 1/2 of the time

C's is between 0 and 1 1/3 of the time, between 1 and 2 1/3 of the time, and between 2 and 3 1/3 of the time.

Each of these six possibilities is equally likely:

{0-1, 0-1, 0-1} - each of the six results is also equally likely

{0-1, 0-1, 1-2} - a < b < c 1/2 of the time, and b < a < c 1/2 of the time

{0-1, 0-1, 2-3} - a < b < c 1/2 of the time, and b < a < c 1/2 of the time

{0-1, 1-2, 0-1} - a < c < b 1/2 of the time, and c < a < b 1/2 of the time

{0-1, 1-2, 1-2} - a < b < c 1/2 of the time, and a < c < b 1/2 of the time

{0-1. 1-2. 2-3} - a < b < c

The probabilities are:

a < b < c = 1/6 x (1/6 + 1/2 + 1/2 + 0 + 1/2 + 1) = 4/9

a < c < b = 1/6 x (1/6 + 0 + 0 + 1/2 + 1/2 + 0) = 7/36

b < a < c = 1/6 x (1/6 + 1/2 + 1/2 + 0 + 0 + 0) = 7/36

b < c < a = 1/6 x (1/6 + 0 + 0 + 0 + 0 + 0) = 1/36

c < a < b = 1/6 x (1/6 + 0 + 0 + 1/2 + 0 + 0) = 1/9

c < b < a = 1/6 x (1/6 + 0 + 0 + 0 + 0 + 0) = 1/36

Hot damn, that was a good solution! Quick too. I agree with your answer.

If you're ever in Vegas I owe you a beer.

It's not whether you win or lose; it's whether or not you had a good bet.

May 28th, 2015 at 1:18:12 PM
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I used excel to do 10k trials

1 44.92%

2 19.21%

3 19.41%

4 2.95%

5 10.72%

6 2.79%

edit to compare with fractions given in Don's solution:

1: 4.043/9

2: 6.916/36

3: 6.988/36

4: 1.062/36

5: 0.965/9

6: 1.0044/36

1 44.92%

2 19.21%

3 19.41%

4 2.95%

5 10.72%

6 2.79%

edit to compare with fractions given in Don's solution:

1: 4.043/9

2: 6.916/36

3: 6.988/36

4: 1.062/36

5: 0.965/9

6: 1.0044/36

"Rule No.1: Never lose money. Rule No.2: Never forget rule No.1." -Warren Buffett on risk/return

May 28th, 2015 at 6:30:52 PM
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I was a little rusty, so thanks Wizard for posting this. And yea, kudos to ThatDonGuy’s creative answer.

The Uniform probability density function is 1/(b-a) where b,a are the maximum and minimum values, respectively. Therefore, the density functions for each of the random variables are:

f

f

f

Since each of the random variable are independent, the joint density function is:

f(a,b,c) = f

Therefore, we integrate 1/6 with respect to a, b and c. It can be tricky to get the correct limits.

So for the first case, a<b<c:

The other cases are done the same way but just changing the limits.

The Uniform probability density function is 1/(b-a) where b,a are the maximum and minimum values, respectively. Therefore, the density functions for each of the random variables are:

f

_{A}(a) = 1

f

_{B}(b) = 1/2

f

_{C}(c) = 1/3

Since each of the random variable are independent, the joint density function is:

f(a,b,c) = f

_{A}(a) * f

_{B}(b) * f

_{C}(c) = 1(1/2)(1/3) = 1/6

Therefore, we integrate 1/6 with respect to a, b and c. It can be tricky to get the correct limits.

So for the first case, a<b<c:

The other cases are done the same way but just changing the limits.

May 28th, 2015 at 7:07:13 PM
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Quote:PeeMcGee

The Uniform probability density function is 1/(b-a) where a,b are the maximum and minimum values, respectively. Therefore, the density functions for each of the random variables are:

f_{A}(a) = 1

f_{B}(b) = 1/2

f_{C}(c) = 1/3

Since each of the random variable are independent, the joint density function is:

f(a,b,c) = f_{A}(a) * f_{B}(b) * f_{C}(c) = 1(1/2)(1/3) = 1/6

Therefore, we integrate 1/6 with respect to a, b and c. It can be tricky to get the correct limits.

So for the first case, a<b<c:

The other cases are done the same way but just changing the limits.

Nice notation! That is pretty much how I did it too.

It's not whether you win or lose; it's whether or not you had a good bet.