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1 vote (12.5%)
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Wizard
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Wizard
Joined: Oct 14, 2009
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May 28th, 2015 at 12:45:47 PM permalink
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:

  1. a<b<c
  2. a<c<b
  3. b<a<c
  4. b<c<a
  5. c<a<b
  6. 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.
ThatDonGuy
ThatDonGuy
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May 28th, 2015 at 1:05:38 PM permalink
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

Wizard
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Wizard
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May 28th, 2015 at 1:13:05 PM permalink
Quote: ThatDonGuy

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



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.
surrender88s
surrender88s
Joined: Jun 23, 2013
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May 28th, 2015 at 1:18:12 PM permalink
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

"Rule No.1: Never lose money. Rule No.2: Never forget rule No.1." -Warren Buffett on risk/return
PeeMcGee
PeeMcGee
Joined: Jul 23, 2014
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May 28th, 2015 at 6:30:52 PM permalink
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:

fA(a) = 1
fB(b) = 1/2
fC(c) = 1/3

Since each of the random variable are independent, the joint density function is:
f(a,b,c) = fA(a) * fB(b) * fC(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.
Wizard
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Wizard
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May 28th, 2015 at 7:07:13 PM permalink
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:

fA(a) = 1
fB(b) = 1/2
fC(c) = 1/3

Since each of the random variable are independent, the joint density function is:
f(a,b,c) = fA(a) * fB(b) * fC(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.

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