(no subject)
May. 4th, 2005 11:15 pmDear Lord...
I was arguing with a friend in another journal about my probability of using 4 unique user icons in 4 comments I made, and here's the reply I made:
Given that I have 15 different icons, there are 15 * 14 * 13 * (4 choose 2) permutations of only 2 of them the same, 15 * 14 * (4 choose 1) combinations [I meant permutations] for [only] 3 of them the same, and 15 combinations for 4 the same. This sums to 15 * 14 * 13 * 3 * 2 + 15 * 14 * 4 + 15 = 17235 possible choices.
The total number of permutations of 15 icons is 15^4, or 50625.
The probability that I choose at least 2 icons the same is thus 17235/50625 = 383/1125 or about .34. Thus probability states that it is more likely that I chose 4 unique icons, and not at least 2 of them the same.
I win.
Too much 251... I think I need to sleep.
(Also, I probably messed up my math somewhere, but don't tell
mogrothir that.)
I was arguing with a friend in another journal about my probability of using 4 unique user icons in 4 comments I made, and here's the reply I made:
Given that I have 15 different icons, there are 15 * 14 * 13 * (4 choose 2) permutations of only 2 of them the same, 15 * 14 * (4 choose 1) combinations [I meant permutations] for [only] 3 of them the same, and 15 combinations for 4 the same. This sums to 15 * 14 * 13 * 3 * 2 + 15 * 14 * 4 + 15 = 17235 possible choices.
The total number of permutations of 15 icons is 15^4, or 50625.
The probability that I choose at least 2 icons the same is thus 17235/50625 = 383/1125 or about .34. Thus probability states that it is more likely that I chose 4 unique icons, and not at least 2 of them the same.
I win.
Too much 251... I think I need to sleep.
(Also, I probably messed up my math somewhere, but don't tell
(no subject)
Date: 2005-05-04 08:21 pm (UTC)(no subject)
Date: 2005-05-04 08:26 pm (UTC)(no subject)
Date: 2005-05-04 08:35 pm (UTC)Thus I win even more.
(no subject)
Date: 2005-05-04 08:32 pm (UTC)Say I use icons A, A, B, and C. There are 15 choices for A, 14 for B, and 13 for C. To count for ordering, I choose where to place B and C, hence 4 choose 2.
Next I use icons A, A, A, and B. I have 15 choices for A and 14 for B. I can place B in 4 choose 1 places.
Lastly, I just use A, A, A, A. Ordering doesn't matter here.
Do I have some faulty reasoning here?
(no subject)
Date: 2005-05-04 08:37 pm (UTC)(no subject)
Date: 2005-05-04 08:27 pm (UTC)OWNED BY THE REAL WORLD.
(no subject)
Date: 2005-05-04 08:32 pm (UTC)(no subject)
Date: 2005-05-04 08:33 pm (UTC)You have no proof to suggest otherwise.