Thread Rating:

Poll

17 votes (50%)
13 votes (38.23%)
5 votes (14.7%)
2 votes (5.88%)
11 votes (32.35%)
3 votes (8.82%)
6 votes (17.64%)
5 votes (14.7%)
10 votes (29.41%)
8 votes (23.52%)

34 members have voted

ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 106
  • Posts: 4975
December 1st, 2020 at 10:32:19 AM permalink
Quote: charliepatrick

Total perms =25*24*23/6 = 2208
For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.
Diagonal 3-length lines - 4 of them, 1 perm = 4 combos
Diagonal 4-length lines - 4 of them, 4 perms = 16 combos
Diagonal 5-length lines - 2 of them, 10 perms = 20 combos
Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos
Total = 140 combos.
So P=140/2208.



...having three in the same "line" that are knight's moves from each other?
i.e. if you label the rows 1-5 from top to bottom and the columns A-E from left to right, then if the pegs are in A1, C2, and E3, that's not a triangle?
There appear to be 12 of these:
A1-C2-E3, A2-C3-E4, A3-C4-E5
A3-C2-E1, A4-C3-E2, A5-C4-E3
A1-B3-C5, B1-C3-D5, C1-D3-E5
A5-B3-C1, B5-C3-D1, C5-D3-E1

USpapergames
USpapergames
Joined: Jun 23, 2020
  • Threads: 18
  • Posts: 807
December 1st, 2020 at 10:47:16 AM permalink
Quote: ThatDonGuy

Quote: charliepatrick

Total perms =25*24*23/6 = 2208
For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.
Diagonal 3-length lines - 4 of them, 1 perm = 4 combos
Diagonal 4-length lines - 4 of them, 4 perms = 16 combos
Diagonal 5-length lines - 2 of them, 10 perms = 20 combos
Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos
Total = 140 combos.
So P=140/2208.



...having three in the same "line" that are knight's moves from each other?
i.e. if you label the rows 1-5 from top to bottom and the columns A-E from left to right, then if the pegs are in A1, C2, and E3, that's not a triangle?
There appear to be 12 of these:
A1-C2-E3, A2-C3-E4, A3-C4-E5
A3-C2-E1, A4-C3-E2, A5-C4-E3
A1-B3-C5, B1-C3-D5, C1-D3-E5
A5-B3-C1, B5-C3-D1, C5-D3-E1



Nice, can't you go a step beyond the knight's move & do a 1 diagonal and 3 spaces up movement or 2 diagonal & 2 spaces up movement to get a triangle as well?

Wait looks like Charlie is only counting the permutations of non-triangle outcomes. So he is accounting for all other outcomes to give a triangle as a result.
Math is the only true form of knowledge
USpapergames
USpapergames
Joined: Jun 23, 2020
  • Threads: 18
  • Posts: 807
December 1st, 2020 at 11:09:11 AM permalink
Ok, so I have the actual answer! Looks like you guys need help with combination analysis.

Total combinations = 25!/(25-3)!3! = 2,300
Total permutations = 25*24*23 = 13,800

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical. It's ironic that Charlie knows this but still got the wrong answer.

Vertical / Horizontal - 10 combos per 10 lines = 100 combos.
Diagonal - 10 combos per 2 (5 hole) lines + 4 combos per 4 (4 hole) lines + 1 combo per 4 (3 hole) lines = 40 combos.

So P = 140/2,300

Last edited by: USpapergames on Dec 1, 2020
Math is the only true form of knowledge
unJon
unJon
Joined: Jul 1, 2018
  • Threads: 14
  • Posts: 2936
December 1st, 2020 at 11:26:14 AM permalink
Quote: USpapergames

Ok, so I have the actual answer! Looks like you guys need help with combination analysis.

Total combinations = 25!/(25-3!)3! = 2,300
Total permutations = 25*24*23 = 13,800

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical. It's ironic that Charlie knows this but still got the wrong answer.


Vertical / Horizontal - 10 combos per 10 lines = 100 combos.
Diagonal - 10 combos per 2 (5 hole) lines + 4 combos per 4 (4 hole) lines + 1 combo per 4 (3 hole) lines = 40 combos.

So P = 140/2,300



You missed the same extra diagonals that Donguy pointed out above. For example, A1-C2-E3
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ChesterDog
ChesterDog
Joined: Jul 26, 2010
  • Threads: 8
  • Posts: 973
Thanks for this post from:
Gialmere
December 1st, 2020 at 11:29:27 AM permalink
Quote: Gialmere

It's toughie Tuesday...



A 5 by 5 square lattice is formed by drilling holes in a piece of wood. Three pegs are placed in this lattice at random.

What is the probability that the three pegs will form a triangle?




I counted 152 combinations that are colinear out of a total of 2300 leaving 2148 combinations to make triangles.

Probability of a triangle = 2148 / 2300 = 537 / 575 or approximately 93.39%

USpapergames
USpapergames
Joined: Jun 23, 2020
  • Threads: 18
  • Posts: 807
December 1st, 2020 at 11:31:42 AM permalink
Quote: unJon

Quote: USpapergames

Ok, so I have the actual answer! Looks like you guys need help with combination analysis.

Total combinations = 25!/(25-3!)3! = 2,300
Total permutations = 25*24*23 = 13,800

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical. It's ironic that Charlie knows this but still got the wrong answer.


Vertical / Horizontal - 10 combos per 10 lines = 100 combos.
Diagonal - 10 combos per 2 (5 hole) lines + 4 combos per 4 (4 hole) lines + 1 combo per 4 (3 hole) lines = 40 combos.

So P = 140/2,300



You missed the same extra diagonals that Donguy pointed out above. For example, A1-C2-E3



It's funny because I thought so also but I believe Donguy is wrong since I am only counting outcomes that don't result in a triangle outcome. So all the knight moves are definitely counted in terms of successful triangle outcomes.
Math is the only true form of knowledge
USpapergames
USpapergames
Joined: Jun 23, 2020
  • Threads: 18
  • Posts: 807
December 1st, 2020 at 11:34:41 AM permalink
Quote: ChesterDog

Quote: Gialmere

It's toughie Tuesday...



A 5 by 5 square lattice is formed by drilling holes in a piece of wood. Three pegs are placed in this lattice at random.

What is the probability that the three pegs will form a triangle?




I counted 152 combinations that are colinear out of a total of 2300 leaving 2148 combinations to make triangles.

Probability of a triangle = 2148 / 2300 = 537 / 575 or approximately 93.39%



Please share your work, this can't be right. Your saying I missed 12 outcomes that don't result in a triangle, and that's just hard for me to believe because from what I have seen on this form, I am the Combo King.
Math is the only true form of knowledge
ChesterDog
ChesterDog
Joined: Jul 26, 2010
  • Threads: 8
  • Posts: 973
December 1st, 2020 at 11:47:10 AM permalink
Quote: USpapergames

...
Please share your work, this can't be right. Your saying I missed 12 outcomes that don't result in a triangle, and that's just hard for me to believe because from what I have seen on this form, I am the Combo King.




I found 12 combinations congruent to A1-B3-C5.
USpapergames
USpapergames
Joined: Jun 23, 2020
  • Threads: 18
  • Posts: 807
December 1st, 2020 at 11:58:04 AM permalink
Here are the remaining outcomes with line shapes.


2 Vertical/Horizantal | 1 Diagonal = 6
1 Vertical/Horizantal | 2 Diagonal = 6
Last edited by: USpapergames on Dec 1, 2020
Math is the only true form of knowledge
USpapergames
USpapergames
Joined: Jun 23, 2020
  • Threads: 18
  • Posts: 807
December 1st, 2020 at 11:58:10 AM permalink
Quote: ChesterDog


I found 12 combinations congruent to A1-B3-C5.



Your right, I completely misunderstood the point Donguy was making. The entire time I thought he was attempting to count all the triangle outcomes and was forgetting to include all the triangles from knight moves lol. So exclusion method doesn't work unless you are aware of all possible straight line combinations. Ok so then there are 12 that we missed :(

It's possible I was too narrow-minded after agreeing with charliepatrick but I think the real culprit was being under the impression the question would be easy. Coming up with the answer in less than 2 minutes felt right to me. No need to overthink a simple question lol. It's times like this that I realize my arrogance can be a huge weakness if I were to ever underestimate a problem that could cost me a fortune.
Last edited by: USpapergames on Dec 1, 2020
Math is the only true form of knowledge

  • Jump to: