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fun math puzzles

Started by leinad May 12, 2003 at 9:34 PM 185 replies 21.4k views
Original Post
leinad
leinad
This is not homework; I am an expert in Probability. I. Use this diagram: A B C D A'' B'' C'' D'' E'' F'' A-F'' are all trivial points. 1. If you randomely select 3 points from the whole group [of 10 points], what is the probability of forming a triangle with 2 points from the bottom and 1 point from the top? 2. What''s the probability of not making a triangle with your point selection of 3? (skill level: medium) II. No diagram for this one: In an orthogonal 8x7-square grid, how many rectangles can be formed? Need a hint? Here: remember that a rectangle can be formed by two of the horizontal lines in the grid and two vertical lines in the grid. And don''t forget that a square is a rectangle. 8 squares by 7 squares equals 56 squares in the grid. (difficulty: hard) If you can answer one of the above problems, then you can contribute one of your own. I''ll be posting the answer in 30 minutes from the live time of this post.
vanillacoke
vanillacoke
II: 588 rectangles.

...now for I...

edit: oh, I thought you meant 8x7 points... for 8x7 squares, the answer is 1008. And that's assuming that all rectangles have sides parallel to the rows & columns... there are other ones besides those ones.



[edited by - vanillacoke on May 12, 2003 10:44:04 PM]
You know what I never noticed before?
vanillacoke
vanillacoke
I.
1. 1/6 is probablity of getting thing you said.

2. 1/5 is probability of getting no triangle.

edit: that's assuming that the points are picked without replacement.



[edited by - vanillacoke on May 12, 2003 10:48:41 PM]
You know what I never noticed before?
leinad
leinad
Your first answer was correct: 588. If you disagree, wait''ll I post the solutions in 15 mins.
vanillacoke
vanillacoke
I''ll bet my method was correct, either way. It was probably just a misunderstanding.
You know what I never noticed before?
leinad
leinad
Re to your second post: 1/6 is incorrect. The probability is actually (for number 1) a 50% chance of forming a square out of 2 bottom and 1 top in a selection of 3 from the group of 10. Again, wait for my answer post.
leinad
leinad
Twelve minutes and counting...
When are you going to try''n stump me? I said you could contribute if you got one right, and you did. Now it''s my turn to have fun!
sQuid
sQuid
It''s not mine but I love this one:

A game show contestant is presented with three doors numbered 1, 2, and 3. Behind one of the doors is the grand prize. The contestant chooses a door. The host, who knows what''s behind each door, opens up one of the two remaining doors which doesn''t have the grand prize behind it. He then asks the contestant, "Do you want to stay with your original choice or would you like to switch to the other remaining door?"

Should the contestant stay with her original choice, should she change to the other door, or does it not make any difference?
vanillacoke
vanillacoke
Ok. You have a regular hexagon, in the cartesian plane. The y-coordinates of the vertices are unique elements of the set {0,2,4,6,8,10}. Find the area of the hexagon.
You know what I never noticed before?
leinad
leinad
The odds of the grand prize being in one of the two unopened doors is 50:50, thus it does not matter which door she chooses.
vanillacoke
vanillacoke
sQuid... I heard that one like 3 years ago, and I can never remember whether the correct answer goes with or against my intuition... I''ll say contestant should change, because there is a 2/3 chance of being wrong on the first guess.
I think the trick is that you''re supposed to think that it doesn''t matter whether you change, because you ignore the third door and wrongly think that there is a fifty percent chance of winning whether you switch or not.
You know what I never noticed before?
sQuid
sQuid
quote:
Original post by leinad
The odds of the grand prize being in one of the two unopened doors is 50:50, thus it does not matter which door she chooses.


Heh. INCORREECT

Don''t worry, Paul Erdos got it wrong too so you''re in good company.
vanillacoke
vanillacoke
Marilyn vos Savant answered this question once, but of course I can''t remember what she said.
You know what I never noticed before?
leinad
leinad
P(I.1)= [(6 C 2) * (4 C 1)]/(10 C 3)
=[(6!/2!4!) * 4]/(10!/3!7!)
=[(6*5/2) * 4 ]/(10*9*8/6)
=60/120
=1/2
=.5
=50 *1/100
=50%

P(I.2)= [(6 C 3) + (4 C 3)]/(10 C 3)
....see I.1 ....
=20% probability

II:
(8 C 2) * (7 C 2) = 588 combinations, or rectangles.
leinad
leinad
sQyui, you are wrong. Your question did not state anything about first guess/second guess. LOOK: You have two choices, A or B. Randomely choose....equals 50 % chance of either one!!! There''s no law of the universe that says otherwise! That is not a trick question because the person who wrote it pulled it out of their ass and didn''t know shit about probability.
sQuid
sQuid
quote:
Original post by leinad
sQyui, you are wrong. Your question did not state anything about first guess/second guess. LOOK: You have two choices, A or B. Randomely choose....equals 50 % chance of either one!!! There''s no law of the universe that says otherwise! That is not a trick question because the person who wrote it pulled it out of their ass and didn''t know shit about probability.


You''re an expert in probability theory right? What do you recall about dependent random variables?

Failing that you can just enumerate all possible outcomes. You''ll see that if you switch you have more chance of winning.
vanillacoke
vanillacoke
daniel, I hope you''re not using equations without backing them up with logical thought...
You know what I never noticed before?
leinad
leinad
Ummmm, nooooo, say you choose box 1, the guy shows you box 3, whose to say which box (1 or 2) the prize is in? You cannot overlap those two events, moron.
cowsarenotevil
cowsarenotevil
quote:
Original post by sQuid
It''s not mine but I love this one:

A game show contestant is presented with three doors numbered 1, 2, and 3. Behind one of the doors is the grand prize. The contestant chooses a door. The host, who knows what''s behind each door, opens up one of the two remaining doors which doesn''t have the grand prize behind it. He then asks the contestant, "Do you want to stay with your original choice or would you like to switch to the other remaining door?"

Should the contestant stay with her original choice, should she change to the other door, or does it not make any difference?


Ten bucks I can write a computer program that says it''s 50%.

-~-The Cow of Darkness-~-
-~-The Cow of Darkness-~-

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