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Everything posted by Prime
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There are few interesting questions herein. (Perhaps, people who study gentics and population have some ready-made formulas.) Won't these stable population proportions (BB:Bg:gg = 25:30:9) be reached in the very next generation? If the initial population distribution was different, we would get a different proportion for stable population. Does that stable proportion always happen in the very first offspring generation? Why?
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My friend and I play in a chess tournament
Prime replied to BMAD's question in New Logic/Math Puzzles
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I suppose, blue eyed child also must play part in figuring out question 1. I only took it into account for figuring out question two, so my previous answer is wrong.
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Hey, I resent that =). I spend some time trying to make the wording clear and lucid. There is no way this puzzle can match the sheer genius of grammatical obfuscation in 'Give monkey enough rope'. I did not mean to imply any grammatical obfuscation in this puzzle. I meant probability obfuscation with backward refrences, like first child. My analysis uses the first blue eyed child as prior information (highlighted in red here). That's how I got probability 3/13 rather than 2/7 for Judy's genotype Bg.
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This way it's more interesting. I understand, one person has 4 times the money of his neighbor at the end of the game.
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Edit: When you say, "as long as possible," what exactly does that entail? That may change the solution. There are more than 2 people. The phrase as long as possible means that each person passed until the last person received money. If the "last" person does not pass money to the "first", why do they sit in a circle and not in row?
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I took a little time to verify my guess. Somehow, to me this problem seems similar to Bonanova's “Give monkey enough rope.” (Some serious untangling to perform one must.)
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The probability of finding an isosceles triangle.
Prime replied to BMAD's question in New Logic/Math Puzzles
What's triangle inequality? So, 0-angle triangles are being discriminated agianst. That's an injustice. Still, my last post gives two different answers even for that case. That depends on how "probability" was meant in the OP. The way OP was worded ("probability of forming"), it looks more like 4/10. In this context, I assume, we form a triangle by randomly picking sides for it. If OP said: "What is the probability that the triangle is isosceles," then I would be inclined to answer: 2/3. -
The probability of finding an isosceles triangle.
Prime replied to BMAD's question in New Logic/Math Puzzles
What's triangle inequality? So, 0-angle triangles are being discriminated. That's an injustice. Still, my last post gives two different answers even for that case. -
In that case, like Sp said. And the number of decks does not matter. well this is embarrassing but for some reason i am getting a higher number, what am i missing? Do your decks have Jokers? (Wouldn't matter, anyway, since Joker helps making 4 of a kind.) I assume, standard deck is 52 cards 2 - 10, J, Q, K, A. If the above assumptions are correct, but the answer is still wrong, I give up and would like to see a list of more cards than Sp posted not having at least 1 four of a kind.
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The probability of finding an isosceles triangle.
Prime replied to BMAD's question in New Logic/Math Puzzles
What's triangle inequality? -
In that case, like Sp said. And the number of decks does not matter.
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The probability of finding an isosceles triangle.
Prime replied to BMAD's question in New Logic/Math Puzzles
Depending on random selection process, there is another possibility. -
In old (Euclid) times, they did not count the number as its own divisor, but they did count 1. Thus, Perfect Number used to be equal to the sum of its divisors (Aliquot Parts.) (Don't quote me on that.)
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Perhaps, OP is looking for this:
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I am only aware of this general rule. That is not to say there aren't any other possibilities for finding perfect numbers.
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The probability of finding an isosceles triangle.
Prime replied to BMAD's question in New Logic/Math Puzzles
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Agree. A fair method cannot guaranty any maximum number of tosses. I see your 1 and I will raise you a 0 =) Not acceptable. The winner must be decided by coin toss.
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Assuming 52-deck card (no Jokers):