Easy odds puzzle
Moderator: Michael Wallace
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Easy odds puzzle
8 players left in a randomly drawn straight KO.
4 top players and 4 muppets.
What are the odds of all the top players drawing the muppets?
4 top players and 4 muppets.
What are the odds of all the top players drawing the muppets?
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Re: Easy odds puzzle
In odds speak, I make it 34/1.
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Re: Easy odds puzzle
Can you show me how you got to that please, I've had differing answers from 'clever' people, It looks easy to me, but it obviously isn't???
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Re: Easy odds puzzle
I make it a lot lower odds than that.
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Re: Easy odds puzzle
I did it sequentially. Maybe I fucked it up. But I'll do it again now. There's four top players: A, B, C, D. Player A is paired against a random player. There are four muppets and three top players to choose from so 4/7 of picking a muppet. B is next. Three muppets left and two top players. So 3/5. C has one top player and two muppets to pick from. 2/3. D is left with a muppet. So 4/7 * 3/5 * 2/3 * 1 = 24/105 = 0.23. So yeah, I probably did fuck it up last time.
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Re: Easy odds puzzle
Correct answer - thick question: how did you arrive at 24/105? 0.23 would have done.
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Re: Easy odds puzzle
And if you're going to add 24/105 into your answer wouldn't you call it 8/35?
Just asking.
Just asking.
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Re: Easy odds puzzle
He's just showing you 24/105 because that's a direct result of the multiplication (probably a better word when it comes to probabilities I'm sure) of the probabilities of each individual component.
i.e. 24 = 4*3*2 and the 105 = 7*5*3
i.e. 24 = 4*3*2 and the 105 = 7*5*3
Re: Easy odds puzzle
I make the probability to be 8!/(4!*2^4)=8!/(24*16)=1/105.
Currently flicking between my own working and Gevin's reasoning to find an error to see which of us is wrong.
Currently flicking between my own working and Gevin's reasoning to find an error to see which of us is wrong.
Re: Easy odds puzzle
Ignore me, I'm wrong.
Re: Easy odds puzzle
Approximating this probability for an knockout tournament with n rounds as n gets large is a good exercise in using Sterling's approximation.
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Re: Easy odds puzzle
Or Stirling's approximation for even better results.
Re: Easy odds puzzle
I'm on fire today.
Oh wait its past midnight. D'oh.
D'oh d'oh.
Oh wait its past midnight. D'oh.
D'oh d'oh.
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Re: Easy odds puzzle
Glad one or two got this wrong at first... Just shows what a tricky little puzzle I came up with
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Re: Easy odds puzzle
Not tricky really. Perhaps a little fiddly and easy to make an error.Dave Preece wrote:Glad one or two got this wrong at first... Just shows what a tricky little puzzle I came up with
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Re: Easy odds puzzle
Tricky then?
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Re: Easy odds puzzle
I got it right with minimal effort. Not tricky.
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Re: Easy odds puzzle
Tricky (for some)?
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Re: Easy odds puzzle
Sorry, I meant to put emphasis on the "I".
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Re: Easy odds puzzle
Then you're obviously clevererer than Jack then? Well done!
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Re: Easy odds puzzle
He's the cleverest person here. You'll have to deal with that.Dave Preece wrote:Then you're obviously clevererer than Jack then? Well done!
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Re: Easy odds puzzle
Bit late to bump this, but 8/35 (27:8, or about 22.857%) is correct:
Without loss of generality, put top player A in slot 1 (basically, randomly assign a slot to A, then draw A's opponent next). The probability of a "muppet" being pitted vs. A is 4/7. If B, C, or D is pitted, the trial in its entirety is a failure.
Without loss of generality, put B in a remaining slot, then draw B's opponent only if A got a "muppet." There are 3 muppets left out of 5. If C or D is selected, the trial is failed.
For the last 4, the odds of C vs. D is 1/3, so the odds of !(C vs. D) is 2/3.
These are independent events, and one failure ruins the lot, so the probabilities are multiplied. (4/7)*(3/5)*(2/3) = 24/105 = 8/35.
Now, for any number 2^n, where half are top players and half muppets, the probability is (2^(n-1))!/([product of all positive odd integers less than 2^n]). This may have an error but I think this is accurate.
Without loss of generality, put top player A in slot 1 (basically, randomly assign a slot to A, then draw A's opponent next). The probability of a "muppet" being pitted vs. A is 4/7. If B, C, or D is pitted, the trial in its entirety is a failure.
Without loss of generality, put B in a remaining slot, then draw B's opponent only if A got a "muppet." There are 3 muppets left out of 5. If C or D is selected, the trial is failed.
For the last 4, the odds of C vs. D is 1/3, so the odds of !(C vs. D) is 2/3.
These are independent events, and one failure ruins the lot, so the probabilities are multiplied. (4/7)*(3/5)*(2/3) = 24/105 = 8/35.
Now, for any number 2^n, where half are top players and half muppets, the probability is (2^(n-1))!/([product of all positive odd integers less than 2^n]). This may have an error but I think this is accurate.
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