Easy odds puzzle

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Dave Preece
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Easy odds puzzle

Post by Dave Preece »

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?
Gavin Chipper
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Re: Easy odds puzzle

Post by Gavin Chipper »

In odds speak, I make it 34/1.
Dave Preece
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Re: Easy odds puzzle

Post by Dave Preece »

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???
Dave Preece
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Re: Easy odds puzzle

Post by Dave Preece »

I make it a lot lower odds than that.
Gavin Chipper
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Re: Easy odds puzzle

Post by Gavin Chipper »

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.
Dave Preece
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Re: Easy odds puzzle

Post by Dave Preece »

Correct answer - thick question: how did you arrive at 24/105? 0.23 would have done.
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Re: Easy odds puzzle

Post by Dave Preece »

And if you're going to add 24/105 into your answer wouldn't you call it 8/35?

Just asking.
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Matt Morrison
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Re: Easy odds puzzle

Post by Matt Morrison »

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
Dave Preece
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Re: Easy odds puzzle

Post by Dave Preece »

Oh I see.
JackHurst
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Re: Easy odds puzzle

Post by JackHurst »

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.
JackHurst
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Re: Easy odds puzzle

Post by JackHurst »

Ignore me, I'm wrong.
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Re: Easy odds puzzle

Post by JackHurst »

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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Charlie Reams
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Re: Easy odds puzzle

Post by Charlie Reams »

Or Stirling's approximation for even better results.
JackHurst
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Re: Easy odds puzzle

Post by JackHurst »

I'm on fire today.

Oh wait its past midnight. D'oh.

D'oh d'oh.
Dave Preece
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Re: Easy odds puzzle

Post by Dave Preece »

Glad one or two got this wrong at first... Just shows what a tricky little puzzle I came up with ;-)
Gavin Chipper
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Re: Easy odds puzzle

Post by Gavin Chipper »

Dave Preece wrote:Glad one or two got this wrong at first... Just shows what a tricky little puzzle I came up with ;-)
Not tricky really. Perhaps a little fiddly and easy to make an error.
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Re: Easy odds puzzle

Post by Dave Preece »

Tricky then?
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Matt Morrison
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Re: Easy odds puzzle

Post by Matt Morrison »

I got it right with minimal effort. Not tricky.
Dave Preece
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Re: Easy odds puzzle

Post by Dave Preece »

Tricky (for some)?
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Matt Morrison
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Re: Easy odds puzzle

Post by Matt Morrison »

Sorry, I meant to put emphasis on the "I".
Dave Preece
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Re: Easy odds puzzle

Post by Dave Preece »

Then you're obviously clevererer than Jack then? Well done!
Gavin Chipper
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Re: Easy odds puzzle

Post by Gavin Chipper »

Dave Preece wrote:Then you're obviously clevererer than Jack then? Well done!
He's the cleverest person here. You'll have to deal with that.
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Re: Easy odds puzzle

Post by Steve Balog »

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.
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