Easy odds puzzle

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Dave Preece
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Easy odds puzzle

Post by Dave Preece » Tue Feb 11, 2014 7:30 pm

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?

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Re: Easy odds puzzle

Post by Gavin Chipper » Tue Feb 11, 2014 8:40 pm

In odds speak, I make it 34/1.

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Re: Easy odds puzzle

Post by Dave Preece » Tue Feb 11, 2014 8:49 pm

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

Post by Dave Preece » Tue Feb 11, 2014 11:14 pm

I make it a lot lower odds than that.

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Re: Easy odds puzzle

Post by Gavin Chipper » Tue Feb 11, 2014 11:30 pm

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

Post by Dave Preece » Wed Feb 12, 2014 4:43 am

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 » Wed Feb 12, 2014 4:44 am

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 » Wed Feb 12, 2014 8:44 am

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

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Re: Easy odds puzzle

Post by Dave Preece » Wed Feb 12, 2014 4:10 pm

Oh I see.

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Re: Easy odds puzzle

Post by JackHurst » Tue Feb 18, 2014 7:27 pm

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.

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Re: Easy odds puzzle

Post by JackHurst » Tue Feb 18, 2014 7:33 pm

Ignore me, I'm wrong.

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Re: Easy odds puzzle

Post by JackHurst » Tue Feb 18, 2014 7:52 pm

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 » Tue Feb 18, 2014 10:33 pm

Or Stirling's approximation for even better results.

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Re: Easy odds puzzle

Post by JackHurst » Wed Feb 19, 2014 12:06 am

I'm on fire today.

Oh wait its past midnight. D'oh.

D'oh d'oh.

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Re: Easy odds puzzle

Post by Dave Preece » Fri Feb 21, 2014 4:14 pm

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

Post by Gavin Chipper » Fri Feb 21, 2014 4:42 pm

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 » Sat Feb 22, 2014 5:22 am

Tricky then?

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Matt Morrison
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Re: Easy odds puzzle

Post by Matt Morrison » Sat Feb 22, 2014 10:47 am

I got it right with minimal effort. Not tricky.

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Re: Easy odds puzzle

Post by Dave Preece » Sat Feb 22, 2014 12:12 pm

Tricky (for some)?

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Matt Morrison
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Re: Easy odds puzzle

Post by Matt Morrison » Sat Feb 22, 2014 1:12 pm

Sorry, I meant to put emphasis on the "I".

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Re: Easy odds puzzle

Post by Dave Preece » Sat Feb 22, 2014 4:12 pm

Then you're obviously clevererer than Jack then? Well done!

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Re: Easy odds puzzle

Post by Gavin Chipper » Sat Feb 22, 2014 7:28 pm

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 » Sat Mar 08, 2014 12:52 am

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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