c4countdown

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?

In odds speak, I make it 34/1.

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???

I make it a lot lower odds than that.

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.

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

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

Just asking.

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

Oh I see.

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.

Ignore me, I'm wrong.

Approximating this probability for an knockout tournament with n rounds as n gets large is a good exercise in using Sterling's approximation.

Or Stirling's approximation for even better results.

I'm on fire today.

Oh wait its past midnight. D'oh.

D'oh d'oh.

Glad one or two got this wrong at first... Just shows what a tricky little puzzle I came up with

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.

Tricky then?

I got it right with minimal effort. Not tricky.

Tricky (for some)?

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

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

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.

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.