## BOF's probability problem...

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BOF
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### BOF's probability problem...

Dear All,
many years ago (quite a big many), I set a probability teaser in one of the early BDGA newsletters.
I cannot find any copy of my original working, either on computers or on paper!
If anyone has an archive of my original problem and the solution, I would be very grateful.

The detail of the problem was thus:

At the tee, X players flip discs to decide the order of play.
On each flip, if a single disc lands different to all the others, that player is 'out' i.e. next in order of play.
If no single disc lands differently (i.e. they all land the same, or a split of, for example, three 'up' and two 'down') then discs are re-flipped.
When down to the last 2 players, one will call 'odd' (discs land differently) or 'even' (discs land the same) to decide the last positions.

Question:
What is the probability, given X players at the tee, that only X number of flips are required- i.e. a player is 'out' on each flip.
So for 4 players only 4 flips are required (including the final odd/even flip).

Answers, on a virtual postacrd, to the forum...

Archive material (if you have any) to me, please.

BOF

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Mark.A.D
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### Re: BOF's probability problem...

Surely if there are X players, only X-1 flips would be required as the final flip would decide the fate of 2 players
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Paul Holden
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### Re: BOF's probability problem...

Always answer the question set in the paper Mark.
Paul Holden
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Paul Holden
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### Re: BOF's probability problem...

Joking aside I think it goes something like this...

The odds that for any one flip between Y players all but one disc land the same way up (assuming an equal chance for each way up) is:
(1/2)^(Y-1) , where I am using ^ to represent "to the power of"

Beginning with X players, and only looking at the flips required whilst there are 3 or more players left, because the last flip will always decide the last two players, the cumulative odds are:

(1/2)^(X-1+... ...+X-n) ,X > 2 ,n = X-2

Evaluating the sum of the arithmetic series in the exponent gives us:

(n/2)(2X+(n-1)(-1))

(n/2)(2X+1-n)

Substituting for n gives us:

((X-2)/2)(2X+1-X+2)

((X-2)/2)(X+3)

So the final odds that the minimum number of throws, which Mark correctly identifies as X-1, would be used is:

(1/2)^((X-2)/2)(X+3)

Am I even close???
Paul Holden
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PDGA No. 34662

Paul Holden
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### Re: BOF's probability problem...

On the archive front the only BDGA newsletter I have is Issue 1 and it isn't in there I have nothing between that and Inflight 2008, nice pic of Chris OB on the front.
Paul Holden
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PDGA No. 34662

BOF
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### Re: BOF's probability problem...

nice try, Paul, but I think it is simpler than that - unless your solution simplifies down to what I have in front of me here...

I have a nice solution just worked out, but am willing to give others a look before I reveal my solution (in case I'm right or wrong!).

I'll PM you my solution.

BOF

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BOF
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### Re: BOF's probability problem...

If only we could use the superscript BBCode -

http://www.phpbbdevelopers.net/archive/ ... ?f=27&t=93

Thanks

B[sup]O[/sup]F
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LostMeow
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### Re: BOF's probability problem...

Some of this maths is beyond me, but Paul, is your initial probability formula correct?

i.e. the probability that one of the discs is different being (1/2)^(Y-1)

I don't know what the answer is, but if, say, three discs were flipped, there would be 6 ways out of 8 for one of the discs to be different, which is a probability of 3/4, but your formula would calculate it as 1/4.

The number of ways of one disc being different should be the 2nd number from each end of Pascal's triangle added together.
Then the probability is that number divided by the total in that row of the triangle, so for:

2 discs = 2/4 = 1/2
3 discs = 6/8 = 3/4
4 discs = 8/16 = 1/2
5 discs = 10/32 = 5/16
6 discs = 12/64 = 3/16
etc...

I'm not sure what formula gives that series though...
Tom
ND

bruce
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### Re: BOF's probability problem...

LostMeow wrote: 2 discs = 2/4 = 1/2
3 discs = 6/8 = 3/4
4 discs = 8/16 = 1/2
5 discs = 10/32 = 5/16
6 discs = 12/64 = 3/16
etc...

I'm not sure what formula gives that series though...
At a glance: 2X/(2^X)
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LostMeow
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### Re: BOF's probability problem...

bruce wrote:
At a glance: 2X/(2^X)
D'oh. Knew it would be deceptively simple! Anyway, is that the right probability, and can it be arithmetically progressed like Paul did with the other one?
Tom
ND

Paul Holden
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### Re: BOF's probability problem...

LostMeow wrote:Some of this maths is beyond me, but Paul, is your initial probability formula correct?

i.e. the probability that one of the discs is different being (1/2)^(Y-1)

I don't know what the answer is, but if, say, three discs were flipped, there would be 6 ways out of 8 for one of the discs to be different, which is a probability of 3/4, but your formula would calculate it as 1/4.
Yeah, ignore that, I am home ill today and I think it was showing... Paul Holden
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Paul Holden
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### Re: BOF's probability problem...

I am out by factor of Y...

(1/2)^(Y-1)

((1/2)^(-1))((1/2)^(Y))

2((1/2)^(Y)

2/(2^Y)

Whereas Bruce has:

2Y/(2^Y)
Paul Holden
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PDGA No. 34662

LostMeow
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### Re: BOF's probability problem...

http://www.wolframalpha.com/input/?i=pr ... rom+3+to+n

Formula is:

(2^(-0.5(n-1)n))n!
Tom
ND

Paul Holden
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### Re: BOF's probability problem...

Looks good, I had just got to:

N! ( (1/2)^( (1/2)((N^2)-N) )

Which I think is the same?
Paul Holden
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PDGA No. 34662

BOF
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### Re: BOF's probability problem...

LostMeow, Paul - well done!

I wrote it as:

n!/(2^(0.5n(n-1))

Ten points to Gryffindor!

Class dismissed.

BOF

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