Comments on: EN. Plinko http://mathfactor.uark.edu/2008/09/en-plinko/ The Math Factor Podcast Site Fri, 08 Aug 2014 12:52:06 +0000 hourly 1 https://wordpress.org/?v=4.9.25 By: strauss http://mathfactor.uark.edu/2008/09/en-plinko/comment-page-1/#comment-388 Wed, 08 Oct 2008 21:20:05 +0000 http://mathfactor.uark.edu/2008/09/25/en-plinko/#comment-388 Tchelyt said that

p(5,5) = (4/9)p(4,5) + (1-5/9)p(5,4)

That’s totally right! But there is something amiss with the rest of the comment— perhaps the starting out values were incorrect. Recall that p(0,n) = p(n,0) = 0 (since one shot was missed, and one made, at the beginning); p(1,2) = p(2,1) = 1/2….

]]> By: tchelyzt http://mathfactor.uark.edu/2008/09/en-plinko/comment-page-1/#comment-385 Mon, 06 Oct 2008 22:03:44 +0000 http://mathfactor.uark.edu/2008/09/25/en-plinko/#comment-385 okay, checked it. Still a different answer to stevestyle :-(

now I have 4*4!*5!/9! = 4/63

Don

]]> By: tchelyzt http://mathfactor.uark.edu/2008/09/en-plinko/comment-page-1/#comment-384 Mon, 06 Oct 2008 21:39:07 +0000 http://mathfactor.uark.edu/2008/09/25/en-plinko/#comment-384 Hi,
I took a different approach:

target is 5-each which I call p(5,5)
now p(5,5) = (4/9)p(4,5) + (1-5/9)p(5,4)
by continuously substituting I do indeed get a result with 9! in the denominator but my answer reduces to 8/105. Perhaps I made a little mistake in there. I’ll check

Great puzzle
Don

]]> By: stevestyle http://mathfactor.uark.edu/2008/09/en-plinko/comment-page-1/#comment-377 Sat, 27 Sep 2008 17:06:04 +0000 http://mathfactor.uark.edu/2008/09/25/en-plinko/#comment-377 I just spotted a typo in my solution.

The total number of steps is A+B-2 not A+B-1. The equations are right.

I had a look at the 2007 Putnam questions. They are quite nice, the one’s I’ve done have solutions which are easy to state but tricky to spot. Perfect!

]]> By: stevestyle http://mathfactor.uark.edu/2008/09/en-plinko/comment-page-1/#comment-376 Sat, 27 Sep 2008 16:57:16 +0000 http://mathfactor.uark.edu/2008/09/25/en-plinko/#comment-376 After many hours on the court I think I have the answer.

Starting with a hits and b misses the chance of going to a+1 misses is a/(a+b) and the chance of going to b+1hits is b/(a+b).
If the target is A misses and B hits then before each shot a will run from 1 to A-1 and b will run from 1 to B-1. a+b will run from 2 to A+B-1.
So the probability of any particular sequence which works will be (A-1)!(B-1)!/(A+B-1)!
Surpisingly the probability is the same whichever sequence we choose.
The number of sequences which work is the number of ways of ordering A-1 misses and B-1 hits. The length of the sequence is A+B-2 so the number of sequences is the number of ways of choosing A-1 steps to be misses from A+B-1 steps. This is (A+B-2)!/(A-1)!(B-1)!.
The probability of getting A misses and B hits is the product of the number of sequences times the probability of each sequence, which is
(A-1)!(B-1)!/(A+B-1)! x (A+B-2)!/(A-1)!(B-1)! = 1/(A+B-1)

The total number of shots taken is A+B.

So the probability of any particular score from n shots is 1/(n-1), regardless of the score.

In the example this is 1/9.

This is a very neat result, it is a flat distribution which is quite different to the bell curve you mentioned for coin tosses. I guess we are always favouring the more extreme result which compensates exactly for the fact there are less ways of reaching the extreme results.

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