EN. Plinko
We settle some business and address the game of Plinko.
September 25, 2008 · answers, math puzzles, numbers, The Mathcast · Permalink
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We settle some business and address the game of Plinko.
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stevestyle said,
September 27, 2008 at 10:57 am
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.
stevestyle said,
September 27, 2008 at 11:06 am
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!
tchelyzt said,
October 6, 2008 at 3:39 pm
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
tchelyzt said,
October 6, 2008 at 4:03 pm
okay, checked it. Still a different answer to stevestyle :-(
now I have 4*4!*5!/9! = 4/63
Don
strauss said,
October 8, 2008 at 3:20 pm
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….