November 2014
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A simple way to solve this month's challenge is by p1=1/17 and p2=1/2. However, this solution requires seven coin tosses, and we asked for six.
The solution we meant, and which many of you found, is p1=8/17 and p2=1/3.
Tossing p1 twice, we split the probability to 64/289, 72/289, 72/289, 81/289 and using up to four more coin tosses we get the desired distribution.
See Oleg Vlasii's picture (PNG, 11KB) of the binary tree to solve it.
1
(1 lambda
(2 (2 alpha beta) (2 zeta epsilon)))
(1 (2 (2 gamma omicron) (2 omicron upsilon))
(2 (2 (2 (2 tau eta) (2 rho mu))
(2 (2 phi nu) (2 xi delta)))
(2 (2 (2 chi theta) (2 theta kappa))
(2 (2 pi sigma) (2 omega iota)))))
1 (1 lambda
(2 (2 alpha beta) (2 zeta epsilon)))
(1 (2 (2 gamma omicron) (2 omicron upsilon))
(2 (2 (2 (2 tau eta) (2 rho mu))
(2 (2 phi nu) (2 xi delta)))
(2 (2 (chi theta) (theta kappa))
(2 (2 pi sigma) (2 omega iota)))))
Another solution uses p1=46/289 p2=1/3.
Philip Kinlen sent us another visualization http://abitofmaths.blogspot.hk/2014_12_01_archive.html
This month's problem was inspired by UYHIP's October's challenge: http://www.brand.site.co.il/riddles/201410q.html
If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com