Sunday Times Teaser 3175 – Blindfold Roulette
by John Owen
Published Sunday July 30 2023 (link)
Our roulette wheel has fewer than 100 equal-sized sectors. Before each game a fixed number (fewer than half) of the sectors are randomly designated winning ones and the others as losing. A player can see which are the winning sectors, then is blindfolded. A ball is then placed at random in a winning sector and the player chooses to spin (S) the wheel or move (M) the ball clockwise by one sector. The game continues from there (the player has the same choices) and ends when the ball lands in a losing sector.
Alf’s longest winning run was MMSM while Bert’s was SSS and Charlie’s was MMMS. Being expert logicians, they had always chosen the option that gave the better chance of the ball landing in a winning sector. Remarkably, the probabilities of achieving each run of success were the same.
How many sectors are there and how many are winning sectors?
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BRG permalink12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061626364656667686970717273747576777879808182838485from fractions import Fraction as RFprint_clumping = False# partition the integer <n> into <k> non-decreasing positive integersdef decompose(n, k, seq=()):if k == 0:if n == 0:yield seqelse:for i in range(1 if not seq else seq[-1], n + 1):yield from decompose(n - i, k - 1, seq + (i,))# M probabilities: if the winning positions are in clumps of length a, b, c ...# only the last positions in each clump lose; hence the probability of an M# winning is sum(a - 1, b - 1, ...) / sum(a, b, ...). Given M wins, the# probability of MM is then sum(a - 2, b - 2, ...) / sum(a - 1, b - 1, ...)# and so on ...# with <n> sectors, <w> winning in clumps <cs>, find the probability# of the sequence <seq>def prob(n, w, cs, seq):p = 1# the probability of a winning SpinS = RF(w, n)# calculate probability of winning sequencesmc = csfor s in seq:# calculate the conditional probability of# the current M in a sequence of M'smp = sum(mc)mc = list(x - 1 for x in mc if x > 1)M = RF(sum(mc), mp)# check that a move wins over a spinif s == 'M':if M > S:p *= Melif S > M:p *= Smc = csreturn p# for <n> sectors with <w> winners, find clumping sequences# <cs> with at least <mnc> clumps giving probability <pt>def pr_clmps(n, w, ss, mnc, pt):for k in range(1, w + 1):for cs in decompose(w, k):if cs[-1] < mnc:continueif prob(n, w, cs, ss) == pt:yield cs# N sectors and W winners, Probability(spin winner) = W / N# Let C(N) = Probability(of a move landing N or more from the end of clumps)## Person Sequence Probability# Bert SSS [W / N]^3# Charlie: MMMS [C(4) / W] x [W / N]# Alf: MMSM [C(3) / W] x [W / N] x [C(2) / W]## [W / N]^3 = [C(4) / W] x [W / N] ==> C(4) = W^3 / N^2# [W / N]^3 = [C(3) / W] x [W / N] x [C(2) / W] ==> C(2).C(3) = W^4 / N^2# consider C(4) = W^3 / N^2for n in range(1, 100):for w in range(1, (n + 1) // 2):C4, r = divmod(w ** 3, n ** 2)if not r:p = RF(w, n)SSS = p ** 3if print_clumping:# find clumping for MMMSfor cl1 in pr_clmps(n, w, 'MMMS', 4, SSS):# find clumping for MMSMfor cl2 in pr_clmps(n, w, 'MMSM', 3, SSS):print(f"N = {n}, W = {w} (P = {SSS}) [{cl1} {cl2}]")else:try:cl1 = next(pr_clmps(n, w, 'MMMS', 4, SSS))cl2 = next(pr_clmps(n, w, 'MMSM', 3, SSS))print(f"N = {n}, W = {w} (P = {SSS})")except StopIteration:continue
