Sunday Times Teaser 3148 – Quiz Probabilities

by BRG on January 22, 2023

by Edmund Marshall

Published Sunday January 22 2023 (link)

Each of four contending couples in a quiz game has equal probability of elimination at the end of each of the first three rounds, one couple going after each round. In the fourth round, the remaining couple has a constant probability p, less than ½, of winning the jackpot, which consists of £1000 in the first game; if the jackpot is not won, it is added to the £1000 donated in the next game. Each couple may enter three successive games of the quiz, except that any couple having played for the jackpot in the fourth round of any game then withdraws altogether, being replaced by a new couple in the next game.

If the probability that a couple, competing from the first game, wins £2000 is 7/96, what is the value of p as a fraction?

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Brian Gladman permalink

I believe that this teaser has a flaw. If it is interpreted as the author intended, it has no solution.

Consequently I don’t have a full Python solution. But here is a program that paves the way to a discussion of possible solutions by giving the equations for the probability of winning the various possible amounts in terms of the probability \(p\) described in the teaser.

Please note that this code uses the latest version of my polynomial library which I have just updated to provide features that help with this teaser.


from polynomial import Poly
from fractions import Fraction as RF
from gmpy2 import mpq as RF

# probability of winning the jackpot in the final round (= p)
win = Poly([0, 1])

# credit:  Jim Randell
# generate the polynomials P[p] that determine the probabilities of
# all possible outcome amounts (in units of the base prize) with a 
# maximum of <n> games; return: <amount>, <probability polynomial>
def outcomes(n, t=1, cp=1):
  if n == 0:
    # finished without a win
    yield 0, cp
  else:
    # there is a 1/4 chance of reaching the final and leaving
    q = cp * RF(1, 4)
    # with a win 
    yield t, q * win
    # or a loss
    yield 0, q * (1 - win)

    # there is a 3/4 chance of getting knocked out
    # and then entering more gaames after 
    q = cp * RF(3, 4)
    # a win and no rollover
    yield from outcomes(n - 1, 1, q * win)
    # or a loss and a rollover
    yield from outcomes(n - 1, t + 1, q * (1 - win))

# accumulate probabilities of winning amounts over three games
cp, ps = dict(), Poly([0])
for v, p in outcomes(3):
  ps += p
  if v not in cp:
    cp[v] = Poly([0])
  cp[v] += p

# output the amounts and their probability polynomials
for v, p in sorted(cp.items()):
  print(f"   £{v}000 ==> P[p] = {p:p}")
print(f"     Sum ==> P[p] = {ps}")
cp[23] = cp[2] + cp[3]
cp[123] = cp[1] + cp[23]
print(f">= £1000 ==> P[p] = {cp[123]:p}")
print(f">= £2000 ==> P[p] = {cp[23]:p}")
print("\nPossible Solutions:")

# consider solutions for other amounts and other
# probabilities that produce rational results
for c, f in ((cp[2], '   £2000'),
             (cp[3], '   £3000'),
             (cp[23], '>= £2000')):
  for n in range(-20, 21):
    # for the polynomial to solve
    r = (c - RF(n, 96)).zq_roots()[1]
    if r:
      r = r[0][0]
      if 0 <= r <= 1:
        print(f"  {f} (p = {n}/96) -> {r}")

from polynomial import Poly

from fractions import Fraction as RF

from gmpy2 import mpq as RF

# probability of winning the jackpot in the final round (= p)

win = Poly([0, 1])

# credit: Jim Randell

# generate the polynomials P[p] that determine the probabilities of

# all possible outcome amounts (in units of the base prize) with a

# maximum of <n> games; return: <amount>, <probability polynomial>

def outcomes(n, t=1, cp=1):

if n == 0:

# finished without a win

yield 0, cp

else:

# there is a 1/4 chance of reaching the final and leaving

q = cp * RF(1, 4)

# with a win

yield t, q * win

# or a loss

yield 0, q * (1 - win)

# there is a 3/4 chance of getting knocked out

# and then entering more gaames after

q = cp * RF(3, 4)

# a win and no rollover

yield from outcomes(n - 1, 1, q * win)

# or a loss and a rollover

yield from outcomes(n - 1, t + 1, q * (1 - win))

# accumulate probabilities of winning amounts over three games

cp, ps = dict(), Poly([0])

for v, p in outcomes(3):

ps += p

if v not in cp:

cp[v] = Poly([0])

cp[v] += p

# output the amounts and their probability polynomials

for v, p in sorted(cp.items()):

print(f" £{v}000 ==> P[p] = {p:p}")

print(f" Sum ==> P[p] = {ps}")

cp[23] = cp[2] + cp[3]

cp[123] = cp[1] + cp[23]

print(f">= £1000 ==> P[p] = {cp[123]:p}")

print(f">= £2000 ==> P[p] = {cp[23]:p}")

print("\nPossible Solutions:")

# consider solutions for other amounts and other

# probabilities that produce rational results

for c, f in ((cp[2], ' £2000'),

(cp[3], ' £3000'),

(cp[23], '>= £2000')):

for n in range(-20, 21):

# for the polynomial to solve

r = (c - RF(n, 96)).zq_roots()[1]

if r:

r = r[0][0]

if 0 <= r <= 1:

print(f" {f} (p = {n}/96) -> {r}")

Here is its output (edited):


   £0000 ==> P[p] = -(37/64)p + 1
   £1000 ==> P[p] = (21/64)p^2 + (1/4)p
   £2000 ==> P[p] = -(9/64)p^3 - (3/64)p^2 + (3/16)p
   £3000 ==> P[p] = (9/64)p^3 - (9/32)p^2 + (9/64)p
     Sum ==> P[p] = 1
>= £1000 ==> P[p] = (37/64)p
>= £2000 ==> P[p] = -(21/64)p^2 + (21/64)p

£0000 ==> P[p] = -(37/64)p + 1

£1000 ==> P[p] = (21/64)p^2 + (1/4)p

£2000 ==> P[p] = -(9/64)p^3 - (3/64)p^2 + (3/16)p

£3000 ==> P[p] = (9/64)p^3 - (9/32)p^2 + (9/64)p

Sum ==> P[p] = 1

>= £1000 ==> P[p] = (37/64)p

>= £2000 ==> P[p] = -(21/64)p^2 + (21/64)p

If we set \(P[p]\) to be \(q/96\) for a win of £2000 and simplify the appropriate polynomial, we obtain the cubic equation \[27p^3+9p^2-36p+2q=0\]

For the teaser \(q\) is 7, for which the polynomial has one negative real root and two complex ones, none of which can solve the teaser. In practice it seems likely that the probability \(p\) set by the author in the teaser is a rational fraction so it makes sense to consider other \(q\) values to see if any provide solutions of this form. When this is done two possible solutions emerge: \(p=1/3\) when \(q = 5\) and \(p=2/3\) when \(q=6\). The teaser hence has solution when the teaser’s probability of \(7/96\) is replaced by \(5/96\).

Jim Randell (on his S2T2 site) has suggested that “£2000” could be interpreted as “£2000 or more”. When we use the polynomial for this value and set it to \(7/96\) it simplifies to give a quadratic equation:\[9p^2-9p+2=0\]with solutions of \(p = 1/3\) and \(p = 2/3\) the first of which meets the teaser constraints.

Of these I prefer the former since it is nice to see a solution involving a cubic instead of the much more common quadratics.

John Z permalink

Here is a linear approach enumerating many of the 512 possible outcomes of the three games and calculating the probabilities of winning £2000 or of winning £2000 or £3000.
Forwarned that there is no solution for ‘7/96’, a solution is considered found if the denominator of the probability = 96.


from itertools import product, combinations
from fractions import Fraction
from math import gcd

# Original couples are A B C D, new couple in game 2 is E; in game 3 is F
cp = set('ABCD')

# win or lose
wl = 'WL'

# number of times one of 'ABCD' wins second game; no win in first game: £2K
pLWX = 0

# times one of 'ABCD' wins third game after a win in game 1, no win in 2: £2K
pWLW = 0

# times one of 'ABCD' wins third game after no wins in game 1 and 2: £3K
pLLW = 0

# consider possible first game outcomes: couple in fourth round; W/L
for c1, g1 in product(cp, wl):

  # replace couple who was in fourth round of game 1
  cp2 = cp - {c1} | {'E'}
  
  # consider possible second game outcomes: couple in fourth round; W/L
  for c2, g2 in product(cp2, wl):
    
    if g1 == "L" and g2 == "W":
      if c2 in cp: # c2 has won £2000
        pLWX += 1

    elif g1 == g2 == "W": # no possibility of £2K or £3K win, only £1K wins
      continue
      
    else: # go on to third game if first 2 are WL or LL
      
      # replace couple who was in fourth round of game 2     
      cp3 = cp2 - {c2} | {'F'}

      # consider possible third game outcomes: couple in fourth round;
      # W/L is indifferent c3 wins half, loses half
      for c3 in cp3:        

        if g1 == "W" and g2 == "L" and c3 in cp: # c3 wins £2000
          pWLW += 1
          
        elif g1 == g2 == "L" and c3 in cp: # c3 wins £3000
          pLLW += 1

print(f'Chances of £2000 win in games 2 & 3: {pLWX}, {pWLW}; £3000: {pLLW}')

# Second stage: try different values of p
# consider numerator and denominator of p
for n, d in combinations(range(1, 100), 2):

  # p must be < 1/2
  if n * 2 >= d or gcd(n, d) > 1:
    continue

  # prob of winning in the fourth round of a game
  p = Fraction(n, d)

  # calculate probabilty of one of the starting couples winning £2000 
  prob2 = (Fraction(pLWX * (1 - p) * p, 64) +
           Fraction(pWLW * p * (1 - p) * p, 512))

  # or £2000 or £3000
  prob23 = prob2 + Fraction(pLLW * p * (1 - p) * p, 512)

  # found solution with correct denominator?
  if prob2.denominator == 96 or prob23.denominator == 96:
    print(f'p = {p}, probability of winning £2000: {prob2}')
    print(f'         probability of winning £2K or £3K: {prob23}')

from itertools import product, combinations

from fractions import Fraction

from math import gcd

# Original couples are A B C D, new couple in game 2 is E; in game 3 is F

cp = set('ABCD')

# win or lose

wl = 'WL'

# number of times one of 'ABCD' wins second game; no win in first game: £2K

pLWX = 0

# times one of 'ABCD' wins third game after a win in game 1, no win in 2: £2K

pWLW = 0

# times one of 'ABCD' wins third game after no wins in game 1 and 2: £3K

pLLW = 0

# consider possible first game outcomes: couple in fourth round; W/L

for c1, g1 in product(cp, wl):

# replace couple who was in fourth round of game 1

cp2 = cp - {c1} | {'E'}

# consider possible second game outcomes: couple in fourth round; W/L

for c2, g2 in product(cp2, wl):

if g1 == "L" and g2 == "W":

if c2 in cp: # c2 has won £2000

pLWX += 1

elif g1 == g2 == "W": # no possibility of £2K or £3K win, only £1K wins

continue

else: # go on to third game if first 2 are WL or LL

# replace couple who was in fourth round of game 2

cp3 = cp2 - {c2} | {'F'}

# consider possible third game outcomes: couple in fourth round;

# W/L is indifferent c3 wins half, loses half

for c3 in cp3:

if g1 == "W" and g2 == "L" and c3 in cp: # c3 wins £2000

pWLW += 1

elif g1 == g2 == "L" and c3 in cp: # c3 wins £3000

pLLW += 1

print(f'Chances of £2000 win in games 2 & 3: {pLWX}, {pWLW}; £3000: {pLLW}')

# Second stage: try different values of p

# consider numerator and denominator of p

for n, d in combinations(range(1, 100), 2):

# p must be < 1/2

if n * 2 >= d or gcd(n, d) > 1:

continue

# prob of winning in the fourth round of a game

p = Fraction(n, d)

# calculate probabilty of one of the starting couples winning £2000

prob2 = (Fraction(pLWX * (1 - p) * p, 64) +

Fraction(pWLW * p * (1 - p) * p, 512))

# or £2000 or £3000

prob23 = prob2 + Fraction(pLLW * p * (1 - p) * p, 512)

# found solution with correct denominator?

if prob2.denominator == 96 or prob23.denominator == 96:

print(f'p = {p}, probability of winning £2000: {prob2}')

print(f' probability of winning £2K or £3K: {prob23}')

Sunday Times Teaser 3148 – Quiz Probabilities

by Edmund Marshall

Published Sunday January 22 2023 (link)

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