Sunday Times Teaser 3086 – Four-Sided Dice Game

by BRG on November 13, 2021

by Howard Williams

Published Sunday November 14 2021 (link)

I have two identical four-sided dice (regular tetrahedra). Each die has the numbers 1 to 4 on the faces. Nia and Rhys play a game in which each of them takes turns throwing the two dice, scoring the sum of the two hidden numbers. After each has thrown both dice three times, if only one of them scores an agreed total or over, then he or she is the winner. Otherwise the game is drawn.

After Nia has thrown once, and Rhys twice, they both have a chance of winning. If Rhys had scored less on his first throws and Nia had scored double her first throw total, then Nia would have had a 35 times greater chance of winning.

What are Nia’s and Rhys’ scores and the agreed winning total?

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Brian Gladman permalink


from itertools import combinations, product

# consider the possible agreed values after three throws
for agv in range(6, 25):

  # compile Nia's totals and chances of winning after one throw
  n2p = {x:0 for x in range(2, 9)}
  for x in n2p.keys():
    for p in product((1, 2, 3, 4),  repeat=4):
      if sum(p) + x >= agv:
        n2p[x] += 1
    
  # find Nia's totals for which one is double the other 
  for nia, p_n1 in n2p.items():
    if 0 < p_n1 < 256 and 2 * nia in n2p.keys():
      
      # Nia's probability of winning in the second scenario
      p_n2 = n2p[2 * nia]
  
      # compile Rhys's totals and chances of winning after two throws  
      r2p = {x:0 for x in range(4, 17)}
      for x in r2p.keys():
        for p in product((1, 2, 3, 4), repeat=2):
          if sum(p) + x >= agv:
            r2p[x] += 1
      
      # consider Rhys's second (lower) and first (higher) scenario totals and 
      # chances
      for (r2, p_r2), (r1, p_r1) in combinations(sorted(r2p.items()), 2):

        # the first and second scenarios give Nia a 35 times higher
        # chance of winning in the second compared to the first
        if 0 < p_r1 < 16 and 35 * p_n1 * (16 - p_r1) == p_n2 * (16 - p_r2):
          print(f"Agreed score {agv}; "
                f"Nia [{nia}:({p_n1}/256), {2 * nia}:({p_n2}/256)]; "
                f"Rhys [{r1}:({p_r1}/16), {r2}:({p_r2}/16)]")

from itertools import combinations, product

# consider the possible agreed values after three throws

for agv in range(6, 25):

# compile Nia's totals and chances of winning after one throw

n2p = {x:0 for x in range(2, 9)}

for x in n2p.keys():

for p in product((1, 2, 3, 4), repeat=4):

if sum(p) + x >= agv:

n2p[x] += 1

# find Nia's totals for which one is double the other

for nia, p_n1 in n2p.items():

if 0 < p_n1 < 256 and 2 * nia in n2p.keys():

# Nia's probability of winning in the second scenario

p_n2 = n2p[2 * nia]

# compile Rhys's totals and chances of winning after two throws

r2p = {x:0 for x in range(4, 17)}

for x in r2p.keys():

for p in product((1, 2, 3, 4), repeat=2):

if sum(p) + x >= agv:

r2p[x] += 1

# consider Rhys's second (lower) and first (higher) scenario totals and

# chances

for (r2, p_r2), (r1, p_r1) in combinations(sorted(r2p.items()), 2):

# the first and second scenarios give Nia a 35 times higher

# chance of winning in the second compared to the first

if 0 < p_r1 < 16 and 35 * p_n1 * (16 - p_r1) == p_n2 * (16 - p_r2):

print(f"Agreed score {agv}; "

f"Nia [{nia}:({p_n1}/256), {2 * nia}:({p_n2}/256)]; "

f"Rhys [{r1}:({p_r1}/16), {r2}:({p_r2}/16)]")

John Z permalink

What a brilliant solution, Brian!

As an exercise I rewrote your code using lists instead of dictionaries. Dictionaries are definitely more intuitive in this case.


from itertools import combinations, product

# consider the possible agreed values after three throws
for agv in range(6, 25):

  # compile Nia's totals and chances of winning after one throw
  n2p = [0] * 9
  for x in range(2, 9):
    for p in product((1, 2, 3, 4),  repeat=4):
      if sum(p) + x >= agv:
        n2p[x] += 1
    
  # find Nia's totals for which one is double the other 
  for nia, p_n1 in enumerate(n2p[2:5], 2):
    if 0 < p_n1 < 256:
      
      # Nia's probability of winning in the second scenario
      p_n2 = n2p[2 * nia]
  
      # compile Rhys's totals and chances of winning after two throws  
      r2p = [0] * 17
      for x in range(4, 17):
        for p in product((1, 2, 3, 4), repeat=2):
          if sum(p) + x >= agv:
            r2p[x] += 1
      
      # consider Rhys's second (lower) and first (higher) scenario totals and 
      # chances
      for (r2, p_r2), (r1, p_r1) in combinations(zip(range(4, 17), r2p[4:]), 2):

        # the first and second scenarios give Nia a 35 times higher
        # chance of winning in the second compared to the first
        if 0 < p_r1 < 16 and 35 * p_n1 * (16 - p_r1) == p_n2 * (16 - p_r2):
          print(f"Agreed score {agv}; "
                f"Nia [{nia}:({p_n1}/256), {2 * nia}:({p_n2}/256)]; "
                f"Rhys [{r1}:({p_r1}/16), {r2}:({p_r2}/16)]")

from itertools import combinations, product

# consider the possible agreed values after three throws

for agv in range(6, 25):

# compile Nia's totals and chances of winning after one throw

n2p = [0] * 9

for x in range(2, 9):

for p in product((1, 2, 3, 4), repeat=4):

if sum(p) + x >= agv:

n2p[x] += 1

# find Nia's totals for which one is double the other

for nia, p_n1 in enumerate(n2p[2:5], 2):

if 0 < p_n1 < 256:

# Nia's probability of winning in the second scenario

p_n2 = n2p[2 * nia]

# compile Rhys's totals and chances of winning after two throws

r2p = [0] * 17

for x in range(4, 17):

for p in product((1, 2, 3, 4), repeat=2):

if sum(p) + x >= agv:

r2p[x] += 1

# consider Rhys's second (lower) and first (higher) scenario totals and

# chances

for (r2, p_r2), (r1, p_r1) in combinations(zip(range(4, 17), r2p[4:]), 2):

# the first and second scenarios give Nia a 35 times higher

# chance of winning in the second compared to the first

if 0 < p_r1 < 16 and 35 * p_n1 * (16 - p_r1) == p_n2 * (16 - p_r2):

print(f"Agreed score {agv}; "

f"Nia [{nia}:({p_n1}/256), {2 * nia}:({p_n2}/256)]; "

f"Rhys [{r1}:({p_r1}/16), {r2}:({p_r2}/16)]")

John Z permalink

Once again a chunk of code has dissappeared from my post. I’ll send it by email.

Reply
- Brian Gladman permalink
  
  Hi John,
  
  Good to see you posting again! I have added your emailed version.
  
  I’m sorry that you are having issues. It is a site issue and not a problem at your end.
  
  I also looked at using lists in place of dictionaries but decided to stay with the latter.
  
  Brian
  
  Reply

Sunday Times Teaser 3086 – Four-Sided Dice Game

by Howard Williams

Published Sunday November 14 2021 (link)

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