Sunday Times Teaser 3210 – Random Rabbit

by BRG on 29 March 2024

by Mark Valentine

Published Friday March 29 2024 (link)

Every year Easter Bunnies must pass a series of demanding tests, the most important being the double jump. Rabbits perform a single jump comprising two hops, with the total distance scoring. For both hops, Max knows he has an equal chance of jumping any distance between two limits. For the first hop, the limits are 80 and 100cm. For his weaker second hop, these limits decrease but keep the same proportion.

However, the instructors have increased the required standard from 152 to 163cm. Max is worried; he can still pass, but his chance is half what it was before.

What, as a percentage, is his chance of passing this year?

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BRG permalink


# last year/this year jump target distances 
tgts = (152, 163)

# find the root of the function <f> in range <a>..<b> using bisection
def root(f, a, b, tol=1e-4):
  while b - a > tol:
    c = (a + b) / 2
    if f(c) * f(a) < 0:
      b = c
    else:
      a = c
  return c

# create a discrete uniform probability density 
# function (PDF) over the interval min_x .. max_x  
def fill(min_x, max_x, nb):
  bs = (max_x - min_x) / nb
  return {min_x + n * bs + bs / 2: 1 / nb for n in range(nb)}

# integrate xp(x) * yp(y - x) for all y >= t 
def integrate(xb, yb, t):
  pb = 0
  for x, p in xb.items():
    for y, q in yb.items():
      if x + y >= t:
        pb += p * q
  return pb

# accelerate series convergence using Aitken's method
aitken = lambda a, b, c: (a * c - b * b) / (a + c - 2 * b)

# compute the probability for two hop success with
# second hop ratio <r> and jump target <t>
def compute(r, t):
  s = []
  # the number of probability bins
  for nb in (40, 80, 160):
    # first hop range PDF
    xb = fill(80, 100, nb)
    # second hop range PDF
    yb = fill(80 * r, 100 * r, nb)
    # integrate the combined PDF
    s.append(integrate(xb, yb, t))
  return(aitken(*s))  

# compute the ratio of the 2nd/1st year probabilities - 1 / 2
def fn(r):
  p1, p2 = compute(r, tgts[0]), compute(r, tgts[1])
  return p2 / p1 - 0.5

# find the second hop ratio <r> for a probability ratio of 0.5 
r = root(fn, 0.64, 1.0)
p, q = 100 * compute(r, tgts[0]), 100 * compute(r, tgts[1])
print(f'This year {q:.0f}% (last year {p:.0f}%, {r = :.1f}).')

# last year/this year jump target distances

tgts = (152, 163)

# find the root of the function <f> in range <a>..<b> using bisection

def root(f, a, b, tol=1e-4):

while b - a > tol:

c = (a + b) / 2

if f(c) * f(a) < 0:

b = c

else:

a = c

return c

# create a discrete uniform probability density

# function (PDF) over the interval min_x .. max_x

def fill(min_x, max_x, nb):

bs = (max_x - min_x) / nb

return {min_x + n * bs + bs / 2: 1 / nb for n in range(nb)}

# integrate xp(x) * yp(y - x) for all y >= t

def integrate(xb, yb, t):

pb = 0

for x, p in xb.items():

for y, q in yb.items():

if x + y >= t:

pb += p * q

return pb

# accelerate series convergence using Aitken's method

aitken = lambda a, b, c: (a * c - b * b) / (a + c - 2 * b)

# compute the probability for two hop success with

# second hop ratio <r> and jump target <t>

def compute(r, t):

s = []

# the number of probability bins

for nb in (40, 80, 160):

# first hop range PDF

xb = fill(80, 100, nb)

# second hop range PDF

yb = fill(80 * r, 100 * r, nb)

# integrate the combined PDF

s.append(integrate(xb, yb, t))

return(aitken(*s))

# compute the ratio of the 2nd/1st year probabilities - 1 / 2

def fn(r):

p1, p2 = compute(r, tgts[0]), compute(r, tgts[1])

return p2 / p1 - 0.5

# find the second hop ratio <r> for a probability ratio of 0.5

r = root(fn, 0.64, 1.0)

p, q = 100 * compute(r, tgts[0]), 100 * compute(r, tgts[1])

print(f'This year {q:.0f}% (last year {p:.0f}%, {r = :.1f}).')

Frits permalink

Wow!

All this (probability density function, probabability bins, Aitken) was already part of your tool kit?

Reply
- BRG permalink
  
  Hi Frits,
  
  Yes, from my youth but very rusty now – it took me a long time to resurrect it! And my reference books are in Worcester while I am in Cambridge at the moment.
  
  The code is fairly brittle since some of the parameters are finely tuned to balance accuracy and speed.
  
  I like Jim’s approach (which is effectively 2D integration) but it is hard to duplicate it without just copying his A(t,f) function.
  
  Brian
  
  Reply
Frits permalink

Hi Brian,

Can you elobarate on the probability bins and the “aitken” function?

If I choose for the number of probability bins (20, 50, 100) an incorrect answer is given.
If I use “aitken = lambda a, b, c: c” with this set up it does give a correct answer.

I also noticed that when I use a tolerance of 1e-5 your “aitken” function sometimes returns values outside 0…1.

Reply

Sunday Times Teaser 3210 – Random Rabbit

by Mark Valentine

Published Friday March 29 2024 (link)

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