S2T2

Jim Randell 9:20 am on 27 June 2023 Permalink Reply
Tags: by: C Higgins ( 35 ), by: H Bradley ( 35 )   

• 

  Jim Randell 9:21 am on 27 June 2023 Permalink | Reply

  If the percentages bet are p1, p2, p3, and the remaining amounts after each bet it lost are r1, r2, r3, then:

  p1 + p2 + p3 = r3
  r1 = 100 − p1
  r2 = r1 − p2 . r1 / 100 = r1(100 − p2)/100 [and r2 is not a whole number]
  r3 = r2 − p3 . r2 / 100 = r2(100 − p3)/100

  So we can choose a value for r3, and split that into the three percentages (p1, p2, p3) and check the bets turn out as required.

  And this approach does find the answer in a reasonable time, but it is not hugely fast.

  This Python program runs in 347ms. (Internal runtime is 237ms).

  Run: [ @replit ]

  from enigma import (irange, decompose, printf)
  
  # generate possible (p1, p2, p3) percentage values
  def generate():
    # consider the number of pounds remaining after the 3 bets
    for r in irange(3, 98):
      # this is equal to p1 + p2 + p3
      yield from decompose(r, 3, increasing=0, sep=0, min_v=1)
  
  # consider values for p1, p2, p3
  for (p1, p2, p3) in generate():
    # bet 1, we lose p1 pounds
    r1 = 100 - p1
    # bet 2, we lose not a whole number of pounds
    b2 = r1 * p2  # = (bet 2) * 100
    if b2 % 100 == 0: continue
    r2 = 100 * r1 - b2  # = (remaining 2) * 100
    # bet 3, we are left with (p1 + p2 + p3) pounds
    b3 = r2 * p3  # = (bet 3) * 100^2
    r3 = 100 * r2 - b3  # = (remaining 3) * 100^2
    if r3 != (p1 + p2 + p3) * 10000: continue
    # output solution
    printf(
      "p1={p1}% p2={p2}% p3={p3}% -> b1={b1:.2f} b2={b2:.2f} b3={b3:.2f}; r1={r1:.2f} r2={r2:.2f} r3={r3:.2f}",
      b1=float(p1), b2=(b2 * 0.01), b3=(b3 * 0.0001), r1=float(r1), r2=(r2 * 0.01), r3=(r3 * 0.0001),
    )
  

  Solution: £2.25 was bet (and lost) on the third race.

  The scenario is:

  start = £100.00
  bet 1 = 25% of £100.00 = £25.00
  bet 1 lost, remaining = £75.00
  bet 2 = 25% of £75.00 = £18.75
  bet 2 lost, remaining = £56.25
  bet 3 = 4% of £56.25 = £2.25
  bet 3 lost, remaining = £54.00
  sum of percentage values = 25 + 25 + 4 = 54.

  ---

  For a faster program we can do a bit of analysis:

  We can write (r1, r2, r3) in terms of (p1, p2, p3) to get:

  r1 = 100 − p1
  r2 = (100 − p1)(100 − p2)/100
  r3 = (100 − p1)(100 − p2)(100 − p3)/10000
  r3 = p1 + p2 + p3
  ⇒
  10000(p1 + p2 + p3) = (100 − p1)(100 − p2)(100 − p3)

  Note that 5^4 divides the LHS, so it must divide the terms of the RHS, so one of them must be a multiple of 25 (i.e. 25, 50, 75), and another must also be a multiple of 5.

  And this gives us a faster way to generate possible (p1, p2, p3) values.

  This Python 3 program uses a different [[ generate() ]] function, and runs in 61ms. (Internal runtime is 89µs).

  Run: [ @replit ]

  from enigma import (irange, div, subsets, printf)
  
  # generate possible percentage values
  def generate():
    # one of the values is a multiple of 25
    for a in (25, 50, 75):
      # an another is a multiple of 5
      for b in irange(5, 95 - a, step=5):
        c_ = div(10000 * (a + b + 100), 10000 + (100 - a) * (100 - b))
        if c_ is None or not (c_ < 100): continue
        # return the values (in some order)
        yield from subsets((a, b, 100 - c_), size=len, select='mP')
  
  # consider values for p1, p2, p3
  for (p1, p2, p3) in generate():
    # bet 1, we lose p1 pounds
    r1 = 100 - p1
    # bet 2, we lose not a whole number of pounds
    b2 = r1 * p2  # = (bet 2) * 100
    if b2 % 100 == 0: continue
    r2 = 100 * r1 - b2  # = (remaining 2) * 100
    # bet 3, we are left with (p1 + p2 + p3) pounds
    b3 = r2 * p3  # = (bet 3) * 100^2
    r3 = 100 * r2 - b3  # = (remaining 3) * 100^2
    if r3 != (p1 + p2 + p3) * 10000: continue
    # output solution
    printf(
      "p1={p1}% p2={p2}% p3={p3}% -> b1={b1:.2f} b2={b2:.2f} b3={b3:.2f}; r1={r1:.2f} r2={r2:.2f} r3={r3:.2f}",
      b1=float(p1), b2=(b2 * 0.01), b3=(b3 * 0.0001), r1=float(r1), r2=(r2 * 0.01), r3=(r3 * 0.0001),
    )
  

  In fact only one set of values (25, 25, 4) satisfies the analysis. And only with the 4% value for the final bet is the condition that the second amount bet is not a whole number of pounds satisfied.

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• 

  Frits 7:08 pm on 27 June 2023 Permalink | Reply

  Manual solution:

  Let the percentages NOT bet at each stage be p1, p2 and p3 and
  let the remainders after each stage be r1, r2 and r3 with the
  initial amount r0. Then r1 = r0.p1 / 100, r2 = r1.p2 / 100 and
  r3 = r2.p3 / 100 giving:
  
  p1 = 100.r1/r0 = 10000.r2/p2.r0 = 1000000.r3/p2.p3.r0
  
     p1.p2.p3 = 10^6.r3/r0
  
  Also:
  
    100.r3/r0 = (100 - p1) + (100 - p2) + (100 - p3)
  
  Eliminating r3/r0 now allows p3 to be derived from p1 and p2:
  
    p3 = (300 - p1 - p2).10^4 / (10^4 + p1.p2)
  
    p1 * p2 * p3 - 10^4 * (300 - p1 - p2 - p3) = 0
  
    p1 * p2 * p3 = (300 - p1 - p2 - p3) * 16 * 5^4
  
  if p1, p2 and p3 are all multiples of 5 then
  (300 - p1 - p2 - p3) is also a multiple of 5 so
  p1 * p2 * p3 must be a multiple of 5^5
  
  so we can conclude at least 2 out (p1, p2, p3) must be equal to 75,
  remaining percentage p* must be a multiple of 16
  and (300 - p1 - p2 - p3) must be a multiple of 9 (3 * 3)
      (150 - p*) must be a multiple of 9
      so p* mod 9 = 6
  the only multiple of 16 and mod 9 = 6 out of 64, 80 and 96 is 96
  so (p1, p2, p3) = (75, 75, 96) in any order
  
  p1 can't be 96 as 0.75 * 96 is a whole number
  p2 can't be 96 as 75 * 0.96 is the same whole number
  so (p1, p2, p3) = (75, 75, 96) in this order
  
  on the third race was lost: (3/4)^2 * 100 * (100 - 96) / 100 = 2.25 pounds
  

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