Acca Arithmeticworked problems for accumulator maths

Chapter 4

Systems and full covers

A system bet splits one set of selections into several smaller accumulators. Counting how many bets that makes, and what each costs, is a combinations problem.

Problems 18–23 · 2 warm-up · 2 steady · 2 stretch

ABCD6 lines = 6 doubles
Every pair of four selections: AB, AC, AD, BC, BD, CD, the 6 doubles of a system 2/4.

Problem 18: Doubles from three selections

Warm-up

How many different doubles can be made from three selections A, B and C?

Selections
3
Bet size
2 legs
Show working for problem 18
  1. List them: AB, AC, BC.
  2. Check with the formula: 3! ÷ (2! × 1!) = 3.

Answer 3 doubles.

Problem 19: A system 2/4

Steady

A “system 2/4” makes every double from four selections. At ₦200 per double, how many bets is that and what is the total stake?

Selections
4
Stake per bet
₦200
Show working for problem 19
  1. Doubles from 4: 4! ÷ (2! × 2!) = 6.
  2. Total stake: 6 × ₦200 = ₦1,200.

Answer 6 bets, ₦1,200 in total.

Problem 20: Covering four selections

Steady

Four selections are covered with every double, every treble and the fourfold. How many bets is that? How many if the four singles are added?

Selections
4
Show working for problem 20
  1. Doubles 6 + trebles 4 + fourfold 1 = 11.
  2. Formula check: 24 − 4 − 1 = 11.
  3. With singles: 11 + 4 = 15.

Answer 11 bets, or 15 with singles. Many slips call these a Yankee and a Lucky 15; check the slip's own description.

Problem 21: A system 2/3 when one leg loses

Stretch

Three selections at 1.50 × 2.00 × 2.50 are played as a system 2/3: three doubles at ₦500 each, ₦1,500 in total. What comes back if all three win, and if the 2.50 selection loses?

Selections
1.50 × 2.00 × 2.50
Stake
3 × ₦500
Show working for problem 21
  1. Double 1.50 × 2.00 = 3.00 → ₦1,500.
  2. Double 1.50 × 2.50 = 3.75 → ₦1,875.
  3. Double 2.00 × 2.50 = 5.00 → ₦2,500.
  4. All win: ₦5,875 back on ₦1,500.
  5. If 2.50 loses, only the 1.50 × 2.00 double pays: ₦1,500.

Answer ₦5,875 if all win; ₦1,500 if the 2.50 selection loses, exactly the ₦1,500 staked.

Problem 22: Three selections, fully covered

Warm-up

Three selections are covered with every double and the treble. How many bets? And with the three singles added?

Selections
3
Show working for problem 22
  1. Doubles 3 + treble 1 = 4.
  2. With singles: 4 + 3 = 7.

Answer 4 bets, or 7 with singles. Many slips call these a Trixie and a Patent.

Problem 23: A system 3/5

Stretch

A system 3/5 makes every treble from five selections at ₦100 each. How many bets, what total stake, and how many selections must win for any return?

Selections
5
Bet size
3 legs
Stake per bet
₦100
Show working for problem 23
  1. Trebles from 5: 5! ÷ (3! × 2!) = 10.
  2. Total stake: 10 × ₦100 = ₦1,000.
  3. A treble needs all three of its legs, so at least three selections must win.

Answer 10 bets, ₦1,000; at least 3 winners for any return.