Acca Arithmeticworked problems for accumulator maths
Chapters

Chapter 3

Void legs

These problems use one common convention: when a slip's rules declare a leg void, it counts at odds of 1.00 and stops multiplying anything. A void is not a loss. A slip's own rules decide how its voids are settled.

Problems 13–17 · 2 warm-up · 2 steady · 1 stretch

1.50leg 1×2.00leg 2void → 1.00×1.80leg 3×1.60leg 4
Leg 2 is void and counts at 1.00: 1.50 × 1.00 × 1.80 × 1.60 = 4.32. It stops multiplying; it does not end the bet.

Problem 13: One void leg

Warm-up

A four-leg accumulator at 1.50 × 2.00 × 1.80 × 1.60, stake ₦1,000. Leg 2 is declared void under the slip's rules and counts at 1.00. The other legs win. What is the return?

Legs
1.50 × 2.00 × 1.80 × 1.60
Void
leg 2 → 1.00
Stake
₦1,000
Show working for problem 13
  1. Without the void: 1.50 × 2.00 × 1.80 × 1.60 = 8.640.
  2. With leg 2 at 1.00: 1.50 × 1.00 × 1.80 × 1.60 = 4.320.
  3. Return: ₦1,000 × 4.320 = ₦4,320.

Answer ₦4,320 (it would have been ₦8,640 with all four winning).

Problem 14: Two voids make a double

Steady

In the same accumulator, legs 2 and 3 are void. Legs 1 and 4 win. What bet is left, and what does it return on ₦1,000?

Left
1.50 and 1.60
Show working for problem 14
  1. 1.50 × 1.00 × 1.00 × 1.60 = 2.40.
  2. Return: ₦1,000 × 2.40 = ₦2,400.

Answer In effect a double at 2.40, returning ₦2,400.

Problem 15: Every leg void

Warm-up

All four legs are void. What happens to the ₦1,000 stake?

Legs
4 × 1.00
Show working for problem 15
  1. 1.00 × 1.00 × 1.00 × 1.00 = 1.00.
  2. Return: ₦1,000 × 1.00 = ₦1,000.

Answer The stake, ₦1,000, comes back where the rules treat every leg this way.

Problem 16: A void and a loss

Steady

Leg 2 is void and leg 4 loses. The other legs win. What comes back?

Leg 2
void
Leg 4
lost
Show working for problem 16
  1. The void leg counts at 1.00 and does not end the bet.
  2. The lost leg multiplies the bet by zero.

Answer Nothing: a void cannot rescue a lost leg.

Problem 17: When the long leg is void

Stretch

A treble at 1.50 × 3.20 × 1.80, stake ₦1,000. The 3.20 leg is void and the others win. How much smaller is the return than if all three had won?

Legs
1.50 × 3.20 × 1.80
Void
the 3.20 leg
Show working for problem 17
  1. All won: 1.50 × 3.20 × 1.80 = 8.64; return ₦8,640.
  2. With the void: 1.50 × 1.00 × 1.80 = 2.70; return ₦2,700.
  3. Difference: ₦8,640 − ₦2,700 = ₦5,940.

Answer ₦5,940 smaller. A void removes that leg's whole contribution.