SA Footy

I was wondering if any of you fellas know the actual mathematical formula for calculating the number of possible combinations in a trifecta bet (or quartet for that matter) ?

Whilst I know you can look at a chart on the wall in most TAB shops to get a costing on trifectas there is a specific mathematical formula that is used. I used to know it and had it written down but lost it a few years ago .

f X (f-1) X (f-2)= x

Punk Rooster wrote:f X (f-1) X (f-2)= x

Cheers Punk.

This equation assumes you take the same number of runners all-ways

What if you did something like this in a 9-horse field when you were anchoring it on a certain number of horses for 1st & 2nd and the remainder to fill the 3rd spot:-

1st - 1,2,3,4
2nd - 1,2,3,4,5
3rd - 5,6,7,8,9

How would you calculate that?

westozfalcon wrote:

Punk Rooster wrote:f X (f-1) X (f-2)= x

Cheers Punk.

This equation assumes you take the same number of runners all-ways

f=10. 10 X 9 X 8= 720.
That covers all possible outcomes (ie a boxing the field achieves this)

westozfalcon wrote:What if you did something like this in a 9-horse field when you were anchoring it on a certain number of horses for 1st & 2nd and the remainder to fill the 3rd spot:-

1st - 1,2,3,4
2nd - 1,2,3,4,5
3rd - 5,6,7,8,9

How would you calculate that?

4 X 4 X 5= 80 (I could be open to correction on this one though)

westozfalcon wrote:

Punk Rooster wrote:f X (f-1) X (f-2)= x

Cheers Punk.

This equation assumes you take the same number of runners all-ways

What if you did something like this in a 9-horse field when you were anchoring it on a certain number of horses for 1st & 2nd and the remainder to fill the 3rd spot:-

1st - 1,2,3,4
2nd - 1,2,3,4,5
3rd - 5,6,7,8,9

How would you calculate that?

Punk is close enough...

I would work it out 4 x 4 x 5 = 80 LESS the combinations of 4 x 1 x 1 (1 combination can't be doubled up 5 - when you have it for 2nd & 3rd = 4
so makes it 80 less 4 = 76 units.

scaffidi can't drive wrote:

westozfalcon wrote:

Punk Rooster wrote:f X (f-1) X (f-2)= x

Cheers Punk.

This equation assumes you take the same number of runners all-ways

What if you did something like this in a 9-horse field when you were anchoring it on a certain number of horses for 1st & 2nd and the remainder to fill the 3rd spot:-

1st - 1,2,3,4
2nd - 1,2,3,4,5
3rd - 5,6,7,8,9

How would you calculate that?

Punk is close enough...

I would work it out 4 x 4 x 5 = 80 LESS the combinations of 4 x 1 x 1 (1 combination can't be doubled up 5 - when you have it for 2nd & 3rd = 4
so makes it 80 less 4 = 76 units.

Yes I can see how scd has calculated that.

Like Punk's example suggests most punters would keep it simple and just box the same runners all-ways.

Just intrigued me as to what the actual mathematical algebraic formula is. They used to print it on trifecta leaflets here in WA but they stopped doing it a few years back. I'm sure some university maths professor would be able to quote the formula in his sleep .

scaffidi can't drive wrote:

westozfalcon wrote:

Punk Rooster wrote:f X (f-1) X (f-2)= x

Cheers Punk.

This equation assumes you take the same number of runners all-ways

What if you did something like this in a 9-horse field when you were anchoring it on a certain number of horses for 1st & 2nd and the remainder to fill the 3rd spot:-

1st - 1,2,3,4
2nd - 1,2,3,4,5
3rd - 5,6,7,8,9

How would you calculate that?

Punk is close enough...

I would work it out 4 x 4 x 5 = 80 LESS the combinations of 4 x 1 x 1 (1 combination can't be doubled up 5 - when you have it for 2nd & 3rd = 4
so makes it 80 less 4 = 76 units.

Yes, that's why I put up the disclaimer- I could work out the first "double up", but not the 2nd

just put it in the machine and have a crack!!!!!!

another grub wrote:just put it in the machine and have a crack!!!!!!

exactly (if your not sure ask the tab/tote to see how much it costs first then they can cancel straight away if too much) mind you its only ever too much when you lose