How would you do a cup trifecta?

[The Spornstar]

I'm not asking for horses here, but rather combinations.

Up in NSW you have the option of choosing your trifecta combinations and then nominating the amount of money you want to spend. So you might do a simple box trifecta of 5 horses - like they’ve done at this website: http://www.racetab.com.au/Flexi/Flexi_piggy_main.htm

So a 5 horse box would usually cost $60 but if you only wanted to bet $6 you can and you get 10% of the $1 dividend.

So given all this, what do you think the ideal combination would be for a Melb Cup trifecta (eg. 4 horses for first, 6 for second, 10 for third)?

 

[rumply]

field for 1st
prepost fav for 2nd
field for 3rd

 

[Pigskin]

I have always favoured a "pyramid" style trifecta myself in all types of races. I see no reason why the Melb Cup is any different.

The logic behind this, is that in any given there are only a certain amount of chances to win. However given a perfect run in the race, an exceptional ride, a good barrier, many more chances even though on face value have very little chance of winning, could run a minor place.

I will be having a trifecta on the Melb Cup. And at this stage whilst trying to keep my cost somewhat under control, I have plotted 2 horses I consider a good chance of winning, a further 7 horses whom I consider could possibly run second, and a further 5 horses that could possibly sneak into third place.

therefore in total i will be taking

for 1st - 2 horses
for 2nd - 9 horses
for 3rd - 14 horses

keep in mind that when a horse appears, it is carried down

total cost for a 50c unit = $96.00

Oink !

 

[harry_hawka]

I agree with Pig on this one.. a pyramid style Trifecta is much more sensible than a Box trifecta, especially in a field of 24 such as the Melbourne cup.

If anyone picks 3 horses and boxes them, and succesfully pulls out the trifecta in a field of 24, they are a genius.

In smaller races I usually try a bit of a roughie to win, then the favourite for second, and a few others for third. However in the Melbourne Cup I like to think the quality horses will win, and therefore try this system in a different order.

Keep in mind that I'm cutting back on costs majorly, and that this isn't the only bet for the race:

1st: Good horse, the one you most fancy, usually one of the favourites.

2nd: Horse at reasonably long odds, usually one that might lead, or has a good record at long distances. i.e. a Mr Prudent who will be sure to be plodding on towards the end. (Note: I will not be backing Prudent, see other thread)

3rd: Another few horses, usually a mixture of favourites and long shots. If you throw in 5 horses here it costs $5 for a $1 unit, likewise with 6 or 7 and so on.

I haven't tried trifectas much in the cup, but this is what I will be doing this year.

 

[Lensen]

They usually have a single ticket up here that costs $15, and you pick three horses that are boxed for a trifecta, boxed for a quinella, and $1 each way on them. They don't have that this year, but I did it manually.

It's probably not the greatest way of doing it, but I can never pick a winner anyway so why not.

 

[Slax]

Originally posted by rumply
field for 1st
prepost fav for 2nd
field for 3rd

Pretty expensive trifecta at $528 for the $1.

I'd take 4 for 1st, 6 2nd & 8 third cost $120 for $1

 

[SpecialBruce]

Try a mystery boxed trifecta. If the computer pulls it off for you, the computer is a genius.

 

[The Passenger]

i'm no gambling expert or anything, so i'd just go the box trifecta and hope for the best.....


how does the pyramid trifecta work in terms of working out how much you have to pay for a $1 return?

 

[Pigskin]

Black Thunder

the trifecta I proposed above was

for 1st = 2 horses
for 2nd = 9 horses
for 3rd = 14 horses

now if you put a horse in for 1st, it must also go in for second and third

if you put a horse in for 2nd, you must also put it in for 3rd

any horse that you put in, must be carried down to lower levels

nb. you don't have to carry your horses down to lower levels, but working out the cost becomes infinitely harder if you just have random numbers on each level but not on all lower levels

therefore you multiply out the numbers

2 (for first) x

(9 - 1 horse that is already taking up the first position) = 8 x

(14 - 2 because 2 horses are now taking up first and second position) = 12

it is physically impossible for a horse to come both first and second for example, so we then rule that horse out of lower levels

so we have finally 2 x 8 x 12

= $192 for $1
or
$96 for 50c


I hope this makes sense, its not easy to explain without a pen and paper in front of somebody

Pigskin

 

[Pigskin]

$21K on UniTab

Oink !