Calculating a multiple P6 ticket

[rmania]

Never was one to play multiple P6 tickets but seeing some posted in the Selections forum, along with the posted cost, has me wondering how these tickets are calculated.
Is there a standard forumula? And, if so, can someone explain it.

Thanks...

[formula_2002]

Never was one to play multiple P6 tickets but seeing some posted in the Selections forum, along with the posted cost, has me wondering how these tickets are calculated.
Is there a standard forumula? And, if so, can someone explain it.

Thanks...

FROM MY POST AT

http://www.paceadvantage.com/forum/member.php?u=573

THE TICKET HAD A PROBABILITY OF 1%.

IT COST $23,520 PER PERCENT PROBABILITY, WHEREAS THE TWINSPIRE WINNING TICKET AT SARATOGA HAS A COST ONLY $1519... A MUCH WISER BET.

I WOULD THINK A WELL STRUCTURED BET WOULD HAVE A LOW $ COST PER PROBABILITY

[rmania]

FROM MY POST AT

http://www.paceadvantage.com/forum/member.php?u=573
Not looking for win probability...
Asking for formula used to determine cost of ticket.

[raybo]

Although I've never played the P6, I do play superfectas with multiple horses on each position line and I believe the calculations would be very similar. Here's my take on P6 calcs:

Let's say you have singled 1 horse in each race, that would be:

1x1x1x1x1x1=1 combination

Now with 2 horse in 1 of the races:

1x2x1x1x1x1=2 combinations

So, with multiple horses in several races it would seem that the same process should work:

1x2x3x2x4x5=240 combinations

Someone correct me if I'm wrong.

[Shelby]

Although I've never played the P6, I do play superfectas with multiple horses on each position line and I believe the calculations would be very similar. Here's my take on P6 calcs:

Let's say you have singled 1 horse in each race, that would be:

1x1x1x1x1x1=1 combination

Now with 2 horse in 1 of the races:

1x2x1x1x1x1=2 combinations

So, with multiple horses in several races it would seem that the same process should work:

1x2x3x2x4x5=240 combinations

Someone correct me if I'm wrong.

You are correct :)

[goforgin]

I find the dollar amount that I'm comfortable with for that day. Like a $36, $48, $72 or $96 ticket. For example:
1x1x2x2x3x4 = 48 x $2 = $96 (13 selections)
1x1x1x3x4x4 = 48 x $2 = $96 (14 selections)
Note the first one has 13 horses on the ticket. The second one has 14 horses on the ticket, but the negative is 3 single selections.
1x1x2x2x2x3 = 24 x $2 = $48 (11 horse ticket)
1x1x1x2x3x4 = 24 x $2 = $48 (12 horse ticket)
There's probably an easier way to do it with a spreasheet formula. But, I did it manually and then have the different tickets and costs written down on a piece of paper.

[formula_2002]

Not looking for win probability...
Asking for formula used to determine cost of ticket.

its really part of the cost..
cost per percent probability

[Greyfox]

Although I've never played the P6, I do play superfectas with multiple horses on each position line and I believe the calculations would be very similar. Here's my take on P6 calcs:

Let's say you have singled 1 horse in each race, that would be:

1x1x1x1x1x1=1 combination

Now with 2 horse in 1 of the races:

1x2x1x1x1x1=2 combinations

So, with multiple horses in several races it would seem that the same process should work:

1x2x3x2x4x5=240 combinations

Someone correct me if I'm wrong.

That would be correct if it were a $1 ticket.
The last time I played a So Cal Pick 6, the minimum ticket was $2.
Multiply your combinations by 2 = $ 480

[Mike at A+]

I wonder how the handle would be affected if tracks were to go to a ten cent minimum on pick-6's instead of the current $2. Would they gain enough bettors to offset the losses incurred by the usual $2 multi-ticket bettors switching to the new minimum? The $2 minimum is what has always kept me from seriously playing the pick-6. Those combo tickets can be very expensive.

[speed]

I wonder how the handle would be affected if tracks were to go to a ten cent minimum on pick-6's instead of the current $2. Would they gain enough bettors to offset the losses incurred by the usual $2 multi-ticket bettors switching to the new minimum? The $2 minimum is what has always kept me from seriously playing the pick-6. Those combo tickets can be very expensive.
Mike the benefit from the $2 minimum is the carryover. With much lower minimum's you would never get the monster carryovers that attract the massive betting pools.

[PhantomOnTour]

Multiply the number of horses you have in each leg and then multiply that total by 2 and you have your ticket cost. Period.

1x1x2x2x3x3=36 combos
36 combos x2 (that's the minimum bet...$2) = $72

[positive4th]

Mike the benefit from the $2 minimum is the carryover. With much lower minimum's you would never get the monster carryovers that attract the massive betting pools.


This is absolutely true, AND absolutely the reason that $2 needs to be the minimum base unit for the P6. Well said, poster.

[rrpic6]

I wonder how the handle would be affected if tracks were to go to a ten cent minimum on pick-6's instead of the current $2. Would they gain enough bettors to offset the losses incurred by the usual $2 multi-ticket bettors switching to the new minimum? The $2 minimum is what has always kept me from seriously playing the pick-6. Those combo tickets can be very expensive.

Beulah Park has the Fortune6 which is a 25 cent play. Its a nice gimmick, since the pool carries over when there is more than one winning pick 6 ticket. Closing day at Beulah this year made for a must payout of over of a pool that ended up with over a million bucks in it.

RR

[raybo]

That would be correct if it were a $1 ticket.
The last time I played a So Cal Pick 6, the minimum ticket was $2.
Multiply your combinations by 2 = $ 480

I'm aware of that, that's why I showed the "combinations" not the cost.

As I said, I'm not a "Picks" player and did not know for sure if $2 was the minimum cost of each combination at all tracks.

[xfile]

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