Quick calc of amountto bet without affecting payout

[podonne]

This is mentioned alot, don't bet enough to upset the pools, but does anyone have a quick method of determining the maximum amount you can bet without disrupting the pool payouts? Theoretically any bet of any size changes the makeup of the pool somewhat, and since the payouts aren't rounded to thier odds ("2 to 1" might pay out 4.05 or 3.95), is it the amount to change the pool payout by a penny?

Cheers,
Phil

[Tom Barrister]

It would be the amount that lowered the payoff to just above the breakage point (the next lowest dime on a dollar or nickel at tracks which offer nickel breakage).

It would probably be possible to work out a formula for it, and it wouldn't be hard to program it into a small application, but I don't see the point, since the ratios would change after the race went off and the late and simulcast monies came in.

[Cratos]

This is mentioned alot, don't bet enough to upset the pools, but does anyone have a quick method of determining the maximum amount you can bet without disrupting the pool payouts? Theoretically any bet of any size changes the makeup of the pool somewhat, and since the payouts aren't rounded to thier odds ("2 to 1" might pay out 4.05 or 3.95), is it the amount to change the pool payout by a penny?

Cheers,
Phil

This is a very simple calculation, but it depends on the size of the pool, the amount wagered, and the take percent.

For example a win pool has $10,000 in it with $3,300 on Horse A; which means Horse A is 2-1 on the tote because of $3,300/$10,000 = .33 or 1/ (2+1) which is .33.

Remember there is a take percent and if we assume 15%, the return to the bettors would be $8,500. This doesn’t affect the calculation because 85% of $3,300 is $2,805 or 33% of $8,500.

If a bettor wages another $500 on Horse A the odds would drop from 2-1 (payoff of $6.00) to 1.86-1 (payoff of $5.60 after breakage)

However the calculation in real life become complicated because bettors are wagering on different horses in the race and the constancy implied in the above calculation doesn’t exists.

[equicom]

You need to start with the amount in the pool.

Say it is 100,000

There are eight runners:

A = 6.40
B = 21.40
C = 14.20
D = 9.40
E = 34.20
F = 57.00
G = 19.00
H = 8.40

If you add all the numbers together you'll get 170.00, and because we are working with a dividend base of $2, it means that you have to divide the number by 2 which gives you 85% so this indicates that the track is taking out 15% as the "take".

Now that we know the take and the "true" amount available in the pool for distribution, we can calculate how much has been invested on each runner.

To begin with, I would suggest simplifying matters by dividing all the payouts by 2. This would give us:

A = 3.20
B = 10.70
C = 7.10
D = 4.70
E = 17.10
F = 28.50
G = 9.50
H = 4.20

Now all you have to do is work out the amount invested on each runner, using this simple formula:

(Pool - Take) / price = investmet

So runner A would be

(100k - 15k) / 3.20 = 26,562

Of course it is approximate, not exact.

If you do this for all runners you will find that the amounts invested are as follows:

A = 26,562
B = 7,944
C = 11,972
D = 18,085
E = 4,971
F = 2,982
G = 8,947
H = 20,238

With these figures we end up about $1700 or so out from the true amount, but it is close enough (getting exact is not worth the trouble).

From here it should be fairly obvious.

Let's see what happens if we put $1000 on Runner F.

Pool now equals 101,000
Take is 15,150
Amount available for distribution is 85,850

A = 27,562
B = 7,944
etc (the others are all still the same)

So now, the amount invested on Runner A has increased by 1000.

If we invert our formula we get:

(pool - take) / investment = price

So:

A: 85,850 / 27,562 = 3.10 (decrease of 0.10)
B: 85,850 / 7,944 = 10.80 (increase of 0.10)
C: 85,850 / 11,972 = 7.10 (no increase)
D: 85,850 / 18,085 = 4.70 (no increase)
E: 85,850 / 4,971 = 17.20 (increase of 0.10)
F: 85,850 / 2,982 = 28.80 (increase of 0.10)
G: 85,850 / 8,947 = 9.60 (increase of 0.10)
H: 85,850 / 20,238 = 4.20 (no increase)

Why doesn't the price of runner C, runner D, or runner H increase while most of the others do? Because the totalizator is a greedy thief and won't pay true odds. They always round the number down to the nearest 10 cents and keep the excess (which is partly why we don't mess about trying to get exact numbers, because it will still be wrong!).

But what you can see from all this is that throwing an extra $1000 into the pool, betting it on the favorite, has changed the prices by approximately 10c on nearly every runner in a $100,000 pool.

If, on the other hand, you were to invest that $1000 on the outsider (runner E), you would cause the price on that horse to change by more than $7.

This is what the payouts would be in that situation:

A = 3.20
B = 10.80
C = 7.10
D = 4.70
E = 17.20
F = 21.50 (decrease of $7.30!)
G = 9.60
H = 4.20

So once again most prices stay the same or change by 10 cents but the odds on runner F drop by a lot. However, it is not as simple as this. You have to time that $1000 bet much more carefully if you are investing on runner F than if you are investing on runner A.

The reason is that people are more likely to notice the odds of runner F dropping, and they will notice it especially on a $2 board (because the drop is more pronounced... $14.60!).

That could cause some people to think that there is smart money going onto that runner and so they will put a little bit on "just in case it beats my horse". If 500 people all decide to put on $2 because you put on $1000, then that means the total investment on the runner has now increased by a total of $2000 and the odds will fall even more sharply.

This starts a viscious chain-reaction where the odds fall, inspiring more investment, making the odds fall again, and so on, until the mug market is expended. Then you get the final price. So in this case it is much better to wait until the last moment to bet in order to avoid such a situation.

Conversely, you can also use mug madness as a strategy if you have a *lot* of money to play with. What you would do there is invest a large sum on an outside chance that looked on paper like it might be a "set-up" horse. So you trickle $1000 onto it and make the odds drop. Then you wait about 4 or 5 minutes and do it again. And then again. People should start to notice and it helps if you go around giving the "tip" to all and sundry. Start a rumor "I overheard this fella that works for Mullins say that they were going to clean up on this one" or whatever.

Result: the horse that should have started at $2.80 goes out at $4.00. Then it loses and teaches you a big lesson about being greedy enough to try this sort of crap.

So in summary, all you need to do is work out the formula as:

(Pool - Take) / price = investment

and

(Pool - Take) / investment = price

[equicom]

The problem with taking the time to give a detailed explanation is someone else comes along and answers the same question first! Sorry about that.

:bang:

[asH]

The problem with taking the time to give a detailed explanation is someone else comes along and answers the same question first! Sorry about that.

:bang:

I thank you for a well explained answer
asH

[podonne]

Wow, great answer. So at what point are you affecting the odds "too much" Is the dime change "too much" or do you measure the distance from your value odds to the actual odds and try not to cross it, or go 50% of the distance?

[rrbauer]

Two principles at play here:

1. The more you bet, the more you win!

2. Since the pool total and individual bet amounts are a moving target and can change significantly between the last amounts displayed to the public (that someone could use to make calculations) and the final totals for the race that determine the payout prices, whatever you calculate is subject to change and there's no opportunity to react accordingly.