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Let’s start with a simple case.
Select two runners and back them in singles and a double.
They both win at 2/1 and 3/1.
Therefore your total return to a £1 stake is (4×5) – 1 = £19Now let’s up the pace by selecting 3 horses and covering them in singles, doubles and a treble.
This time their odds are 3/1, 4/1 and 5/1
Therefore your total return to a £1 stake is (5x6x7) -1 = £209
Had you not had the singles then the total would have been (5x6x7) – (1 + 4+ 5 + 6) =£194The final lesson before the real thing.
You have a £1 yankee and get winners at 2/1, 3/1, 4/1 and 9/1
Your total return to a £1 stake is (4 x 5 x 6 x 11) – (1 + 3 + 4 + 5 + 10) = £1297Now the Goliath.
You get 8 winners at Evens, 6/4, 2/1, 3/1, 4/1, 5/1, 6/1 and 7/1
Therefore your total return is ( 3 x 3.5 x 4 x 5 x 6 x 7 x 8 x 9 ) – ( 1 + 2 + 2.5 + 3 + 4 + 5 + 6 + 7 + 8 )
= £635,001.50 eh voilaThanks, simple enough but by no means obvious and nothing to do with the binomials used to calculate combinations!
So, from your examples I reckon the general formula for calculating multiples including singles is:
[(p1+2) x (p2+2) x…x ( pn+2)] -1
and for multiples excluding singles:
[(p1+2) x (p2+2) x…x (pn+2)] – [1 + (p1+1) + (p2+1) +…+ (pn+1)]
where p = odds and n = number of selections
e.g. Lucky 15 [(p1+2) x (p2+2) x (p3+2) x (p4+2)] -1
Yankee [(p1+2) x (p2+2) x (p3+2) x (p4+2)] – [(1+ (p1+1) + (p2+1) + (p3+1) + (p4+1)]
Presumably, if a selection loses then it is just removed from the equation
e.g. if selection p3 in the Yankee above loses, the calculation becomes [(p1+2) x (p2+2) x (p4+2)] – [(1+ (p1+1) + (p2+1) + (p4+1)]
Have a feeling I’d get a ‘5 out of 10, see me’ from teacher for the bracketing, but you get my drift
Is this how bookmakers worked out returns on multiples before automation?
Congratulations; that’s a brilliant way of summing it up. Yes, bookmakers would have used this method before their online calculators did it for them.
You now fully understand the mathematics and have extended it to suit your own example.
If I were your teacher I would give that 10/10 and a house point.Right,
A tricky one,
43 selections
11 winners,5/6,1/1,9/4,7/2,7/2,
9/2,6/1,13/2,9/1,20/1,
22/1.Would my goliaths get me a profit?
I have written a spreadsheet to show all perms and i reckon that I have 165 goliaths up.
Need to work out how to calculate the returns now
I haven’t ventured into the discussion on this one because, although you are correct in calculating that you have 165 full Goliaths, you also have many more in which you have 7 correct, thousands in which you have 6 correct, and so on and so forth, with billions in which you have just a double for instance.
So, in your lifetime, you will not be able to work out exactly how much your return is and whether you actually make profit or not.Just a thought on the ‘add a point’ method highlighted by DC: it’s fine for straightforward prices as illustrated. Wouldn’t it become a shade less accurate with prices like 4/6 in there?
(I’m useless at maths, by the way, but used to manually settle bets in the days before even the little Sporting Life Bet Calc)
No, it is still fine even with weird prices. So 4/6 is 0.66666666 and the figures you would use would still be 2.6666666 and 1.6666666 (if deducting the singles). These numbers look daunting, but an excel spreadsheet will be provide you with an answer to the nearest penny.
I haven’t ventured into the discussion on this one because, although you are correct in calculating that you have 165 full Goliaths, you also have many more in which you have 7 correct, thousands in which you have 6 correct, and so on and so forth, with billions in which you have just a double for instance.
So, in your lifetime, you will not be able to work out exactly how much your return is and whether you actually make profit or not.Have I over simplified it?
165 x28 doubles 4620
165X 56 trebles 9240
165×70 4’s 11550
165×56 5’s 9240
165×28 6’s 4620
165×8 7s 1320
165 8’s 165Not a full perm
First of all congratulations on now being able to calculate the returns on your best ‘all winners’ Goliath.
Yes, you have underestimated the difficulty of the task by a gigantic factor.
Let me give you just one example.
You have selected 43 horses and asked to cover them in Goliaths. I think you are now able to work out what that costs. It is:-
(43x42x41x40x39x38x37x36x247)/(8x7x6x5x4x3x2)
In billions of those Goliaths you have selected 8 losers. But that does not necessarily make you a loser!?
The point is that in some of those bets you will have a Goliath in which you have selected 6 losers and your 20/1 and 22/1 winners. In fact I have calculated below exactly how many of your bets include this winning double.
(32x31x30x29x28x27)/(6x5x4x3x2x1) = 906,192
Now, each of these successul doubles returns 21×23 = £483
So your total return from these doubles alone is £437,690,736
You have to make a similar calculation for all the other times you just get a double, of which there are 54 more individual occasions.
Then there are the trebles, the fourtimers, fivetimers, sixtimers and seven timers. Finally, we reach the times when you get successful Goliaths.
If you did not have a headache before you started reading this, enjoy the one you have now!Flipin heck, I didnt take into account the losing bets. Thanks
I still think there might be a profit there. LOL
Don’t speak too soon for the wheel’s still in spin. Consider the following:-
If you are going to cover 43 selections in Goliaths at 10p per bet (i.e £24.70 per Goliath) then there are 145,008,513 Goliaths costing you an initial outlay of £3,581,710,271.
On 10,518,300 of those Goliaths you are not even going to get one winner, that’s £259,802,010 down the swanee straight away. What is worrying is that my recycling bin holds just 50,000 losing slips at any one time, and is only collected every fortnight, so will take me 8 years just to get rid of the losing slips.
It does mean that I have 134,490,213 slips on which I am going to get some sort of return, and whilst I am always satisfied to get some sort of return, it takes the ladies in my betting shops more than 10 seconds to satisfy me each time. Therefore, even if they pay me out at 10 seconds per slip, it is going to take me over 43 years, being there all day and night every day and night for 43 years, before I get paid. Now while my wife likes me to give her a bit of peace every now and then, she might suspect something if I am in the company of the betting shop lady for 43 years!
I grant you that, even on the Goliaths where you only get a double up at 20/1 and 22/1, your total cost for those Goliaths was £22,382,942 and your total return is £43,769,073, giving you a total profit on those bets of £21,386,131, before you get too cocky think of this:-
On 906192 Goliaths you only got two winners at 5/6 and evens. You would have only got a return of £398,724 from these, meaning you made a loss of £21,984,217 on these bets.
Believe me, overall you are a loser!!!
Over and out.or would i be better off with, 8,9,10 horse accumulators?
As accas are dependent on consecutive winners e.g. 8 winners from 8 selections or nowt, I suspect the maths determining the returns from perming 8-horse accas from 43 selections is somewhat different from Goliaths
It seems to me that the probability of 8 consecutive winners from a sample of 43 is akin to the probability of getting 8 consecutive Heads over 43 coin tosses, with the added complication that unlike the 43 tosses each having an evens chance, the 43 horses do not
Unnecessary complication?
Over to teacher
Given 43 selections, the number of 8 horse accumulators is:-
(43x42x41x40x39x38x37x36)/(8x7x6x5x4x3x2x1) = 145,008,513
This is much more feasible than having Goliaths.
Furthermore, if you got the 11 winners you outlined, then you would actually have 165 successful 8-horse accumulators.
I could even be tempted to work out whether you would make a profit, given the stated odds of your winners.
Hope this helps and I haven’t put the student off too much.
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