HOW MANY TREBLES

80 trebles

Sorry 560 trebles

Its 56 with just one runner in each race.

ta

How many trebles...

Posted on Virgin's Terrible Treble Thread?

At the last count: 223,542 of which 3 have been successful!

56 * 32 = 1792

231

Mafamatickal genie arses on ere

..... Oh and my answer........ Not a clue onlooker....  Apologies

231   is the correct totol

total

thanks

231

Pyramid block the numbers as you would to settle but do not add a point.

2
2 4
2 8  8
2 12 24
2 16 48
1 10 40
1 11 50
1    61
--------
     231

2 12 24
2 16 48
1 10 40
1 11 50
1    61



1



2
2
1
1
1

the easiest way to work out any perms doubles trebles four timers etc
say you have ten selections and you want to know how many trebles
take the first three numbers 1x2x3 = 6  then the last three numbers 10x9x8 =720
then divide 6 by 720 =120 simples hope this helps you in the future

whatyoulookingat,we all know that,but this was totally different,as there were 5 races with 2 qualifiers.As you implied,simples!

sorry unbiased never read that bit

its 120 trebles   if they came in your account with any of the big 3.....would be closed  but hey you won a million   ..enjoy

Can someone walk me through this slowly as that pyramid block is very confusing and although a couple of people have arrived at the answer of 231 I'm not sure how they've got there.

Clearly, it's not a case of multiplying by 56 by 32 (2x2x2x2x2) as all the options aren't included in every treble.

8 selections (one in each race) = 56 trebles and with an extra selection in ONE event, that makes another 21 bets as there
are 35 trebles in 7 selections - can someone confirm if that is correct so far?

I'm lost here and I'd like to know where I've (very likely) gone wrong

I make it;

2x2x2 trebles = 8x10 = 80

2x2x1 trebles = 4x30 120

1x1x1 trebles = 3x10 = 30

1x1x1 treble = 1

TOTAL = 231

Thanks Dixie...that's a bit clearer than the pyramid block though I still don't know how you've worked 4x30 and 3x10 but I get the 8x10 for 5 coupled selections = 10 doubles x (2 x 2 x 2) - it's hardly straightforward

gibberish 

  my block is not that clear
  but if you look at where I have got the answers by x
  you might see where its coming from
  i.e  1st line are the singles
       2nd line would be the doubles
       3rd line are the trebles
       if I was comp. lit. I could make it clearer
        look to see how I have x across to get the answer

Sorry Pete but that's gibberish to me
Thanks anyhow for trying.

It's worked out with factorials.

3 factorial is 3 * 2 * 1
6 factorial is 6 * 5 * 3 * 2 * 1

If
n = no. of selections
a = multiple (treble = 3 double = 2 etc)
r = remainder

n! / a! r!


e.g. for trebles in 10 selections, it's :-

10! / 3! 7!

(The 7 comes from 10-3)

10! / 7! 3!

cancelling out the 7! from 10! leaves 10 * 9 * 8

so the answer is

10 * 9 * 8 / 3 * 2 * 1

720/6 = 120

I wish I'd read your post properly

It's 56 trebles for one per race

8! / 3! 5!

8*7*6/6

56 trebles

Gibberish- my mistake.  It's not 4x30, it's 3x40.

It is simplicity itself to explain a Pyramid Block if one can use a pen and paper to do so but on screen it gets messy.
Try asking at your local bookmaker's. If the manager is worth his salt he will explain it to you in no time.

It is simplicity itself to explain a Pyramid Block if one can use a pen and paper to do so but on screen it gets messy

   not arf

The Betfair Sportsbook agrees on 231 trebles.

8 events with 2 selections in 5 events and 1 selection
in the remaining 3 events throws up the following
multiples :-


8 Folds (x32)
7 Folds (x176)
6 Folds (x416)
5 Folds (x552)
4 Folds (x450)
Trebles (x231)
Doubles (x73)

the answer is 191:-

10 x 2x2x2 treble = 80
20 x 2x2x1 treble = 80
15 x 2x1x1 treble = 30
1 x 1x1x1 treble =  1

total = 191

^^^^^^^  ^^^^^^^

201

201

for f uck sake    ITS 231

yes agree 231

sorry yeah, in trying to make 56 all in all I mistakenly totalled 46 so there are 10 extra 2x2x1 trebles so:

10 x 2x2x2 trebles =  80
30 x 2x2x1 trebles = 120
15 x 2x1x1 trebles =  30
1 x 1x1x1 treble =    1

total = 231

an easy way to do it on a settler is to have 5 even money winners and 3 non runners in trebles for £1 and see what it returns

I'm not doubting the answer a few have given as 231 but I AM questioning WHY - I hate being confused about things like this (as I should know better) and I have asked several smart geezers I know and NONE of them has got 231 as their answer.

Furthermore, they can't (like me) see HOW that answer has been arrived at...all I want is an explanation that doesn't look like the bones of the bet have been lifted from a manual!

It's irritating to see 4x30 & 3x10 bets with NO reasoning!


PS - Dixie (or anyone else) AGAIN - I still don't know how you've worked 3x40 and 3x10 but I get the 8x10 for 5 coupled selections = 10 doubles x (2 x 2 x 2) - a full & simple explanation would be nice.

PPS - is that reasoning I used correct about 8 selections (one in each race) = 56 trebles and with an extra selection in ONE event, that makes another 21 bets (totalling 77 trebles) as there are 35 trebles in 7 selections.

Help me out here guys...this is pissing me off

Gibberish - at the risk of confusing further, it can be explained in terms of combinatorials - ie mathematical notation.

If you write nCr, or C(n,r), for n!/r!(n-r)!, or the number of ways of selecting r from n.


Then the number of 2x2x1 trebles, for instance, is 5C2 * 3C1 * 2^2 = 10 * 3 * 4 = 120


You have to treat the '2 per race' and the '1 per race' races separately.

There are 5 races of the first type
There are 3 races of the second type

So, in a '2x2x1' treble, you're taking 2 races from the first type (there are 5C2 ways of selecting the races, ie 10), 1 from the 2nd (there are 3 ways of doing that) - so 30 race combinations.

Then you mult by 2^2 = 4, to get the number of trebles, because you can double up selections on each of the 'type 2' races.

Hope that helps!

..sorry, next to last line of my post should have said "each of the 'type 1' races".

Thanks for your help Ged though at the risk of making myself look even more thick, that's of little assistance to me...I think I've got a mental block on this conundrum

I'm usually more than OK at arithmetic and working out puzzles like this but I'm well & truly stumped with this one!

I'll probably have to wait a while before someone tackles the two questions I've already asked;


"Dixie (or anyone else) AGAIN - I still don't know how you've worked 3x40 and 3x10 but I get the 8x10 for 5 coupled selections = 10 doubles x (2 x 2 x 2) - a full & simple explanation would be nice.

&

Is that reasoning I used correct about 8 selections (one in each race) = 56 trebles and with an extra selection in ONE event, that makes another 21 bets (totalling 77 trebles) as there are 35 trebles in 7 selections"


PS - that Pascal bloke (of the triangle fame) didn't put such variables into his equations

So in that notation, the full answer would be:

The Sum of

8 * 5C3
4 * 5C2 * 3C1
2 * 5C1 * 3C2
1 * 3C3

which come to

80
120
30
1

as several others have said above.

Gibberish - yes I'm sorry, my explanation would only help if you're familiar with that notation.

I'll step aside and let someone else explain.

But whatever, you do have to break it down into 4 separate calculations, as others have done above.

Thanks squire...I'll pass that info over to one of my chums later tonight who hopefully will hold my hand through the labyrinth as it's too tough for me to get my skull around - I only got 'O' Level Maths so I'm not familiar with how some of you guys have explained things.

PS - just so I can get my head around things...is that business about ONE extra selection being 77 trebles correct?
AND am I putting myself on a bum steer using that rationale?

PS - that 'O' Level was Grade 'A' though...in the good old days (1981) when they were hard

A
B D
C E F

A*B=D
(A+B)*C =E
D*C =F
The first column when summed gives you the number of singles,in this case for 3 events,A,B and C
The sum of the second column gives you the number of doubles and the third gives you the number of trebles.
To start with let A,B and C all equal 1.
You will arrive at
1
1 1
1 2 1
Are you with me so far?

Dunlaying - sorry mate but I've now got a mental block and the use of letters instead makes me even more confused
Maybe one of my mates will help me out later with that pyramid block that was mentioned earlier.

PS - is that 77 trebles I mentioned right? I can't seem to get that way of calculating things out of my brain.

Gibberish - yes, your 77 is correct. But the problem, as I see it, is that making that first step is ok, but making the next step is not so easy because you then have to be careful not to count the 'same' treble more than once - you might need to start drawing pictures (not that there's necessarily anything wrong with that!)

Thanks Ged...I was aware of how complicated my method was becoming though I figured my first step was correct - that's probably the reason for my mental block as because (the differences between the extra selections) 21 + 15 + 10 + 6 + 3 and the original 56 add up to 111 though I'm unsure about that and a few of you shrewdies seem settled on 231 with reasoning that you're confident about.

Jesus wept.

I've given myself a headache squire...following my own wild goose chase!

When the scales fall from your eyes you will kick yourself.

Dunlaying - I'm getting there slowly and the way I'm doing it (what I'm comfortable with) is quite complicated.
I'm a bit of a mong that struggles with the letters

If we had just ONE extra selection in various combinations with TWO options in the last leg (totalling 8) making it 1-1-1-1-1-1-1-2
am I on the right lines so far with;

3 events 1-1-2 makes 2 trebles
4 events 1-1-1-2 makes 7 trebles
5 events 1-1-1-1-2 makes 16 trebles
6 events 1-1-1-1-1-2 makes 30 trebles
7 events 1-1-1-1-1-1-2 makes 50 trebles (the straight 6 making 20 trebles plus the other 15 (x 2) for the extra selection)
8 events 1-1-1-1-1-1-1-2 makes 77 trebles (the straight 7 making 35 trebles plus the other 21 (x 2) for the extra selection)

I realise that's very long-winded and another variable (never mind 5) is gonna be absolute havoc but that's my starting point that I'm trying to make progress with

Gib  You get the 80 trebles.  That is 2x2x2x10.  Knock off one of the 2s, that gives you 40 doubles.

You make up the trebles by inserting a 1 where the 2 was.  There are 3 1s. 3x40=120.

Thanks again Dixie...that's making a bit of sense to me now.

I'm in the middle of doing the calculations for a combination of 1-1-1-1-1-1-2-2 and I've come up with an answer of 102 - can you confirm if that's right as then I'll know whether I'm making SLOW progress.

I suspect that might be a handful out and I've miscalculated as the differences between the number of events (starting from 3,4,5,6,7 & 8) with TWO selections TWICE are constants and mine aren't.

I reckon my methods will be utterly bewildering to some of you guys but the way you've arrived at your calculations (probably the correct way) is alien to me.

Thanks to some of you **** patiently trying to help...this is hurting what's left of my brain
If it's remotely consoling, I have a mate that's quite 'confident' the answer to the original poser is 1792 (56 x 32 but that's ludicrous and redundant thinking) and juggling that is getting me down as well!

**** = guys

Maybe I wrote 'gays' instead?

2
1 2
1 3 2
1 4 5    sum the third column and you get 7

Look at your  figures and compare. Do you see now?

Ah...obviously NOT

Using 1 1 1 1 1 1 2 2

Combinations with one selection, six races 20 trebles.

Combinations involving 1 1 2 selections, 4 (2x2) x 15 (no of doubles with 6 selections) = 60

Combinations involving 1 2 2 selections, 4 (2x2) x 6 = 24

Total 104

Thanks a lot for that Dixie. I thought I was close with 102 and as I'm a bit flustered with this business, I didn't think to confirm it in the simplified way you've done there...a method that I can follow.

As you seem very conversant with this, can you be arsed spelling it out for me (with the above methods) for 3,4 and especially FIVE (the original question) double entries?
If you ain't got the time then no problem...you've already been helpful.

It's taking its time but I'm slowly cottoning on - I'm gonna give this a rest for now and I might RE-tackle it later after I've been inspired with a few drinks...that's always beneficial for me

PS - I tried calculating the amount of trebles with 2 double selections from 3 then 4,5,6,7 & 8 - I got answers of 4,12,25,44,71 & 102 but as I said, the differences between the numbers weren't 'constants' as they should be so I'm wondering where I went wrong - I've spent too long on it so I'm not thinking straight...time for a rest

I will work it out in the AM when I am sober.

The calculations may be best explained by using a spreadsheet and utilising the VLOOKUP function based on...

Race No.	Selections
1	1
2	2
3	2
4	2
5	2
6	1
7	2
8	1

The results of doing so are, with a bit of luck, displayed below.

Race No.	Race No.	Race No.	Calculation	No. of bets
1	2	3	1 x 2 x 2	4
1	2	4	1 x 2 x 2	4
1	2	5	1 x 2 x 2	4
1	2	6	1 x 2 x 1	2
1	2	7	1 x 2 x 2	4
1	2	8	1 x 2 x 1	2
1	3	4	1 x 2 x 2	4
1	3	5	1 x 2 x 2	4
1	3	6	1 x 2 x 1	2
1	3	7	1 x 2 x 2	4
1	3	8	1 x 2 x 1	2
1	4	5	1 x 2 x 2	4
1	4	6	1 x 2 x 1	2
1	4	7	1 x 2 x 2	4
1	4	8	1 x 2 x 1	2
1	5	6	1 x 2 x 1	2
1	5	7	1 x 2 x 2	4
1	5	8	1 x 2 x 1	2
1	6	7	1 x 1 x 2	2
1	6	8	1 x 1 x 1	1
1	7	8	1 x 2 x 1	2
2	3	4	2 x 2 x 2	8
2	3	5	2 x 2 x 2	8
2	3	6	2 x 2 x 1	4
2	3	7	2 x 2 x 2	8
2	3	8	2 x 2 x 1	4
2	4	5	2 x 2 x 2	8
2	4	6	2 x 2 x 1	4
2	4	7	2 x 2 x 2	8
2	4	8	2 x 2 x 1	4
2	5	6	2 x 2 x 1	4
2	5	7	2 x 2 x 2	8
2	5	8	2 x 2 x 1	4
2	6	7	2 x 1 x 2	4
2	6	8	2 x 1 x 1	2
2	7	8	2 x 2 x 1	4
3	4	5	2 x 2 x 2	8
3	4	6	2 x 2 x 1	4
3	4	7	2 x 2 x 2	8
3	4	8	2 x 2 x 1	4
3	5	6	2 x 2 x 1	4
3	5	7	2 x 2 x 2	8
3	5	8	2 x 2 x 1	4
3	6	7	2 x 1 x 2	4
3	6	8	2 x 1 x 1	2
3	7	8	2 x 2 x 1	4
4	5	6	2 x 2 x 1	4
4	5	7	2 x 2 x 2	8
4	5	8	2 x 2 x 1	4
4	6	7	2 x 1 x 2	4
4	6	8	2 x 1 x 1	2
4	7	8	2 x 2 x 1	4
5	6	7	2 x 1 x 2	4
5	6	8	2 x 1 x 1	2
5	7	8	2 x 2 x 1	4
6	7	8	1 x 2 x 1	2
			Total no. of bets	231