Win Racing Post Annual – Competition Boxing Day/King George

[Nathan Hughes]

The Racing Post have put up a prize in the shape of an RP annual
To be in with a chance of winning you simply have to find the correct answer or be the closest to it, in the event of a tie the winner will be the person with the highest answer. No duplicate answers, first come first served.

So here we go

you have to guess number of finishers in the King George
multiplied by winning horse racecard number
multiplied by winning distance

for example 7 finish the races multiplied by the winner number 10 = 70 who wins by 2 lengths = 140

Any winning distance under one length becomes 1 (so a short head, for multiplication purpose, would be a 1)

Blackbeard to conquer the World

[goodfellow]

There is a lot of room for error but I will go with 120 which I think would be a cosy victory for the Don

Merry Christmas

[DiamondGeezer]

Cue Card (no 3), six finishers, under half a length 0.5 = 9

[Nathan Hughes]

I think to keep it simple anything under a length will be score as 1

I put you down for 18 DG

Blackbeard to conquer the World

[chalk jockey]

110

If you go to back a certainty always buy a return ticket.

[raymo61]

I will go for thirty six .
Please don’t ask me how I come to that because it has become very complicated

It all revolves around Vaitour and Conti but that is all I am goning to say!! LOL

36

[roosterbooster]

54

[roosterbooster]

54 please

[yeats]

70

[BigG]

Good seasonal bit of fun Nathan

Don Cossack No 4, 6 finishers, wins by 5 lengths = 120

[Triptych]

7 to complete the race
Silviniaco Conti to win – No. 6
= 42 x 1 lengths winning distance

42

Jac

Things turn out best for those who make the best of how things turn out...

[cormack15]

112 for me.

[robnorth]

I’ll try 150

[Seasider]

It’ll be a procession – 216.

[rocky91]

7 finishers x the don No 4 x4 wins by 4 lengths = 112

[Tonge]

28

[Chivers1987]

7x4x7=196

[Author]

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