listener 3970

[sohcahtoa]

Is there any chance of solving this if one knows nothing about betting? How can I start?

[workshy]

solving what?

[cluelessJoe]

Re-read the preamble, there's plenty of instruction there.

It may even help if you ignore the fact that this has anything to do with betting, and just view it as three (win/each way place/each way win = rule 1, rule 2, rule 3) basic mathematical rules for solving equations.
Hence, for 1 across, Agio 70p win you are looking at Rule 1 which is that at 1 to 1 the return is 1x70p +70p at 2 to 1 the return is 2x70p + 70p and so on. You may well therefore need to work out the return for all items from 1 to 1 to 100 to 1 for each possible bet.
Rule for place only simply says divide the odds (1 to 10) by 4 and do the same math.
Rule for each way win says first do the win math, then with the other bet do the place bet and then add the two parts.
Sorry I'm no maths teacher, but you only need to understand and apply these 3 simple rules.
The hard bit is applying these rules to a range of odds from 1 to 100 to one, hence your first calculations pre-elimination are manifold ... .
A spreadsheet helps if you can muster an electronic version.
Good Luck

[sohcahtoa]

Many thanks for your suggestions Clueless Joe. I am a Maths teacher but it doesn't seem to help! Will use spreadsheet as you say.

[midazolam]

the way i started sohcahtoa was:

win = bet x odds + bet
e/w win = (bet x odds + bet) + (bet x odds/4 + bet)
e/w place = bet x odds/4 + bet

only one possible ending for 14 down and 2 down
therefore only one answer for 18 across
use the odds for 18 across to work out 1 down
8 across (which you have the beginning and end must only be one of three odds
using these odds 13 down must begin with a certain number

however where i get stuck is that 12 across cannot end with this number???

can all you bright sparks see if i have started right

oh and i presume (although it does not state) that where the horse is duplicated e.g. 1 down and 18 across it is the same odds

[cluelessJoe]

Let's try to help you both ...
Picking a clue at random, let's say 18 across, "Ciao 25p e/w place" .. you need to consider the formula
(25xZ/4) + 25
for all values of Z from 1 to 100 (where Z is the horse's odds)
Hence you have 100 potential solutions for that - and every other - clue.

You can immediately cut those potential solutions down to a lesser number of possible solutions by considering the length of each answer.

Now, answering Mid's query, yes, all solutions to clues for a given horse are calculated at one single set of odds (i.e. one value of Z) ....
so ... now solve the other clue for horse Ciao which is 1 down.
Look at both sets of answers in tandem and you will find that there are only 9 possible values of Z which will give you the required 2 digit & 3 digit answer

Repeat for all horses to reduce possible answers for each clue ... considering these, you should find that certain spaces in the grid can only house one possible digit - then start to look at how clues intersect with others and you can proceed to eliminate further

Mid, you seem to start off ok but I don't know where you reach your statement about 12 across

Good luck both

[midazolam]

aaah...that is because the algorithm i set up rounded to the nearest number not always up

will have another look

roll on friday for a more "normal" listener crossword

[cluelessJoe]

Had the same challenge on my spreadsheet program .. if your rounding function heads for the nearest integer, you can force the return to round upwards by adding a constant very close to a half into your formula - i.e.
round((Stake x odds/4) + 0.499 + Stake)