Calling Relkeel

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SlimChance

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Statistics Question:

20/1
16/1
6/1
12/1
11/10

What do I need to do to work out the betting for the number of winners from the above selection? You've answered this question before for me but I can't find the thread.
 
Bar the Bull said:
Can I have a go, please?

What do you mean by "number of winners"?
Click to expand...

They are the prices for 5 horses. I want to bet on how many winners there will be. I know it's simple statistics but I've searched online and I can't find the answer. I've been meaning to get the answer for ages and it annoys me that I don't know.
 
Here is how you price it up:

5 winners 10,000/1 (real odds 68,000 to 1)
4 winners 500/1 (real odds are 1237/1)
3 winners 33/1 (real odds are 61/1)
2 winners 9/2 (real odds are 6.2 to 1)
1 winner evens (real odds are 1.12 to 1)
0 winners 6/4 (real odds are 1.69 to 1)
 
Bar the Bull said:
Here is how you price it up:

5 winners 10,000/1 (real odds 68,000 to 1)
4 winners 500/1 (real odds are 1237/1)
3 winners 33/1 (real odds are 61/1)
2 winners 9/2 (real odds are 6.2 to 1)
1 winner evens (real odds are 1.12 to 1)
0 winners 6/4 (real odds are 1.69 to 1)
Click to expand...

Appreciate that Barry. Is there a website I can run this through or is it blindingly easy to work out? I hate asking for help on here as it shows weakness but none of you respect me anyway...


.
 
Last edited:
It is a bit fiddly.

All you need is a spreadsheet and a bit of patience.

As I said, 5 winners and 0 winners are easy. Just the cumulative probabilities of 5 losses and 5 wins.

For 1 winner, you need to work out the prob of Horse A winning, in an accumulator with losses for Horses B through E. Then add this to the prob of Horse B winning in an accumulator with a loss for Horse A and Horses C through E. And so on until you have all 5 permutaitons of 1 winner.

Then convert the prob into odds.

4 winners is similar. Just 5 different accumulators of 4 losers and 1 winner.

3 and 2 winners....for each of these there are 10 different possibilities. For example, 2 winners would be AB, AC, AD, AE, BC, BD, BE, CD, CE, DE.

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I suggest you set the spreadsheet up for 10 eventualities....if there are only 5 races or 5 matches, you can just set the odds for the last 5 events to be 100% chance of a win....

Not sure if that makes sense....as always with statistics, ability to understand is heavily dependent on how well it is explained.
 
Apologies for delayed response Slim, but it seems you've got what you need anyway. I wrote a spreadsheet a few years ago, using similar logic to Bar The Bull.

If you're dealing with a large number of events, you might need to start thinking about distribution patterns and drawing bell curves, but I'd be very rusty on all that these days.
 
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