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Dr Z’s formula is based on tote SPs and he recommends betting at the last possible moment so the tote board price is the closest to the tote SP and you can make your judgement call then.
It is much better suited to places like the USA and HK and France where there are comparatively huge tote pools.
Over here, putting more than a few quid on would have a significant impact on the dividend in most races, especially now it is almost impossible to get on at tote prices without bet into the pool.
Hi all
successful horse race betting requires the ability to work within, reasonable, probability estimates.  Securing a worthwhile sustainable profit is the ultimate goal, super accuracy is fools gold.
However a good logical base is vitally important.
I am not the finished article, yet, so I am willing to learn.
I have thought about this thread, and is it fair to say the chance of the horse being placed is it’s win chance + it’s chance of being 2nd.  The latter being an easier task.
byefrom<br>carlisle
ps I refer to the 5 runner example.
The DrZ thing won’t work, for the reasons stated above and there is no guaranteed minimum payout in the UK.
I’ve written a place price calculator since you started this thread and I am putting up place priced in the systems work shop if anyone wants to see … I’m not happy with it yet but I think it’s on the right track .. the place price forecast against the win price is an S-curve, does anyone know if it should be ??
Hi non vintage
do you agree that the "latter task is easier"?
ie. place chance = 0.20 + 0.25
byefrom<br>carlisle<br>
Carlisle,
how do you make it pay? You’re still insisting that a horse’s theoretical chance of coming second in a five runner race is 3/1. Its chance of filling any specific place where there is no place bias is always the same, ie 4/1.
The place book on a five runner race is 200% as there are two places. Each horse has a 40% chance of being placed 1st or second (0.2 + 0.2 as you nearly put it). This is demonstrated perfectly in NV’s spreadsheet.
Once again, I don’t claim that this is true in reality, but it is the model to work from.
(Edited by rory at 10:59 am on June 8, 2007)
Is carlisle saying that a horse has a 3/1 chance of coming second, GIVEN that the winner is known?
This is not the same as saying that a horse has a particular chance of being placed in a five horse race, which is 2 out of 5 (40%).
The first example encompasses two events – the race for first place and the race for second. The second example is one event – whether or not a horse finishes in the first two.
Five differently coloured balls are placed in a cloth bag. What is the probability that you pull out a red ball given one pick of two balls and what is the probability that you pull out a red given two consecutive picks?
In the first instance, the probability is clearly 1/5 + 1/5 =2/5 or 40%.
In the second case, there is a 1/5 chance that you pick the red ball first, hence no second pick is required. There is therefore a 4/5 chance that you will need to make a second pick. The second pick is a 1/4 chance of pulling out the red ball.
I’m not sure of the maths, but the two situations appear to be different.
The second example is the same as the first, as in the second example, you’re making the (unstated) assumption that you’re not replacing the first ball drawn out.
(Edited by psychosis at 1:04 pm on June 8, 2007)
I think you are right, psychosis. The probability is the same. Just looked different, which still puzzles me, because carlisle’s probabilities seemed to make sense.
probabilty of A winning =1/5(20%)
probability of A coming second given that the winner(not A) is known =1/4(25%)
probability of A being first or second looks as though it should be 20% + 25%, but this is clearly wrong. Maybe if all possibilities were listed, it might be clearer.
I have to forgive carlisle, because its got me scratching my head: these things usually do.
I think you are right, psychosis. The probability is the same. Just looked different, which still puzzles me, because carlisle’s probabilities seemed to make sense.
probabilty of A winning =1/5(20%)
probability of A coming second given that the winner(not A) is known =1/4(25%)
probability of A being first or second looks as though it should be 20% + 25%, but this is clearly wrong. Maybe if all possibilities were listed, it might be clearer.
I have to forgive carlisle, because its got me scratching my head: these things usually do.
Well there’s two different situations.  One simple, one slightly more complex.
Simple one is the simple probability of A finishing 1st, 2nd, 3rd, 4th or 5th, without prior knowledge of where other horses have finished.  This is 1/5, or 0.2, for each (analagous to the probability of a red ball being drawn on the 1st, 2nd, 3rd, 4th or 5th draw, if you replace the ball following each draw)
Slightly more complex one is the probability of A finishing 2nd, 3rd, 4th or 5th, given prior knowledge (e.g. the probability it finishes second, given we know it doesn’t finish first, the probability it finishes third, given we know it doesn’t finish in the first two, etc…)
The respective probabilities of this are 1/4, 1/3, 1/2 and 1.  This is analagous to not replacing balls – i.e. the probability of the second ball drawn being red, if we know the first one wasn’t, and the first one wasn’t put pack in the bag is 1/4.  And obviously, if we’ve drawn 4 balls out, none of which are red, and we haven’t put any balls back in, then the last ball must be red.
The probability of A being first or second is simply 1/5 + 1/5.
A further, similar, exercise to this is how to price up "without favourite" races.  e.g what would the win odds be if you took one of the 5 co-favourites out of the race?  What about if the favourite was 2/1 and one of the other four was 14/1 to maintain the 100% book?
(Edited by psychosis at 2:27 pm on June 8, 2007)
Good grief – I actually enjoy this stuff…
1. The probability of A coming 1st is 0.2 (or 1 in 5 or 20%)
2. If A doesn’t win, the probabilities of it coming 2nd or being drawn 2nd are as follows:
1st=B, 2nd=A > 0.2*0.25=0.05<br>1st=C, 2nd=A > 0.2*0.25=0.05<br>1st=D, 2nd=A > 0.2*0.25=0.05<br>1st=E, 2nd=A > 0.2*0.25=0.05
3. These probabilities need to be added up to give the chances of A coming 2nd…<br>so 0.05+0.05+0.05+0.05=0.2
<br>In this example (assuming equal chances), you could equally accurately say:
1. The chance of A not winning is 0.8<br>2. The chance of A being next after something else wins is 0.25<br>3. Thus the chance of A being 2nd is 0.8*0.25=0.2
I think that’s a very long winded way of getting there, but yes, you’ve proven the probability of A finishing second is 0.2. YOu could do similar, much more involved calculations to show the probability of A finishing 3rd is also 0.2, which would involve listing all the combinations of the first two…
I’m itching for the debate to get into Bayesian probability and normal functions…
psycho,
I only went into the nitty-gritty because it makes things much clearer in this instance. The 25%/20% thing needed putting to bed.
And don’t talk to me about Bayesian Probability Theory. I had to read a huge amount on it last year to be able to assess a project at work, and ended up (as I’d thought I would at the outset) saying no because it was impractical.
;)
Thanks chaps,
I thought my maths were ok, but this one had me puzzled. I knew what the answer should be, but couldn’t prove it to myself. Post tenebras lux(after darkness, light. – my old school motto).
Hi rory
I was not insisting anything, I just needed some help, you barstead.  I sent that with love.
Thanks fellows you have done me proud.  You have shot me down, and I deserved it.  And I didn’t feel a thing.
If they all have an equal chance, then winning is no more difficult than finishing last.
cheers from<br>carlisle
"you learn more from your mistakes than your successes"<br>
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