Silly, and I'm sure its very simple question for those maths savvy members....

How do I work out the chances of say hitting 5 winners in a row given X% strike rate?

Also what would be the likely longest run of outs...?

e.g. for :
* 50% Strike Rate
* 35% S/R

Thanks very much in advance.

__________________
Stix
.......Giddy Up..... !!

Easy to do:

for 50% it

1*0.50 = 50%
1*0.50*0.50 = 25%
1*0.50*0.50*0.50=12.5%
1*0.50*0.50*0.50*0.50=6.25%
1*0.50*0.50*0.50*0.50*0.50=3.125%

Hope that explains it.

The longest run of outs is a bit different. It depends on what your definition of "likely" is. Is it 5% of the time, 1% of the time ?

Chance of outs is the similar. To determine the chance of 6 outs in a row at a 30% strike rate it is :


1*0.70 = 70% of 1 out
1*0.70*0.70= 49% chance of hitting 2 outs
1*0.70*0.70*0.70= 34.30% chance of hitting 3 outs
1*0.70*0.70*0.70*0.70= 24.01% chance of hitting 4 outs
1*0.70*0.70*0.70*0.70*0.70= 16.807% chance of hitting 5 outs
1*0.70*0.70*0.70*0.70*0.70*0.70= 11.7649% chance of hitting 6 outs

Stix
I recently pulled a breakdown off of the betbotpro forum. I can't find it now. It had the instances of single wins double wins treble wins etc for the fav. 5 wins were a very rare animal. even 4 wins were rare. double wins were regular and triple wins were serendipity. I would not put any faith in scoring multiple wins. In this there was note that some run of outs were 22. otherwise there are several threads that have given runs of outs on this forum. runs of ins are the inverse of these. Beton

Thanks guys, had a mate ask me the question, so passed info on to him. I'm flat out getting one in a row !

__________________
Stix
.......Giddy Up..... !!

Theoretically , the longest run of outs to expect with 50% SR is 10 in a row.

__________________
Cheers.

There is a 0.097% chance of this happening. That is approx 1 in 1024 that you will get a run of 10 outs in a row.

I have given this formula before but I don't know how to direct you there so I have presented it again below:

Let C=the prob of at least 1 win from n races, called the certainty, expressed as a decimal.
Let q=the prob of a LOSS in a race, expressed as a decimal (i.e. 1-win strike rate).

Now to be 100% certain you need an infinite number of races so this theory expects you to decide on an acceptable certainty usually 90% or 95% of getting at least one winner from n races.

The number of expected outs is given by n from

n=log(1-C)/log q

You can use log or ln on your calculator or spreadsheet but it has to be the same throughout the calculation.

For example, consider top place percent which Knowles suggests has a 22% strike rate (q=1-0.22=0.78) and deciding on 95% certainty (C=0.95) then,

n=log(1-0.95)/log 0.78
n=log 0.05/log 0.78
n=12.06 so you can be 95% sure of at least one winner in about 12 races.

To be 99% sure it takes a run of about 18 or 19 races.

gunny72

Found it when I wasn't looking.
Study done on the fav
12539 wins by the fav were followed by a loss.
3445 wins were followed by another win ie 2 in a row
462 wins were followed by a second win ie 3 in a row
50 wins were followed by a third win ie 4 in a row
3 wins were followed by a fourth win Ie 5 in a row
1 win was followed by a fifth win Ie 6 in a row.

I don't the sample size but it has to be circa 50,000. using the standard adage of the fav winning a third of their starts. Thus the sample size is enough to be consistently applied to everyday racing.

What you can take from this is that it is easier to lose than win, It is easier to string losses together than to string wins together. Multiple wins are a very rare breed. Beton