Roulette Forum
Roulette Forum => Sports Betting => Topic started by: Sputnik on December 02, 2017, 02:39:37 PM

Testing binomial probability equal even money on Horse betting ... first place winner ...
Odds around 3.0
W & LW has 50% chance to win
LL has 50% chance to win
VODDS 3.81 W
VODDS 3.63 W
VODDS 3.29 L
VODDS 3.77 L
VODDS 3.75 L
VODDS 3.62 L
VODDS 3.15 L
VODDS 3.24 W
VODDS 3.15 W
VODDS 3.57 L
VODDS 3.34 W
VODDS 3.06 W
VODDS 3.97 L
VODDS 3.73 L
VODDS 3.88 W
And i also test higher odds around 6.0 with binomial calculation
VODDS 6.18 L
VODDS 6.76 L
VODDS 6.54 W
VODDS 6.30 L
VODDS 6.00 L
VODDS 6.75 L
VODDS 6.07 W
VODDS 6.91 L
VODDS 6.98 L
VODDS 6.55 W
This is pretty amazing :)

The people opinion and money rule what kind of odds to expect and you can not trust them.
I took a pretty common favorit odds and look at the output using STDS to measuring the accuracy.
The odds range is between 2.0 and 2.99 and i find 3.80 STDS sequence in during four weeks.
That is 3 win and 22 loses ... pretty cool to measuring the horse market that way and see how wrong the public eye is when they estimate a horse with relativ low odds.
Cheers

For those of you who do not know what he is talking about, here is a good link to teach yourselves Binomial Probability Distribution.
http://stattrek.com/probabilitydistributions/binomial.aspx
It is pretty simple math and you can easily (unless you do not use Mathematica software) put in the formula in your excel for future instantenous use.

Cumulative Binomial Probability
A cumulative binomial probability refers to the probability that the binomial random variable falls within a specified range (e.g., is greater than or equal to a stated lower limit and less than or equal to a stated upper limit).
For example, we might be interested in the cumulative binomial probability of obtaining 45 or fewer heads in 100 tosses of a coin (see Example 1 below). This would be the sum of all these individual binomial probabilities.
Pure gold! Thanks Thomas!! :D
Heretofore I have solved problems like this by resorting to "bean counting" which is not always a practical solution...