Calculate the probability of each number in 3 columns

[bilbon]

Hi,

Is there anyone who can help me figure out the probability of each number on the different A-B-C so that it will be like the answer to the right.
Preferably with VBA.


[TABLE="width: 1096"]
<colgroup><col span="5"><col span="3"><col><col><col span="3"></colgroup><tbody>[TR]
[TD]Match[/TD]
[TD]%[/TD]
[TD]%[/TD]
[TD]%[/TD]
[TD][/TD]
[TD]Probability % A[/TD]
[TD]Probability % B[/TD]
[TD]Probability % C[/TD]
[TD][/TD]
[TD]Number of Right[/TD]
[TD]Probability % A[/TD]
[TD]Probability % B[/TD]
[TD]Probability % C[/TD]
[/TR]
[TR]
[TD]1[/TD]
[TD]14[/TD]
[TD]25[/TD]
[TD]61[/TD]
[TD][/TD]
[TD]61[/TD]
[TD]25[/TD]
[TD]14[/TD]
[TD][/TD]
[TD]13[/TD]
[TD]0,06[/TD]
[TD]0,00[/TD]
[TD]0,00[/TD]
[/TR]
[TR]
[TD]2[/TD]
[TD]92[/TD]
[TD]6[/TD]
[TD]2[/TD]
[TD][/TD]
[TD]92[/TD]
[TD]6[/TD]
[TD]2[/TD]
[TD][/TD]
[TD]12[/TD]
[TD]0,60[/TD]
[TD]0,00[/TD]
[TD]0,00[/TD]
[/TR]
[TR]
[TD]3[/TD]
[TD]49[/TD]
[TD]30[/TD]
[TD]21[/TD]
[TD][/TD]
[TD]49[/TD]
[TD]30[/TD]
[TD]21[/TD]
[TD][/TD]
[TD]11[/TD]
[TD]2,86[/TD]
[TD]0,00[/TD]
[TD]0,00[/TD]
[/TR]
[TR]
[TD]4[/TD]
[TD]50[/TD]
[TD]27[/TD]
[TD]23[/TD]
[TD][/TD]
[TD]50[/TD]
[TD]27[/TD]
[TD]23[/TD]
[TD][/TD]
[TD]10[/TD]
[TD]8,26[/TD]
[TD]0,01[/TD]
[TD]0,00[/TD]
[/TR]
[TR]
[TD]5[/TD]
[TD]69[/TD]
[TD]20[/TD]
[TD]11[/TD]
[TD][/TD]
[TD]69[/TD]
[TD]20[/TD]
[TD]11[/TD]
[TD][/TD]
[TD]9[/TD]
[TD]16,01[/TD]
[TD]0,07[/TD]
[TD]0,00[/TD]
[/TR]
[TR]
[TD]6[/TD]
[TD]48[/TD]
[TD]30[/TD]
[TD]22[/TD]
[TD][/TD]
[TD]48[/TD]
[TD]30[/TD]
[TD]22[/TD]
[TD][/TD]
[TD]8[/TD]
[TD]21,93[/TD]
[TD]0,39[/TD]
[TD]0,04[/TD]
[/TR]
[TR]
[TD]7[/TD]
[TD]48[/TD]
[TD]30[/TD]
[TD]22[/TD]
[TD][/TD]
[TD]48[/TD]
[TD]30[/TD]
[TD]22[/TD]
[TD][/TD]
[TD]7[/TD]
[TD]21,80[/TD]
[TD]1,66[/TD]
[TD]0,26[/TD]
[/TR]
[TR]
[TD]8[/TD]
[TD]49[/TD]
[TD]28[/TD]
[TD]23[/TD]
[TD][/TD]
[TD]49[/TD]
[TD]28[/TD]
[TD]23[/TD]
[TD][/TD]
[TD]6[/TD]
[TD]15,87[/TD]
[TD]5,24[/TD]
[TD]1,29[/TD]
[/TR]
[TR]
[TD]9[/TD]
[TD]56[/TD]
[TD]25[/TD]
[TD]19[/TD]
[TD][/TD]
[TD]56[/TD]
[TD]25[/TD]
[TD]19[/TD]
[TD][/TD]
[TD]5[/TD]
[TD]8,42[/TD]
[TD]12,25[/TD]
[TD]4,70[/TD]
[/TR]
[TR]
[TD]10[/TD]
[TD]28[/TD]
[TD]29[/TD]
[TD]43[/TD]
[TD][/TD]
[TD]43[/TD]
[TD]29[/TD]
[TD]28[/TD]
[TD][/TD]
[TD]4[/TD]
[TD]3,20[/TD]
[TD]20,95[/TD]
[TD]12,39[/TD]
[/TR]
[TR]
[TD]11[/TD]
[TD]64[/TD]
[TD]22[/TD]
[TD]14[/TD]
[TD][/TD]
[TD]64[/TD]
[TD]22[/TD]
[TD]14[/TD]
[TD][/TD]
[TD]3[/TD]
[TD]0,84[/TD]
[TD]25,54[/TD]
[TD]23,12[/TD]
[/TR]
[TR]
[TD]12[/TD]
[TD]67[/TD]
[TD]19[/TD]
[TD]14[/TD]
[TD][/TD]
[TD]67[/TD]
[TD]19[/TD]
[TD]14[/TD]
[TD][/TD]
[TD]2[/TD]
[TD]0,14[/TD]
[TD]21,02[/TD]
[TD]28,93[/TD]
[/TR]
[TR]
[TD]13[/TD]
[TD]19[/TD]
[TD]30[/TD]
[TD]51[/TD]
[TD][/TD]
[TD]51[/TD]
[TD]30[/TD]
[TD]19[/TD]
[TD][/TD]
[TD]1[/TD]
[TD]0,01[/TD]
[TD]10,48[/TD]
[TD]21,79[/TD]
[/TR]
[TR]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD]0[/TD]
[TD]0,00[/TD]
[TD]2,39[/TD]
[TD]7,47[/TD]
[/TR]
[TR]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD]100%[/TD]
[TD]100%[/TD]
[TD]100%[/TD]
[/TR]
[TR]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD="colspan: 3"]This is the correct answer[/TD]
[/TR]
</tbody>[/TABLE]

 

[shg]

I believe that shg's algorithm reverses the order, Pr(k) should be Pr(13-k)

Don't think so, Eric. The probability of getting all 13 correct is the product of the probabilities, which is the 0.0564% shown at the bottom of my col K. No?

I'm a bit confused as to what the percentages represent though.

I would interpret it to mean, "If I bet on the most/mid/least likely outcome for each game, what's the probability of winning 0 to 13 of my bets?"

 

[Eric W]

Oops, sorry about the order. I now notice your numbers run from 0 to 13 down the column, whereas mine ran in the opposite order. So our results match.

Still confused about the percentages though. There's a stat often used in American football. Say team A plays teams B though N. Based on various criteria, someone predicts that A has a 75% chance of beating B, 66% chance of beating C, etc. Even though A might be favored in every game, the odds of winning every game are quite low, and this is a way of predicting the most likely number of wins for the season. Nevertheless, why 3 percentages for each game? Unless there are 3 sources providing estimates per game. And if that's the case, then sorting by high to low wouldn't make a lot of sense since the estimates would be based on different criteria/algorithms.

 

[shg]

My random assumption is that it represents 13 games on a given weekend, and each row is the probability or win/lose/draw for the home team.

 

[Eric W]

Ah, that makes more sense.

 

[bilbon]

Hi shg and Eric


Thanks for your suggestions which I will test both tomorrow.


The reason that I sort the odds from left to right is to get all the highest numbers in a column and then the second-highest in a column, and the smallest in a column.
But with your knowledge, you may not need any sorting but you could calculate it from
original located at the left and at the top of this thread.


We have a game where you have to predict the outcome of 13 matches and where we get odds on every game such as Leicester-Manchester C 7,95-5.15-1,38 these I count on to the percentage which becomes 12-19-69, and so on which 12 matches and these are the numbers I would figure out said nnolikheten for how many right things get on those 13 matches.

 

[bilbon]

Hi, shg and Eric


I will benefit from both proposals and they work very very well.
Many thanks to you that you took the time to help me.


/Bilbon

 

[Eric W]

Glad we could help. It turned out to be enlightening for me too.

 

[bilbon]

Hi shg


Would there be much change in your code to get 13 at the top and 0 at the bottom?

/Bilbon

 

[Kishan]

Hi, bilbon, Eric and shg </SPAN>

</SPAN>I am interested if any one cloud explains when % A, % B and % C is define that says which team is strong to bet on it (A B or C) then why does require the Probability % A Probability % B Probability % C. I really cannot follow up please need your help</SPAN></SPAN>

Thank you </SPAN></SPAN>

Regards,</SPAN></SPAN>
Kishan</SPAN></SPAN></SPAN></SPAN>

 

[bilbon]

Hi Kishan


I make an attempt to explain a little.
The game is to predict 1, X or 2 (winner, tie, away win) in 13 football matches,
I use it, among other things, to make a theoretical assessment of how the pros (those who set the odds) assess the weekly round.
Then there are many ways to limit the number of rows (totaling 1 594 323 rows) by setting filters to remove rows that you don't want to include in their system using these % numbers.
Just your own imagination that sets the limit.

/Bilbon