Roulette A little math help

Lets say there are odds of an event happening at 1:8,239
70,000 people played. Estimate, how many winners should there be?

Ken

 

5770 people winners

 

What?? No way, too high. I'm GUESSING around 13?

 

4 ,200 people are net winners . ( generally applied estimate )

 

Math applies to school room not casino!!!!!!lol!!!!!!!

 

An average of 8 or 9 players

 

Sharp tracker ,


6 % of the people present being winners is a reasonable estimate .

 

i can be wrong on this one but i just checked on our lotery websites... For an event of this kinf there should be 0.012% of winners (average)... Where did you get your 6%?

 

This is what I was told at another board >>
Divide 70,000 by 8,239.....8.5 winners (est)

Ken

 

Yes but there can be a big derivation at this stage, there could be easily 20 winners. Less probable the event is less derivation will be.

 

Sharptracker ,


Based on information among US gambling gurus and their side kicks like Casino junket operators the consensus is that
60 % of the players are ahead at one time or another and 90% give it all back .Hence the net 6 % .

 

The 70,000 is tickets (chances) sold. If it were 45,000 tickets sold instead, we would get a different number.

 

Yes of course but what i meant is if you repeat the game there won't always be 8/9 winners, it could be 20 winners or 5 winners (what i called derivation). Now if you chose an event of 1:5millions there will be quasi no derivation, always 0 1 or 2.

It says it starts to stick to the theoric number of winners when the odd reachs 1:600K (no more derivation)

 

Agreed. Just lookin for an average, thats all.

 

Average, sure!!!!!!!sure thing amig!!!!!!

 

Thanks Sharp & Nate.