Is it fair to say that you are a winning player is you win more than 33.33% of the games, as long as you take down some of the pot, that are bigger than the double up? Or is it to simple?
My last calculation for this month, I have a win procent of 43%, and is up with $51.
Originally posted by MichFran
Is it fair to say that you are a winning player is you win more than 33.33% of the games, as long as you take down some of the pot, that are bigger than the double up? Or is it to simple?My last calculation for this month, I have a win procent of 43%, and is up with $51.
Hi MichFran
That is a little too simplified if you could breakeven at 33.33% there could be no rake for the house. So you need to include rake and also factor in volume to allow for the jackpot prize effect.
For example:
3 players $1 buy-in are playing for a $2 prize, so the house takes $1, lets say it deducts rake at 7% then $1 minus $0.21 leaves $0.79 to pay for the higher prizes. Therefore you pay $0.26 to the jackpot or lottery fund, 26% of all your buy-ins.
Volume:
At 33% winrate You play 3 games at $1 buy-in, all games land at 2X you win 1 $2, you lose 3 X rake 7c and 3 X jackpot fare 26c so 33% winrate there gives you minus 33%
Lets say all the games above land at 4X in that case you profit $1 so 33% winrate there gives you plus 33%
As volume increases at 100 games you will play
78 X2 14 X4 and 8 X6
A 33% winrate there would yield a loss of 7% approx.
My numbers here are very approximate and just to give a basic idea. Calculating the effects of volume are complicated there are different ways to do it, also it changes for every different buy-in. But 1 thing for sure is that 43% winrate will be very profitable for you, although mixing your buy-ins can create very unusual variance.
Which buy-in do you play?
Sorry Antwo, I am not very good at explaining.
Lets say there was no rake and no spin 3 people play and the winner takes all at a $1 buy-in there would be $3 in the pot so if you win 33% of the time you break even.
The Formula: (the prize) / (your buy-in) gives you a fraction which you then convert to a percentage.
300 / 100 = 3 fraction is 1/3 convert to percentage 1/3=0.333 X100 = 33.33%
With rake the prize is less than 300 so at 7% rake, 300-7%=279 prize pool.
279/100 = 2.79
convert 2.79 to a percentage 1/2.79 = 0.3584 x 100 = 35.84%
So because in spin n go the prize is not steady it is variable in a way released in increments over volume played. With a 35.84% win rate playing a low number of games you could either win a lot or lose a lot depending on where the spin stops, but over a large number of games that variance or inconsistency is reduced to a point where 35.84% will be a break even solution eventually.
To break even on 2X spin n go you need a 50% win rate so to compensate for the amount your win rate is less than 50% you need to play and win some higher multipliers.
Calculating the exact volume needed is very complicated for example you will on average only win 1 10X game in every 300 games you play, and 1 top level game in every 60,000 you play.
Originally posted by hoagy
Which buy-in do you play?
I played:
$1 = 159
Won = 65
$3 = 23
Won = 10
$7 = 4
Won = 3
$15 = 1
Won = 0
Win ratio ended on 41.71%
Thank you for your mathematical explanation. I'll have to look at it closer 
Another factor to consider would be your winrate for different prizes. It will generally be higher when the prizepool is 2x BI for the times people decide to play around around: shoving every hand, playing funny, wandering off because their friend turns up etc. This wouldn't happen at higher prizepools
