Freestyle solitaire win percentage question

[michael1947]

I like to play Freestyle soltaire and the app I use shows my win percentage. To date, I have played 1732 games and won 1639. This computes to .9463048 of the games played, which my app shows as 95%, due to rounding.

My question is: Is there a way to calculate how many more games I must win to get to 96%.

 

[HallsofIvy]

How many more won out of how many more played? Let n be the number of additional games played and m the number of additional games won. Then you want \(\displaystyle \frac{1639+ m}{1732+ n}= 0.96\). That is the same as \(\displaystyle 1639+ m= 0.96(1732+ n)= 1662.72+ 0.96n\) or \(\displaystyle 0.96n= m- 23.72\). That is still one equation in two unknowns and has an infinite number of solutions. Requiring that n and m be positive integers reduces that but there still might be more than one possible solution.

If your question is about how many more wins (you win all additional games), so that m= n, \(\displaystyle 0.96n= n- 23.78\) so that \(\displaystyle 0.04n= 23.78\) and n= 594.5. At minimum, you would have to play 596 more games and win all of them!

 

[michael1947]

You're right; I'm presuming I win 100% of all future games.

I'm wondering where the 23.78 came from and how that produces 594.5?

 

[michael1947]

I can see now that 23.78/.04 = 595.

Another question related; would there be a formula that allows "plugging in" a win percentage of less than 100%