Given a Risk to Reward ratio how to figure break even %

[Jay108]

Hello All,

If I am risking 1 to gain 1, I would need to win 50% of the time to break even.


What is the formula to figure this percentage out so I can play with other ratios.


For instance what if I am risking 1 but gaining 1.35 what percentage of the time would I need to be right in order to break even?

or risking 1 but gaining .72 etc.

Am interested in the formula as well as what kind of math this is.

Thank you

 

[HallsofIvy]

If you win "A" p percent of the time and lose "B" 1- p percent of the time, your net winnings will be pA+ (1-p)(-B)= pA- (1- p)B. "Breaking even" means net winnings are 0 so you have the equation pA- (1- p)B= 0. pA- B+ pB= 0 so pA+ pB+ B= 0. p(A+ B)= B, \(\displaystyle p= \dfrac{B}{B+ A}\)
That is, if you are "risking 1 but gaining 1.35", the probability of breaking even is \(\displaystyle \dfrac{1.35}{1+ 1.35}= \dfrac{1.35}{2.35}= 0.5745\) or 57.45%.

If you are "risking 1 but gaining .72", the probability you will break even is \(\displaystyle \dfrac{.72}{1.72}= 0.419\) or 41.9%.

 

[Jay108]

Thank you

This is great, thank you.

I am not sure how you begin to figure out the equations in the first line, if there is an easy way to explain it that would be great otherwise no worries.

It seems that 'p' always equals 'B' is that true?

Also to get to the number I am looking for (which is to know the percentage of wins needed in order to break even) am I right to say that I need to subtract the final number from 1?

Thank you.

 

[HallsofIvy]

This is great, thank you.

I am not sure how you begin to figure out the equations in the first line, if there is an easy way to explain it that would be great otherwise no worries.

It seems that 'p' always equals 'B' is that true?

That doesn't even make sense. "p" is probability of winning so is always beteen 0 and 1. "B" is the amount lost when you do lose. There is no reason to think that they are equal. The first equation I give was \(\displaystyle p= \frac{B}{B+A}\). How do you get "'p' always equals 'B' from that?

Also to get to the number I am looking for (which is to know the percentage of wins needed in order to break even) am I right to say that I need to subtract the final number from 1?

Thank you.

 

[Jay108]

That doesn't even make sense. "p" is probability of winning so is always beteen 0 and 1. "B" is the amount lost when you do lose. There is no reason to think that they are equal. The first equation I give was \(\displaystyle p= \frac{B}{B+A}\). How do you get "'p' always equals 'B' from that?

Ok that makes more sense now. I had it mixed up I thought "A" was the % of the time you win and "B" was the % of the time you lose.
So if "A" was 60% and "B" was 40% the the equation would always produce "B" as the answer. That's also why the rest did not make sense to me.


Now that I know what 'A' and 'B' are I think you got the equation examples mixed up which is why I wanted to subtract your answer from one.


If you risk 1 to gain 1.35 then A=1.35 and B=1 so p= 42.55%


And risking 1 to gain .72 would give a p of 58.14%

Just to be clear p=Probability of Breaking Even like you mentioned in your first response (rather than winning) correct?



Please let me know if I am on the right track now. Thank you.

 

[Jay108]

Ok, I think I see now.

p is the probability of winning.

Or in this case if we want to find the break even point it equals the % of the time we need to win in order to break even.


Can you please confirm that the "A" and "B" in your first examples were mixed up and that my examples above are correct?
If so then it all makes sense to me.
If not then I am missing something.


Thanks again this has been very helpful.