SAS == k*E SAS(A) = (.55/407 - .45/433)*(.55*433 - .45*407) = 0.017165 SAS(B) = (.10/50 - .90/1000)*(.10*1000 - .90*50) = 0.0605 System B is much more attractive.
For the umpteenth time you are just plain wrong. And evidently you are unteachable as well. You are thus exposed as a crank. There's not a bigger waste of time in humanity than cranks, although trolls are frequently just as useless. Welcome to my ignore list, crank.
I find the second set up to be more attractive because the odds are that we are making more profits with the small risk taken. Thus, on long term if we are able to make such trades its really good for the traders to scale up the profits and make huge portfolio.
Here is the final version of the System Achievement Score : SAS == 4*k*max[ 0, E ]*OZ*min[ 1, N/1000 ] I will apply it to all three setups -- designated A, B and C by abattia -- discussed in this thread. First we have to pick a constant trade size for all three setups. $500 is a nice round number. Looking at the expectations: E(A) = E(B) = E(C) = $55/$500 = 0.11 ⢠Obviously the expectations are no help for ranking these setups. Next we look at the Omega Zeroes: OZ(A) = (.55*(433/500))/(.45*(407/500)) = 1.300 OZ(B) = (.1*(1000/500))/(.9*(50/500)) = 2.222 OZ(C) = (.001*(100,000/500))/(.999*(45.05/500)) = 2.222 ⢠Setup B and setup C are now tied for first place. Finally we look at the Kelly fractions: k(A) = .55/(407/500) - .45/(433/500) = 0.156 k(B) = .1/(50/500) - .9/(1000/500) = 0.55 k(C) = .001/(45.05/500) - .999/(100,000/500) = 0.0061 ⢠Setup B wins this comparison by a wide margin. Putting it all together: SAS(A) = 0.0893 SAS(B) = 0.538 SAS(C) = 0.00597 ⢠Setup B is the best by a wide margin. ⢠Setup C is the worst by a wide margin.
kut2k2 is (IMO) making a mistake by multiplying the win rate (0.55, 0.1) by the *losing* payoff ($407, $50), not by the winning payoff (433, 1000), so he is getting the wrong Kelly values, and thus answering the question incorrectly. Since he responded to my prior post pointing out his error, by putting me on ignore (and making a few rude remarks), he is unable to read this. I would therefore appreciate if other people familiar with the Kelly formula would send him a PM explaining the mistake in his calculations - I would not want someone to risk inadvertent losses by misunderstanding a position sizing formula. For clarification, here is the Kelly formula: http://www.investopedia.com/articles/trading/04/091504.asp 'The Basics There are two basic components to the Kelly Criterion: ⢠Win probability - The probability that any given trade you make will return a positive amount. ⢠Win/loss ratio - The total positive trade amounts divided by the total negative trade amounts. These two factors are then put into Kelly's equation: Kelly % = W â [(1 â W) / R] Where: W = Winning probability R = Win/loss ratio The output is the Kelly percentage,' Using the examples in this thread: Setup A: W (winning probability) = 55%. R (win/loss ratio) = 433/407. Therefore Kelly % = .55 - [(1-.55)/(433/407)] = 0.127 = 12.7% (the answer I gave in my previous reply) Setup B: W = 10%. R = 1000/50. Therefore Kelly % = .1 - [(1-.1)/(1000/50)] = 0.055 = 5.5% (the answer I gave in my previous reply). Another source: https://en.wikipedia.org/wiki/Kelly_criterion - you can plug in the numbers there and get the same result. Setup A is therefore superior according to Kelly.
Btw, several people have said they have the same EV. Whilst this is true in $ terms, EV in $ is totally irrelevant. What is relevant is the EV PER UNIT RISKED. Setup A generates $55 per $407 risked. Setup B generates $55 per $50 risked. Setup B thus has a vastly higher expectation per trade (1.1 for B versus 0.135 for A). Despite this, the far lower win rate of setup B means its Kelly fraction is quite a bit lower, and thus it is less profitable. Also, in the real world of unclear and shifting trade odds, setup B is FAR more dangerous to trade. For example, there is a 34% chance of experiencing a 10 trade losing streak with setup B. For setup A the chance of a 10 trade losing steak is 0.03%. If a 10 trade losing steak is experienced, setup A is highly likely to have degraded and no longer work, and you can safely abandon the system. Whereas setup B will see a 10 trade losing streak often, so you can draw no conclusions as to its robustness from a 10 trade drawdown. An equivalent 0.03% occurrence for setup B would be a 54 trade consecutive losing streak. Are you willing and able to continue a system for 54 consecutive losers before abandoning it? No. Therefore system B would suck in the real world even if it were theoretically more profitable, which it isn't.
Just for laughs, let's try this the contrary way. SAS' == 4*bk*max[ 0, E ]*OZ*min[ 1, N/1000 ] , where bk == the Bad Kelly fraction bk(A) = .55 - .45/(433/407) = 0.127 bk(B) = .1 - .9/(1000/50) = 0.055 bk(C) = .001 - .999/(100,000/45.05) = 0.00055 E(A) = $55/$407 = 0.1351 E(B) = $55/$50 = 1.1000 E(C) = $55/$45.05 = 1.2209 OZ(A) = (.55*(433/407))/(.45*(407/407)) = 1.300 OZ(B) = (.1*(1000/50))/(.9*(50/50)) = 2.222 OZ(C) = (.001*(100,000/45.05))/(.999*(45.05/45.05)) = 2.222 ⢠Putting it all together: SAS'(A) = 0.0893 SAS'(B) = 0.538 SAS'(C) = 0.00597 ⢠No change from the previous results ⢠Setup B is the best by a wide margin. ⢠Setup C is the worst by a wide margin.