How Win Rate, Payoff Size, and Costs Determine Strategy Expectancy
Summary
A prediction model that is correct half the time can still support a profitable strategy when its average winning trade is larger than its average loss. Conversely, equal-sized wins and losses at that accuracy produce zero expectancy before costs, and transaction costs make the expectation negative. The central method is to calculate expected value from the win rate, average gains, average losses, and trading costs, then solve for the accuracy needed to meet a profitability target.
The discussion also illustrates that a strategy can win less often than it loses and still have positive expectancy if its winners are sufficiently large. Examples include an asymmetric payoff game and a simple numerical calculation. The document cautions that costs vary with market conditions and estimated accuracy is uncertain. It gives an expectancy framework, not evidence that any specific prediction model will achieve the assumed outcomes; risk, drawdowns, and implementation constraints also require separate analysis.
Key ideas
- Win rate alone does not determine whether a trading strategy is profitable.
- Expected value depends on win frequency, average win size, average loss size, and trading costs.
- A model with 50 percent accuracy needs larger average wins than losses to overcome costs.
- A strategy can be profitable while winning less often if its payoff is asymmetric.
- Estimated accuracy and transaction costs can vary, so expectancy calculations rely on uncertain inputs.
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# Implementing A 50/50 Prediction Model Strategy # Implementing A 50/50 Prediction Model Strategy Reworded the question for clarity (see edits for original post): How can one knowingly foresee where a 50/50 prediction model will be profitable? For previous posts: I understand that if I have a 50/50 chance of winning the grand prize for a lottery, it should take two ticket purchases to win (assuming the there is a pattern from an algorithm that hasn't changed between purchases). Obviously this is not a realistic scenario. I am assuming that each movement is independent of the last, so the only way I see a 50/50 model as profitable is when predicting large moves is successful half the time and stop losses are implemented to cap the losses. A 50.1-60 / 49.9-40 can be profitable if transaction costs are low, but only under in a high-frequency environment... Although it takes a long time to get from London to Paris taking 6 steps forward and 5 steps back. ## Answer by vanguard2k (score 3) https://quant.stackexchange.com/a/12860 A prediction model that is correct $50\%$ of the time can be profitable if the model gains more when it is right than it loses when it is wrong. You could simplify it like this: A trading strategy is profitable if your trades have positive expected value. Now suppose that your gains when your model is right equals the losses when your model is wrong. If your model is correct $50\%$ of the time your expected value is $0$. If you take the costs of opening and closing your positions into account you have to subtract these transaction costs form your gains and add them to your losses. Now your expected value of your trading stragey is negative. On average, you lose the transaction costs on each trade. So if you want to know how much prediction power a model should have to be profitable you could do a simple analysis of the average gains when your model is right and of the losses when your model is wrong. You can back out the prediction power needed from this equation. Then calculate how much prediction power you need in excess of $50\%$. Of course you could also take into account that transaction costs change with the market environment, that prediction accuracy is subject to estimation error as well as varous other issues. ## Answer by user1165686 (score 0) https://quant.stackexchange.com/a/14506 Imagine this: you roll a fair die. If you roll a 5 or a 6, you get 3 dollars. If you roll a 1-4, you lose a dollar. Positive expectancy, less than 50% correct. The simplest trend followers have such a profile. ## Answer by clearlyMakingFunOfQstAsked (score -1) https://quant.stackexchange.com/a/12867 Expected = win rate * avg winner + (1 - win rate) * avg loser - trading costs. if win rate = 1/2; avg winner = 10; avg loser = -5; trading cost = 1 E = 5 - 2.5 - 1 = 1.5 ## Answer by user3264325 (score -1) https://quant.stackexchange.com/a/12895 Turns out you can make money where you lose most of the time with Parrondo's paradox! https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=parrondo's%20paradox
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