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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.

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    $\begingroup$ A model that's right <50% of the time can be profitable. Many trend-following trading systems are only right 20-40% of the time, but are profitable because the winning trades are so much larger than the losing trades. $\endgroup$
    – Joshua Ulrich
    Jul 1, 2014 at 12:41
  • $\begingroup$ Yes, but how would such a strategy be foreseen and planned on, if not simply occurring by chance? I'm not asking for golden formulas. I asked about stop losses for helping with something as you mentioned (see below), but someone else suggested that moves should be considered independent of each other, rendering stop-losses as not very helpful in a prediction scenario. $\endgroup$
    – poorly_built_human
    Jul 4, 2014 at 2:22
  • $\begingroup$ OP, you're putting a lot of emphasis on knowingly forseeing the future. That's just not possible. Your question seems to imply a lack of understanding about basic probability. Perhaps you should take this kind of question over to stats.se. $\endgroup$
    – ch-pub
    Aug 26, 2014 at 4:59

4 Answers 4

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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.

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  • $\begingroup$ Apart from fixed transaction costs, it is not possible to gain more than lose with 50/50, because you would invest the same amount when you dont know the outcome? $\endgroup$
    – emcor
    Jul 1, 2014 at 9:34
  • $\begingroup$ Thank you for your answer. I will work on using your suggestion in backing out the average gains/losses. But just to confirm, the Sharpe and Information ratios are not applicable in any way? Also, I imagine implementing stop-losses will decrease the losses while allowing the gains to ride out as far as possible. $\endgroup$
    – poorly_built_human
    Jul 1, 2014 at 9:54
  • $\begingroup$ The problem I think every next move is independent. So when you stop a loss at some point, the next move might as well have been a gain that is stopped, and having had a gain in the past does not mean it will "ride out" in the next step aswell. Mathematically 50/50 corresponds to an arbitrage-free martingale, so no riskfree profit possible.. $\endgroup$
    – emcor
    Jul 1, 2014 at 12:11
  • $\begingroup$ @emcor The 50% rate is only the percentage of right classifications. Just imagine an algorithm that gets the lottery numbers right 50% of the time. See also the comment by Joshua Ulrich $\endgroup$
    – vanguard2k
    Jul 2, 2014 at 6:34
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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.

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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

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  • $\begingroup$ how do you define avg winner? and how can you expect to gain more than loose on average, if you know your prediction is equally right and wrong? $\endgroup$
    – emcor
    Jul 1, 2014 at 19:27
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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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  • $\begingroup$ Unfortunately: "Parrondo's games are of little practical use such as for investing in stock markets as the original games require the payoff from at least one of the interacting games to depend on the player's capital." -Wikipedia $\endgroup$
    – poorly_built_human
    Jul 3, 2014 at 13:01

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