This is by no means an attempt to promote high frequency trading (HFT). In fact, the HFT scene has become so crowded that even real pro shops are having a hard time, let alone retail traders with their home brew system (By the way, here is an interesting read. Judge for yourself how much to believe). So why should a trader trade more?
Because it's about how confident you can be in making a conclusion. If you make one trade in your life and it turns out winning, your rate of winning is 100% and yet the only legitimate conclusion you can make is...well, that you don't have enough data point. In the spirit of Bayesian statistics as was used here, we can formulate a simple analysis that hopefully would provide some insights.
Suppose that you are a manager about to hire a new trader for your team. There are two candidates, Andrew Flook who had 5 winning trades out of 5 last year; and Bob Luckson who had 70 winning trades out of 100 last year. Who should get the job? Since Andrew has a 100% winning rate and Bob has only 70%...are we missing something?
Let's cast this in the light of Bayesian statistics. Of course I would have to make some assumptions:
Winning probability of a fluke trader = p = 0.5
Winning probability of a skilled trader = q = 0.7
Total number of winning trades = W
Total number of losing trades = L
Unconditional probability that any trader is a skilled one = s = 0.2
The last assumption about a trader being a skilled one is based on the anecdotal evidence that only about 20% of all traders are profitable in the long run. So, if $R$ stands for the trading Results of last year, $S$($F$) stands for the trader being a Skilled(Fluke) one, we have
$$ P(R|S)P(S) = q^W (1-q)^L s $$
$$ P(R|F)P(F) = p^W (1-p)^L (1-s) $$
$$ P(S|R) = \frac {P(R|S)P(S) }{P(R|S)P(S) +P(R|F)P(F) } $$
Plugging in the numbers, we find that Andrew (the 100% guy) has a 57% chance of being a skilled trader, while Bob enjoys a 99.9% chance.
