## Probability of a String of Profits or Losses

Jun 21, 2008: 2:15 PM CSTHow does flipping a coin relate to trading probabilities? Simple, we can look at different probabilities of trading outcomes based on the distribution that results from a pure 50/50 chance outcome.

What do I mean?

Although the odds of a successful trade are rarely exactly 50/50, let’s assume that – for the moment – they are. Let’s say you take ten trades in a row. What can you expect to happen?

You can expect that 5 out of the 10 trades (or coin tosses) will ‘win’ (or, come up heads). What is the actual distribution like, though?

Did you know that the odds of having exactly 5 of the 10 trades (or coin flips) win is only 24.61%? That means that roughly 75% of the time, you’ll have a different result than what you expect!

To be fair, the probability of having 4, 5, or 6 winning trades (plus or minus 1 from the expected value) is 65%, which is in line with the standard “bell curve.”

What is the probability of having a run of 10 out of 10 trades be winners (or losers)? It’s only 0.10%. Rare, but it can happen.

Let’s look at the results from a randomized study I completed in Excel, based on flipping a fair coin 10 times, for 10,000 trials (something that might take months in the real world) and count the number of times ‘heads’ comes up in each of the 10 series of trials.

This will be similar to what you can expect if you took say 100 sets of 10 trades each and asked “how many trades can be expected to be winners?”

Let’s look at the distribution:

Remember, with odds of 50%, we would expect 5 throws out of 10 to come up heads.

We would expect the next most common values to be 4 or 6 throws out of ten to come up heads.

We would not expect many at all to turn up a series of 10 heads in a row (or 10 tails in a row).

Despite that, out of 10,000 series of 10 tosses, we had 10 runs of 10 tails in a row and 11 runs of 10 heads in a row.

**What does this mean for trading?**

If you can expect to win 50% of your trades, this would be similar to the distribution of possible outcomes, and their respective probabilities based on a series of winners or losers. Most of the time, you’ll fall one or two values away from the expected value (exactly 5 out of 10), but sometimes, you’ll get a string of losers or winners, just through probability alone.

What will happen to the distribution when I shift the odds up to having higher odds of a winning trade than losers? Recall that few situations produce exactly a 50/50 chance of producing a win over a loss.

Check back!

June 23rd, 2008 at 8:32 pm

Corey, what is the equation that is used to calculate the probability of a streak (X number of consecutive winners/losers) in a fair game of chance? Thanks in advance.

June 24th, 2008 at 8:59 am

AC,

I programmed in .50 for both 0 and 1 (tails and heads) and then used Excel’s random number generator.

I also compared the results with the expected values from the Binomial Distribution expected values (not shown) created by Excel with each of the 10 numbers and found extremely comparable values, only off by a very slight amount, which would be expected.

Shifting the probability to, say .60 for heads and .40 for tails only shifts the distribution up one number, and the resulting distribution is equivalent.

Thank you for the comment.