Scalping and The Winning Combination

I had an interesting conversation with someone about scalping. By now most readers will know what I think about scalping and would rather stick a knife in the side of their heads than hear me repeat myself.

Part of the discussion centred around a trade that tries to take advantage of a volatile market that has no particular direction. The discussion led to combinatorics (the calculation of combinations), profit and loss stops, and trend trading.

Their argument was as follows

With regards to scalping being 50/50, let's say 10 minutes before the off I offer £10 at 6.0 and my offer gets accepted, and I place a back order at 6.2. Would you agree with me that there is a greater than 50% chance of my trade being completed successfully before the off if I don't have a stop loss in place.

Intuitively, I knew this to be 50/50. My answer was

If you put in a random lay at 6.0 and a back (at) 6.2 then in the long run 50% of the backs will be taken and with no stop loss you will lose money. If people are backing at 6.0 and taking your lay then I would think there is a slightly greater chance of the market going down than up. You are hoping for market volatility and that the spread marches up and down the ladder. The chances of that are again 50/50.

If they had asked, "In an upwardly trending market." Then I might have said something different. However, because I thought the question was about a volatile market then my answer is correct.

A further email was received

The market's possible next 4 up/down movements are as follows:


The first 4 scenarios result in my trade failing (assuming that I'm not matched fully when the market is at 6.0 - 6.2). The latter 8 scenarios are ones where my back order would be taken out. Does that not suggest that, if the market decides to go on a random walk, I'm far more likely to have a successful trade than not?

Do you see the error? How many combinations should there be? I make it 2^4 (2 directions to the power of 4 successive time periods) but I only see 12, which is why the questioner thought that they had a greater chance of the market being up after 4 moves. Here is the full list of combinations with statistics that will be explained later.

DDDD - 5.4, 5.3, 5.2, 5.1 - lose 8 (-1)
UDDD - 5.6 - win 1 (+1)
DUDD - 5.4, 5.5, 5.4, 5.3 - lose 4 (-1)
DDUD - 5.4, 5.3, 5.4, 5.3 - lose 4 (-1)
DDDU - 5.4, 5.3, 5.2, 5.3 - lose 6 (-1)
UUDD - 5.6 - win 1 (+1)
DUUD - 5.4, 5.5, 5.6 - win 1 (-1)
DDUU - 5.4, 5.3, 5.4, 5.5 - scratch (-1)
UDDU - 5.6 - win 1 (+1)
DUDU - 5.4, 5.5, 5.4, 5.5 - scratch (-1)
UDUD - 5.6 - win 1 (+1)
DUUU - 5.4, 5.5, 5.6 - win 1 (-1)
UDUU - 5.6 - win 1 (+1)
UUDU - 5.6 - win 1 (+1)
UUUD - 5.6 - win 1 (+1)
UUUU - 5.6 - win 1 (+1)

The questioner also assumed that each move would be equal (i.e. 1 tick) so I have additionally marked the combinations with the consequences of 1-tick profit taking and no stop. We also assume that the trader offered 5.5 and was taken and then offered to back at 5.6 rather than taking prices on the other side of the spread.

As we can see the trader had 10 wins of 1 tick but lost 22 ticks without the stop. If there had been a stop (let's say 1 tick) then the result in parenthesis would have been the outcome and we have the 50/50 that I expected; 8 ticks of profit cancelled out by 8 ticks of losses. The mathematics of probabilitity must be fully understood otherwise some serious mistakes can be made.

If you are thinking of scalping (and I don't recommend it) then you must understand that in a volatile market you will lose money (break even minus commission or worse if you are manually trading). In a trending market you shouldn't be tick scalping at all but trading the trend for multiple ticks. Feel free to print this article out and take it to one of those useless trading courses you are about to waste your money on.