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Holding a loser until it gets back to entry: it returns 60% of the time, still costs $80

A long on 1 lot EURUSD at 1.0850, account $12,000, planned stop at 1.0838. That is 12 pips, $120, which is 1% of the account. Price is now 1.0842, so the trade is 8 pips down, $80 floating, and the stop is 4 pips below.

The tempting move is to shift the stop to 1.0830 and wait for the trade to come back to 1.0850. Here are the numbers, assuming price from here has no drift in either direction, a fair assumption for a short window.

For a driftless walk the chance of touching one level before another is the distance to the far level divided by the total distance between them. With the stop at 1.0838, price needs 8 pips up or 4 pips down, so the chance of reaching entry first is 4 / 12 = 33%. With the stop at 1.0830 it is 12 pips down and 8 up, so 12 / 20 = 60%.

Now the dollars. Original stop: 33% of the time you get out at entry for $0, 67% of the time you lose $120. The average is minus $80. Moved stop: 60% of the time $0, 40% of the time minus $200. The average is also minus $80, exactly what you have floating now.

So the wider stop raises the hit rate from one in three to three in five, and changes the average result by nothing. What it changes is the loss when you're wrong, up 67% from $120 to $200, and the amount of the account tied to one idea.

Real markets have costs, which make both averages a bit worse than minus $80. A spread or commission is paid on any exit, so closing at entry gives a few dollars less than zero, and the stop level loses a little more than planned if there is slippage. Neither rule cancels that. Over 20 trades in this spot, the first rule produces about 13 losses of $120 and 7 exits at entry, and the second produces 8 losses of $200 and 12 exits at entry. Both add up to minus $1,600 before costs, so the cost line is the only thing left that separates them.

Does a bigger target change it? Put a 2R target at 1.0874, 24 pips above the original entry, with the original stop at 1.0838. From 1.0842 price needs 32 pips up or 4 down, and the chance of the target first is 4 / 36 = 11.1%. The average is 0.111 x $240 minus 0.889 x $120 = minus $80 again. A tight stop with a far target wins rarely and makes a lot when it does, and the average stays the same. Under a driftless walk, no combination of stop and target changes the average, only its shape.

The one thing that does change the average is drift. If you think price has a reason to rise from 1.0842, say the level that triggered the entry is intact, then holding has a positive average. But that is a judgment about the market now, and it has nothing to do with where you bought. A fair test is to ask whether you would open a new long right here, at 1.0842, with a stop at 1.0838. If the answer is no, the trade shouldn't be held, since the entry price is a fact about your account and not about the chart.

A rule I'd use is that a stop can only move in the direction that reduces risk, and I write the stop in before the order goes out. The numbers above explain why. After entry, the average outcome of any stop and target pair is fixed by where price is, so the only choices that help are cutting size or taking the loss at the planned level. If I ever want a wider stop, I treat it as a new trade with a smaller size, so the dollar risk stays at the original $120 or less.