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Five same colour M5 candles in a row: how often chance alone produces it in 200 candles

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Fairman
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Five same colour M5 candles in a row: how often chance alone produces it in 200 candles

Post by Fairman »

Five same colour M5 candles in a row: how often chance alone produces it in 200 candles

How many times have you seen five green M5 candles in a row and called it a trend? A fair coin gives a baseline to compare against.

Treat each candle as green or red with a 50% chance, independent of the others. Any five given candles share one colour with probability 2 x (1/2)^5 = 1/16, which is 6.25%. A new run of five or more starts at a particular candle when the one before it was the opposite colour and the next five match: 1/2 x 1/16 = 1/32. Over 200 candles, about 17 hours of M5, that gives roughly (200 minus 5) / 32, plus a small start-of-sample term, which is 6.2 runs of five or longer.

Longer runs are rarer but they aren't rare. Eight or more turns up about 0.76 times in 200 candles, and the chance of at least one such run in that stretch is near 54%. Ten or more appears in about 17% of 200 candle stretches. I ran a simulation of 20,000 sequences to check the algebra and it gave 6.15 runs of five and 0.75 of eight.

Real candles aren't coin flips. Some closes are equal, volatility clusters, and the sessions have drift. But it means seeing five or six of one colour a few times a day isn't evidence of anything on its own. A rule such as "enter after five green candles" would trigger about six times in those 200 candles even if the market had nothing in it.

What matters is what the sixth candle does, counted across your own history.
It’s Fairman :geek:
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Fairman
Posts: 6757
Joined: Tue Jul 21, 2026 7:11 am
Location: Abuja

Re: Five same colour M5 candles in a row: how often chance alone produces it in 200 candles

Post by Fairman »

The sixth candle is the entire test. In the coin model it's green 50% of the time after five greens. So a trend rule needs the real figure to be clearly above 50%, enough to pay costs as well, say 56%. With 40 occurrences, 56% is 22 greens, and the standard error of a proportion near 50% is the square root of 0.25 / 40, which is 7.9 points. So 56% sits less than one standard error from 50%, and tells you nothing. With 400 occurrences the standard error is 2.5 points, and the 95% range is about plus or minus 5. That's why this kind of question needs history in bulk, not a week of charts.
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Fairman
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Re: Five same colour M5 candles in a row: how often chance alone produces it in 200 candles

Post by Fairman »

Getting 400 occurrences isn't hard. One pair on M5 has about 288 candles a day, so 22 trading days is 6,336 candles. At 6.16 runs of five or more per 200 candles, that's around 195 triggers a month on a single pair, and several times that across a handful of pairs. A year of M5 history for one symbol holds well over 2,000 of them. The slow part is deciding the definition, not finding data. Write it down before looking: colour by close against open, equal closes count as no colour or break the run, runs counted once at the moment they reach five.
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Fairman
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Re: Five same colour M5 candles in a row: how often chance alone produces it in 200 candles

Post by Fairman »

The fade version of the idea has the same arithmetic, so don't assume that betting against the streak is any different. A rule "sell after five greens" also triggers about six times in 200 candles, and it also needs the sixth candle to be red clearly more than 50% of the time to beat costs. If the real figure is 51% green and 49% red, the trend rule loses to costs and the fade rule loses to costs too. Both can look good on a short sample, which is exactly the sort of sample where a chart review is most convincing and least reliable.
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Fairman
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Re: Five same colour M5 candles in a row: how often chance alone produces it in 200 candles

Post by Fairman »

Here's the sheet layout, with prices in B (open) and C (close), starting on row 2. Column D: =IF(C2>B2,1,-1) for colour. Column E: =IF(D2=D1,E1+1,1) for run length so far. Column F: the colour of the next candle, =D3 on row 2. Then filter column E for exactly 5 and count how many of those have F equal to D. Divide by the number of rows with E = 5. Compare it with the share of green candles overall, which isn't exactly 50%, because the baseline is that overall rate and not a coin. The whole job is about ten minutes once you've pasted the data in.
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