The Gambler's Fallacy is one of the most common misconceptions in roulette:
the assumption that past random events influence future probability.
A mathematical analysis based on patterns of simple chances using the example of a coin toss:
A coin is tossed — 4 times.
Intuitively one would expect that each side probably comes up 2 times.
So let's verify this:
The following 16 combinations are possible (H = Heads, T = Tails):
HHHH (0 times T), 1 combination, probability = 1/16
THHH HTHH HHTH HHHT (1 time T), 4 comb., prob. = 4/16
TTHH THTH THHT HTTH HTHT HHTT (2 times T), 6 comb., prob. = 6/16
HTTT THTT TTHT TTTH (3 times T), 4 comb., prob. = 4/16
TTTT (4 times T), 1 comb., prob. = 1/16
This can also be calculated differently:
The prob. of getting T 4 times in a row is (1/2) ^ 4 = 1/16. Same result.
( ^ = to the power of, exponent)
So in fact it is most probable that T comes up 2 times and H 2 times.
Or in other words:
There are more "mixed" than "uniform" sequences, i.e. more combinations with 2 T than with 0 T etc.
Why past tosses are irrelevant
So far, so clear. Now the "believers" — a classic case of Gambler's Fallacy — argue that this could be used to one's advantage.
The idea is to play 2 times, and if H comes up twice, one bets that H will not come up again.
This is because it was originally calculated that H twice is more probable than H three times.
The problem with this reasoning is that 2 tosses have already been made.
Of the combinations listed above, only those starting with HH remain:
HHHH (0 times T), 1 combination, prob. = 1/4
HHTH HHHT (1 time T), 2 comb., prob. = 2/4 = 1/2
HHTT (2 times T), 1 comb., prob. = 1/4
So it is by no means the case that T twice, i.e. HHTT, would be more probable than HHHH.
In fact, both events are equally probable. Most probable now is one H and one T.
This can also be calculated differently.
The prob. of getting T 2 times in a row is (1/2) ^ 2 = 1/4. Same result.
Above all, one can see that it makes no difference at all how many times H has come up before —
this is the core of the Gambler's Fallacy: the independence of random events.
Perhaps one thinks now:
OK, then one more toss, and if H comes up, one bets on T and vice versa.
Assume H comes up.
Of the combinations only those starting with 3 times H remain:
HHHH, 1 combination, prob. = 1/2
HHHT, 1 combination, prob. = 1/2
Now the combinations HHHH and HHHT are exactly equally probable — namely exactly equal to the prob. of tossing H or T, i.e. 1/2.
This can also be calculated differently: The prob. of getting H in one toss is (1/2) ^ 1 = 1/2.
The same can be done with 30 tosses.
It takes longer to write down, but leads to the same result.
Of course, the prob. of getting H 30 times in 30 tosses is very low, namely (1/2) ^ 30 (= 1/1,073,741,824 = 0.000,000,001)¹
But if H has already come up 28 times,
of the many originally possible combinations only those starting with 28 times H remain:
28H+HH, prob. = 1/4
28H+HT and 28H+TH, prob. = 1/2
28H+TT, prob. = 1/4
28H+HH and 28H+TT are therefore equally probable.
This can also be calculated differently:
Two tosses are made (what happened before is completely irrelevant),
so the prob. for 2 times H = (1/2) ^ 2 = 1/4. Same result again.
The roulette example
Now one might say, roulette is not a coin toss.
The principle of the independence of random events is however exactly the same — one just writes longer when listing all possible combinations
(that is why the coin example is used here), and the prob. is (1/37) instead of (1/2).
Otherwise everything is as before. Previous spins do not influence roulette probability.
Anyone who does not believe this can write down all possible combinations in 10 spins.
And briefly regarding the argument that one plays e.g. 5 spins and not just one:
Of course it is very unlikely that 28 comes up 5 times in a row. But it always is, namely (1/37) ^ 5 (= 1/69,343,957 = 0.000,000,014)¹
Regardless of how many times 28 has come up before.
There is absolutely nothing to be gained by waiting until 28 has come up five times in a row — a typical Martingale fallacy.
One can just as well bet from the outset that 28 will not come up 5 times in a row.
Conclusion
It makes no difference. It is exactly the same.
Past roulette results do not influence future probabilities — this is mathematically provable and the core of the Gambler's Fallacy.
With kind permission of the author Michelangelo. ¹ Note by the editor.
The original version has been revised. Source: Board DC's Campus (R.I.P.)