In roulette, is the frequency of getting long sequences of reds lower than that of shorter sequences?

Question

In a roulette game that continues indefinitely, is it correct to say that achieving a sequence of 100 consecutive reds will occur far less frequently (perhaps once in every million spins) compared to a sequence of 10 consecutive reds?

If the probability of getting a long sequence of reds decreases as the sequence gets longer, why is it considered a gambler's fallacy to believe that black is more likely after a long sequence of reds? If a sequence of 101 reds is less likely than a sequence of 100 reds, why is it incorrect to assume that the current sequence is more likely to be 100 reds long and therefore the next spin will be black?

Edit:

I'm aware that those are independent events and the probability has no memory and thus must stay the same. I'm asking how is that fact compatible with stating that longer sequences occur less often.

Moreover, I'm not saying that according to the law of large numbers it should balance out and therefore it is more likely to be black on the next spin. I am saying that it is more likely that we are on a squence of 100 reds than a 101 sequence, since it occurs at a higher rate, and if it is more likely to assume that now it is the shorter sequence rather than the longer one, then doesn't it logically follows that the next spin is more likely to be black?

Answer 1

Given that you're asking here, I assume you are talking about a mathematically ideal roulette wheel that produces red with probability 18/38 and black with probability 18/38, independently each time.

The key questions are when are you making the calculations, and what precise outcomes are you considering.

After you've see 100 reds in a row, the chance of the next spin producing red is 18/38 and the chance of producing black is 18/38. So after you've seen 100 reds in a row, the chance of 100 reds in a row and 101 reds in a row are exactly the same. When you started, the chance that the first 101 spins will all produce red and that the first 100 will be red and the 101st will be black are also equal.

So what is it that's less likely? If you take a sequence of roulette spins, it is less likely for that sequence to contain a run of 101 red than a run of 100 red. It's less likely because any subsequence of 101 spins can only be all red in one way, but it can have a run of 100 red in two ways (Brrrr or rrrrB). But if you know you have had 100 red already after the 100th spin, you can't be in the (Brrrrr...) scenario, so there is one possibility each way. The numbers increase as you allow more non-red: a run of 99 red can happen in five ways (BrrrB, rBrrrr, BBrrrr, rrrrBB, rrrrrBr).

If you were betting on "some time tonight, there will a run of $N$ red" then the chance of that being true will be higher for $N=100$ than for $N=101$ (though still very very very small). If you bet on "the next $N$ spins will all come up red", the chance of that is higher for $N=100$ than for $N=101$ (though still very very very small).

All that was for an 'ideal' roulette wheel, where the spins are independent and the probability for each slot is the same. An ideal roulette wheel is a very unnatural device, which is one reason their behaviour is unintuitive. Even in casinos, real roulette wheels are meaningfully imperfect. If you saw 100 reds in a row, you would be well advised to find a different casino. 100 reds in a row from a fair roulette wheel is enormously less likely than a crooked casino.

Answer 2

In a roulette game that continues indefinitely, is it correct to say that achieving a sequence of 100 consecutive reds will occur far less frequently (perhaps once in every million spins) compared to a sequence of 10 consecutive reds?

Yes

If the probability of getting a long sequence of reds decreases as the sequence gets longer, why is it considered a gambler's fallacy to believe that black is more likely after a long sequence of reds?

because "A sequence of 100 consecutive reds is very unlikely" does not imply that "the probability that the next one is red given the previous 100 were red is lower than the probability that the next one is black given the previous 100 were red".

(same for any number of consecutive reds)

You're conflating joint probability (the event "101 reds", specified before seeing any outcomes is unlikely) with conditional probability ("the probability of one more red, given you just saw 100 reds" is not small)

If a sequence of 101 reds is less likely than a sequence of 100 reds, why is it incorrect to assume that the current sequence is more likely to be 100 reds long and therefore the next spin will be black?

Let's assume for simplicity that there are no greens (we just re-spin if we get one). A very similar analysis works if you include the green zeroes, the individual outcome probabilities just go to $\frac{18}{37}$ or $\frac{18}{38}$ instead of $\frac12$. Let us further assume that the outcomes are independent (casinos work hard to assure that this is the case; they want a balanced 'fair' wheel whose outcomes are not dependent from spin to spin).

P(101 reds, starting from now) = $\frac12^{101}$ = P(100 reds starting from now) x P(1 red)
$\, =\frac12^{100} \times \frac12$

So it's half the probability of 100 reds.

But now, P(one more red given you just saw 100 reds) = $\frac12$.
The same as P(black, given you just saw 100 reds).

Indeed, P(RRRR.....RB) = P(RRRR.....RR) and P(R|RRR....R) = P(B|RRR...R)

(here read "|" as "given we just saw")

This is all perfectly consistent (and matches very well with actual behavior).

Note that every other specific sequence of 100 R's and B's (there are $2^{100}$ sequences of R's and B's of length 100 in all) are just as unlikely as all-R's.

The fallacy comes in going from "P(RR...RR) is unlikely" to "P(R given an unlikely event happened) is lower than P(B given an unlikely event happened)"

NOTHING compensates for the excess count of reds. NOTHING. It's still as probable as black. If you doubt the wheel is fair, however, you should bet on red.

If you find the present discussions insufficient, the search tool will turn up many explanations of the gamblers fallacy with very similar questions to yours.

Answer 3

Because, after you did your 100 spins (which all came up red), the probability of getting all these 100 spins red is ... $1.0$. The fact that 100 red events occured in a row just before has no influence on the 101th event; the probability of red is still 48.6% (in Europe, and the ROW), and 47.4% in the USA. That is, before you spin; after you spin, that event will have probability 0, or 1.
This is known as the independence property of random events, i.e. previous events do not influence the probability of upcoming events.

Note that not all random events are independent. Examples of independent events are coin toss (even if unfair; the coin will remain just as unfair after 100 events, or after 1), or roulette even/odd or red/black, or toss of a die (again, even if unfair), or drawing a heart from a deck of card (if you replace the card in the deck; will still be 1/4). Examples of non-independent events are drawing a heart from a deck of card, if you do not replace the drawn card in teh deck. That probability is 12/52=25% for the first card, but 11/51=23.5% for a heart on the 2nd draw.

In your case, the fact that the roulette ball landed in red does not in any way change its probability of landing again in red at the next spin. So the 2 events are independent. So, even after 100 red spins, the probability of yet another red remains the same. Hence it is a gambler's fallacy.

Now, you may ask, "but the law of large numbers means that the proportion of reds should average to 48.6%" (let's play in Europe, the odds are better), "so we must be getting more black in the subsequent spins?". If you expect that the roulette will "make it up" over the next few hundred spins, it will not. You just have to wait for more spins for this to happen; it will still converge to the expected mean, just a little slower because of this fluke (and it is a fluke; the probability of 100 reds in a row is ${4.6}$ $10^{-32}$). After only, e.g., 200 spins, this fluke will still heavily influence the overall average, which will be different from 48.6%. But after, e.g., 10,000 it will not anymore (that sequence is now only 1% of all the spins).

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Edited 9/15/24 to answer OP's edits to his/her question

Instead of using 100 vs. 101 spins, let's use, for simplicity, 1 vs. 2 spins. The odds of you being able to correctly "predict" a sequence of 1 are 0.486 (whether you predict red or black). Now, the odds of correctly predicting a sequence of 2 (where the order matters) will be $0.486^2=0.236$, indeed lower. But this probability applies to all 4 possible sequences of 2 (rr, bb, rb, br): all these 4 sequences have the same likelihood to occur. So, the probability of being able to predict a sequence of length $n$ goes down exponentially as $n$ goes up. But once a sequence of $n$ has occured (no matter what it is, no matter how random), then the probability for that sequence of $n$ is $1$ (because it has occured), and the probability for another red (or, for that matter a black) is now just 0,486. So if you happened to have correctly predicted the first spin, your chances of correctly predicting now a sequence of 2 are now $~2.06$ times smaller ($\frac 1 {0.486}=2.06$), because your chance of correctly predicting a spin is 0.486. So the fact that spins are independent, and not influenced by previous spins is completely "compatible with stating that longer sequences occur less often". Every time you add 1 spin to your sequence, you cut your odds by ~about 2.
Note that you would a somehow similar situation even if your random events were not independent. Take the example of drawing (w/o replacement) cards out of a deck; correctly predicting a sequence of n draws is more likely that a sequence of n+1. But actually calculating these odds becomes very messy, because they depend on the n previous draws, not all sequences have the same odds, etc...

Answer 4

You are right: once you have reached a sequence of 100 reds, the probability of extending the sequence is the same as ending it. But that doesn't mean you are equally likely to get a sequence of 100 or 101. Because if you extend it, you might not stop there. You might get 102, 103, etc. The probability of getting a sequence of exactly 100 is the same as the probability of getting a sequence greater than 100, but the latter is split among many subgroups, such that the probability of getting exactly 101 is less than the probability of getting 100.

You can see this pan out clearly in a simulation, albeit looking at much shorter sequences. I simulated drawing red or black with equal probability 10000 times. Of course, the numbers don't match up precisely, but they are close enough.

 x  runs with length = x  runs with length > x 
 1         1266                   1257
 2          628                    629
 3          301                    328
 4          154                    174
 5           89                     85
 6           43                     42
 7           17                     25
 8           13                     12
 9            8                      4
10            1                      3
11            2                      1
12            1                      0

(I was counting only red runs)

Answer 5

I'm asking how is that fact compatible with stating that longer sequences occur less often.

They're compatible in the most boring way imaginable. Imagine you've already observed N reds in a row, for any N (including 0. 0 is fine.)

The probability of the current sequence being at least N reds is 1, because you already know it's true.

The probability of the current sequence being at least N+1 reds is equal to the probability that the next spin is red (if it's not, the sequence ends at N).

The probability that the next spin is red is, by assumption, less than 1 (on a standard roulette wheel it's 18/38).

Therefore the longer sequence is less likely than the shorter sequence. This doesn't require "memory", it's always true unless your roulette wheel only has reds (in which case it produces one sequence of reds which is infinite).

Answer 6

You are saying that:

• a longer run of red is less likely than a shorter run of red, and thus 101 reds is less likely than 100 reds
• at any point, getting either red or black on the next spin are equally likely

and that these two points are contradictory, as after 100 reds, having equal chances of red or black results in equal chances of getting a streak of 100 or 101 reds. At first glance, this does seem like a contradiction. However, the contradiction disappears when you realize you are comparing the situation of getting exactly 100 reds and no more, to the situation of getting at least 101 reds and maybe more. If we make the comparison fair by comparing exactly 100 reds to exactly 101 reds, both your premises can be true without any contradiction.

Let's say you've spun 100 times and gotten red every time. Let's also assume this is a roulette wheel with only red and black, so the chance of getting red next is 50% and the chance of getting black next is 50%. What's the chance of ending up with a streak of 100 reds? That would require your next spin to be black, so you have a 50% chance of ending with 100 reds. What's the chance of ending up with a streak of exactly 101 reds? That requires you to spin a red and then a black (the second spin must be black or you'd be on a streak of at least 102 reds). Having the next two spins go correctly has a 50% × 50% = 25% chance.

Thus, we've assumed one of your premises to be true (equal chances of getting red and black on each spin) and your other premise naturally worked out to be true (50% chance of being on a 100 red streak, and 25% chance of being on a 101 red streak, so longer streaks are less likely than shorter streaks). No contradiction!