First some context, I'm not a mathematician, not even close (as you will soon see) I do grasp some things about it but in a need to know basis, so plain english answers are appreciated (too).

I can't figure out this one even though some have tried very hard to explain/convince me.

So let's say I'm on a roulette (let's leave out the zero, or make it a coin for that matter).

The roulette hits 5 (or any amount of) times red on a row.

My question is why should't I bet to black (or better said why it is the same the one I pick).

I DO UNDERSTAND THE BASIC PRINCIPLE, that there is a material reality which makes the ball randomly fall on any of the two colors (which gives the options a 50-50 chance).

But, I also understand that if you "say": I bet red will come out six times on a row, you do have a very low probability of that happening so:

How is it that if you have seen the ball fall 5 times on a row on red, and you bet on red, you are NOT betting on 6 times on a row on red (that does have lower probability).

Thanks in advance!

I hope I was clear enough with my question, if not, please ask for any clarifications needed.

Sorry for the VERY plain English, feel free to modify or suggest a change to anything that may be misleading.

  • $\begingroup$ This is my first question here and I could not find any other suitable tag, sorry! $\endgroup$ – Trufa Dec 22 '10 at 17:27
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    $\begingroup$ A side remark: If you know that the roulette is fair, then it doesn't matter if you bet on red or black. However, if all that you have observed is that the ball has fallen on red five times in a row, then perhaps the most rational thing to do is to suspect that the roulette is biased towards red, and keep betting on red. $\endgroup$ – Hans Lundmark Dec 22 '10 at 20:16
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    $\begingroup$ Meat for though: Question 1: Which of these sequence of six roulette balls is more probable: RRRRRR RRRRRB RBRBRB ? Answer: all are equally probable. Question 2: Which is more probable, that in six balls we get 3 red/3 black balls or that we get all 6 red? Answer: 3 red/3 black is much more probable. Think until you see no contradiction here. $\endgroup$ – leonbloy Oct 1 '12 at 21:38
  • $\begingroup$ wheels are biased, so every wheel has a memory. $\endgroup$ – user48195 Nov 4 '12 at 20:40
  • $\begingroup$ @leonbloy I don't get the answer for Question 2. why are they not equally probable? both being 1/64? $\endgroup$ – olala Jun 19 '15 at 17:18

Lots of people have trouble with this. If you bet before any spins on six in a row, the chance of winning is 1/2^6=1/64. But having seen the five reds, as you say, the chances are 1/2 each on red and black for the next spin. It is also true for any specific series of six spins, at the start the chance of that series is 1/64. If you were to bet on RBBRRB, that is also 1/64. Now that you have seen five reds, there are only two series of six that are possible: RRRRRR and RRRRRB. They each started out 1/64 and are now each 1/2.

Hope this helps

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    $\begingroup$ I can't believe it, you've made me seen the light!! :) no, but seriously you can't imagine how much "trouble" I had with this!! Thank you very much! I was missing a little piece of the puzzle, but it made it incomprehensible, I could understand that the probability was less (when betting to six) and the got bigger and bigger, which is actually a concept I (of course) undertand but could not apply it here. Bottom line, thanks a lot! $\endgroup$ – Trufa Dec 22 '10 at 17:39
  • $\begingroup$ I'm more more of an old hand at probablilty than the OP, and this answer was enlightening to me (I'm slightly embarassed to admit). Very well put. I'd upvote twice if it were possible. Thanks! $\endgroup$ – Todd Wilcox Dec 21 '12 at 8:30
  • $\begingroup$ Thanks for the question and explanation. Just to make sure I have it right.... I've always wondered why I rarely see 15 reds in a row, for example. The answer sounds like, yes, it is rare to see 15 reds in a row, but it is equally rare to see a combination of any 15 color arrangements. The fact that I target "all red" is simply easier to remember than a combination of various colors. The catch is the next color is equally improbable as the16th color in a given series. Is that right? $\endgroup$ – Sean Nov 6 '16 at 18:47
  • $\begingroup$ That is correct. Intuitively all jumbles of about half red and half black look the same, so we lump them all together. If you focus on one particular run of 15 results, it is just as unlikely to arise as 15 reds in a row. $\endgroup$ – Ross Millikan Nov 6 '16 at 20:37

There's an analogy that I like. Suppose I'm afraid to take planes as it's possible that some crazy terrorist brings a bomb onto the plane. Now it's really improbable that there are two bombs on the plane that I'm going to take. So I'll bring a bomb myself (which I of course won't detonate); then it's virtually impossible that anything happens. Well ...

I don't know, but maybe this helps.

  • $\begingroup$ Ok I (sort of) see your point, but I already understood this. What I mean is I knew what what I was thinking was incorrect, I just wanted to understand why. Thanks :) $\endgroup$ – Trufa Dec 22 '10 at 17:43

The events "given that the roulette hits $5$ reds, the next hit is red" and "the roulette hits $6$ reds" are different. The first has probability $\frac{1}{2}$ because different spins of the roulette are independent and the second has probability $\frac{1}{64}$ by the multiplication principle. (This is the same as the probability of "the roulette hits $5$ reds and a black," which is exactly the point of the first computation.)

For questions like these the general principle for getting to grips with how your intuition is failing is "when in doubt, list out all the possibilities."

  • $\begingroup$ Thanks for the answer, but I sort of new that, I knew I was wrong to think the probability would be less than 1/2 but I needed to understand the why. I do understand now :) $\endgroup$ – Trufa Dec 22 '10 at 17:48

The best explanation I have heard for this is: The roulette wheel has no memory. When you spin the wheel it does not know that there have been 5 reds in a row and it is due for a black. The odds are 1/2 each time.

  • $\begingroup$ Exactly, so if you make a second bet with double money your odds are still the same. The only thing in your favour is that you are saying "I can't be wrong twice in a row so I will double my money" $\endgroup$ – John Magnolia Mar 15 '13 at 17:00

Why is it so unlikely to get 6 reds in a row? Well, you have to cross your fingers and say "Hope the first one's red!" And if you get lucky and that one comes up red, then you have to cross your fingers and say "Hope the second one's red!" And if you get really lucky and that one also comes up red, then you have to cross your fingers and say "Hope the third one's red!" And so on.

On the other hand, if five reds already came up, all you have to do is cross your fingers and say "Hope this one's red!" a single time. In one case, you're hoping for "red red red red red red"; in the other, you don't care about the first five (they're already done with); you're just hoping to get "red" once.

To put it another way: would you rather go for six reds from scratch, or would you rather go for six reds after five have already come up red?


If the roulette is "fair" like China power said, no matter how many Red hit on a row it doesn't matter because the chance of hitting red/black is 50-50 and it's all pure luck when playing this game.


Why wouldn't a series have an increasingly lower probability of extending than a single event has of happening once?

Obviously, each unique spin has a probability of 50% (academically, by excluding the greens) for either color.

However, the overall proportion of total reds vs blacks should be 50%. So every time another of the same color is added to the group, the probability of the group/sample containing ONLY that one color decreases. Is there really a 50% probability that a sample size of 100 (e.g.) will contain 100 reds or blacks? When a person gambles on the roulette spin, he may be betting FOR a single spin, or AGAINST a series.

Isn't saying that the wheel has no memory like saying the wheel hasn't had a haircut in (x) weeks? A gambler isn't betting on the wheel, he's betting on the statistics. Past data would appear to be relevant due to the sampling.

The trick is to be in the right place at the right time: at what point will the group or sampling attempt to correct itself to ensure an equal distribution of reds and blacks? Because obviously it is inevitable.

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    $\begingroup$ You have either not read or not understood the other answers. $\endgroup$ – Rahul Aug 21 '12 at 1:12
  • $\begingroup$ «at what point will the group or sampling attempt to correct itself to ensure an equal distribution of reds and blacks? Because obviously it is inevitable.»: not really. Or to start from your premise of that there's no reason for «Why wouldn't a series have an increasingly lower probability of extending than a single event has of happening once?»: the series already happened. And while «the overall proportion of total reds vs blacks should be 50%», there's an infinite amount of plays to correct that, so any finite sequence of repeating blacks or reds is possible. $\endgroup$ – JMCF125 Feb 5 '14 at 11:37

It is correct that the different spuns are independent from each other when playing the roulette - however it is not entirely correct that roulette has no memory.

If the wheel was spun a million times you would see red and black occur the same amount of times.

If you have a case of five consecutive reds (and these are all your observations) the most rational choise would be to play black - HOWEVER - it is not the rationel choise because of the unlikely occurance of 6 reds in a row, but because you bet on the assumption that the roulette is out of its long term equilibrium.

6 reds in a row is no more unlikely to happen than five reds followed by a black - so that argument is not valid - both is 0,473^6 (47,3% chance of both black and red, and only one possible combination in both cases).

Now if you see five consecutive reds after ten consecutive blacks, and these are all your observations, the most rational choice would be to play red - all we know is that on the long term they will occur the same amount of times!...

You do not beat the roulette this way though ;)

  • $\begingroup$ Did you really intend to write your second sentence? $\endgroup$ – Rick Decker Sep 20 '12 at 14:47
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    $\begingroup$ This doesn't make any sense. You yourself said that the spins are independent of each other. Therefore, what has already happened has no affect on what will happen in the future. Even if there were 100 reds in a row, the odds of red and black are exactly the same on the next spin. So, there would be no inclination to choose one over the other. So, your sentences that say things like "... the most rational choice would be to play red ..." are not at all true. The fact that, over the long term, the number of reds and blacks should be even does not mean you can predict what comes next. $\endgroup$ – Graphth Sep 21 '12 at 12:39

protected by Pete L. Clark Feb 18 '13 at 9:00

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