When the roulette has hit 5 reds why shouldn't I bet to black?

Question

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!

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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.

Answer 1

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

Answer 2

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."

Answer 3

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.

Answer 4

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?

Answer 5

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.

Answer 6

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.

Answer 7

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 ;)

Answer 8

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.