Roulette probability based on last 10 rolls

Question

Let me preface by saying I'm not good at math. This question is purely from a mathematical standpoint. Thanks very much in advance!

Ok, so lets say I want to calculate the odds of a roulette wheel landing on red or black based on the last $10$ rolls. I realize that every roll is independent of each other and each roll has a $50\%$ chance of landing on red or black (assuming no $0$). But how could I calculate the probability of rolling a red or black on the $11$th roll if I know the previous $10$ rolls?

For example if the last 10 rolls had $5$ reds and $5$ blacks, I'd assume the $11$th roll would be a $50-50$ chance since there were an equal number of reds and blacks for the previous $10$ rolls (please correct me if this isn't correct). What if there were only $1$ red and $9$ blacks in the last $10$ rolls? How would I calculate the probability of getting a red or black on the $11$th roll? I'm assuming there's some kind of formula I could use to calculate this right?

Answer 1

A simple experiment will help convince you.

Assume a fair coin, and a sufficiently random tossing mechanism.

Suppose you believed that $10$ heads in a row induced a bias for the $11$-th toss.

But then it would be only natural to assume (in a new experiment) that there would be a bias for the $10$-th toss, assuming the first $9$ tosses came up heads.

Similarly, an experiment which waits for $8$ heads in a row should reveal a bias for the $9$-th toss.

You can see where this is leading . . .

Thus, suppose we start a new experiment where we require just $1$ head, and test the next toss for a bias. If there was a bias for $10$ heads, for $9$ heads, for $8$ heads, . . ., then surely there must be some bias for just $1$ head. Yes?

But now it's easy to get actual data. Toss a coin repeatedly. If one roll is a head, can you detect a bias for the next roll?

Hopefully, that experiment will dispel any pseudo-probabilistic superstition you may have had.

Bottom line: Coins have no memory. From the point of view of the coin, every roll is the first roll.

Answer 2

As you alluded to, these events are independent. The previous $10$ rolls have no bearing on whether the $11^{th}$ roll will be a red or black.

To give a more extreme example, if we flip a fair coin $100$ times and obtain $100$ heads, we still have a $50-50$ chance of getting heads on the next toss.

For two events $A$ and $B$ to be independent, the following must hold

$$P(A\cap B)=P(A)\cdot P(B)$$

We can rewrite this as

$$P(A)=P(A|B)$$

$$P(B)=P(B|A)$$

where $P(A|B)$ means the probability of $A$ occuring $given$ that $B$ happens.

Answer 3

"I realize that every roll is independent of each other ..."

Based on the rest of your question, respectfully I don't think you understand.

What this means is that the previous outcomes do not matter. At all. As in you could have a hundred reds in a row, and the probability of red on the next roll doesn't change.

Answer 4

As you say, an spins on a roulette wheel are independent events. The wheel has no memory, and your next spin will have nothing to do with your last spin or your last 10 spins.

Unless, you want to wander into Bayesian statistics. In which case, you would use the result from the previous 10 spins and say to your self, "this wheel has come up red 9 of the last 10 spins. There is about a 1% chance of that happening if this wheel is fair. Which is more likely, the wheel is rigged, or I have just witnessed an unlikely event?" Whereupon you must evaluate what kind of establishment you have walked into, and whether there is a greater than 1% chance that it is run by cheats. And if so, you might want to update your priors.

Answer 5

Science will tell you that the next roll is independent and previous rolls have no affect, yet we see 5 blacks in a row all the time, but never see 33 in a row our whole lives. This is because when you're looking at the probability of the next outcome INDEPENDENTLY, its true it has equal chance of being anything. But when you are looking at the probability of something being CONSECUTIVE, it is a whole different formula. The probability of 2 blacks showing up back to back WILL happen. But the probably of 33 consecutive blacks is such a low %, you could bet your life it wouldnt happen. So YES, you can calculate the probability of the current streak continuing. The calculation will give you a % chance of the sequence continuing black, which will get exponentially lower after each consecutive black, to the point that its decimal points of a % (time to bet on red).

If a meteor falls to earth, for this independent event we can all say that it could literally land anywhere, right? Right. So it lands and makes a hole in the ground.

But if another meteor comes to earth afterwards, what are the chances it lands in the same exact hole the 1st one did? Out of all the places on earth, the 1st hole is the least likely spot since it has already been struck. And probability agrees, because now we are looking at the outcome of a CONSECUTIVE event.

You cannot argue that the 2nd meteor has as good of a chance landing in the same hole as the 1st, instead of landing in the middle of the ocean.