# What is the probability of the roulette ball landing in a black slot?

There have been $15$ consecutive red. What is the probability of the roulette ball landing in a black slot next?

What I did --

probability of landing on red is $\dfrac{18}{38}$

so there have been $15$ red.

$P(\text{landing on red})^{15} \cdot P(\text{landing on black}) =\left(\dfrac{18}{38}\right)^{15} \cdot \left(\dfrac{18}{38} \right) = \left (\dfrac{18}{38}\right)^{16}$

but it can be just reduced to $\dfrac{18}{38}$

so it's just $\dfrac{18}{38}$ (??)

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Is there any data about the probabily of that roulette landing here or there if we already know what it landed in the last play(s), or are these independent events? – DonAntonio May 8 '13 at 10:10
@DonAntonio : Hello. I don't think there is other data. The problem is "you walk up to roulette table and see there have been 15 consecutive red winners before you got there, and you decide to place a bet on black. what is the probability of ball landing in a black slot for this bet?" – hibc May 8 '13 at 10:16
Irrespective of the conceptual problems (see my answer), shouldn't it be $18/37$ instead of $18/38$? What's the $38$-th slot? – joriki May 8 '13 at 10:24
@joriki. 0 and 00 for green. 1 - 36 for black and red. So I thought 2+36 = 38 total possible ways. – hibc May 8 '13 at 10:28
@hibc: Interesting -- it turns out there are different European and US versions of roulette. – joriki May 8 '13 at 10:30

Ok @hibc, then we must, I guess, exercise a "common sense" probability thing, assuming the roulette is fair, of course. Define

$$R_i:=\text{the roulette lands in red at the i-th spinning}\;,\;\;B_i=\text{similar but with black}$$

We want to calculate the probability

$$P\left(B_{16}/\left(\bigcap_{k=1}^{15}A_k\right)\right)=\frac{P\left(B_{16}\cap\left(\bigcap_{k=1}^{15}A_k\right)\right)}{P\left(\bigcap_{k=1}^{15}A_k\right)}=\frac{\left(\frac{18}{38}\right)^{16}}{\left(\frac{18}{38}\right)^{15}}=\frac{18}{38}$$

which, of course, doesn't surprise since if the roulette is fair then the outcome of each spinning is, or should be, completely independent of what happened before...

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Why are you also assuming $38$ slots? Am I missing something about roulette? I thought it has $18$ black slots, $18$ red and one green zero. – joriki May 8 '13 at 10:26
As far as I remember, the roulette I used to play with as a kid had zero and double zero and the usual 36 numbers... – DonAntonio May 8 '13 at 10:27
Interesting -- it turns out there are different European and US versions of roulette. – joriki May 8 '13 at 10:30
Thank you very much! – hibc May 8 '13 at 10:31
Any time, @hibc – DonAntonio May 8 '13 at 10:32

This question cannot be unambiguously answered using the given data. It can be interpreted in two different ways. In one interpretation, the roulette is an ideal roulette, and in this case the point of the question is that the probability of the ball landing in a black slot is $18/37$ regardless of what happened before, because the results of different spins of an ideal roulette are independent and uniformly distributed over the $37$ slots, of which $18$ are black.

In another interpretation, the roulette is a real roulette and you're trying to find out how much it's biased towards red. In this case, the answer isn't determined by the given data, since one would have to know your prior probability distribution for the bias. If you suspect the casino owners of being crooks, $15$ reds in a row would lead to a high likelihood of the roulette being very biased, whereas if you know them personally and know that they've done everything they could to make the roulette unbiased, $15$ reds in a row are probably just a statistical fluke. Without knowing your prior assessment of the situation, nothing can be said about this mathematically.

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THank you very much – hibc May 8 '13 at 10:32
yeah, this sounds like the answer. agreed! – olala Jun 19 '15 at 17:11

A ball will not 'remember' that in the past it has picked out red for its landing for 15 times (or whatever). The probability of landing in black will be 18/38 again (if that was the case and no changes are applied). No calculations are necessary.

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This is a physical statement, not a mathematical fact. Mathematically speaking, this is an assumption of independence that we usually make in treating situations like roulette; it cannot be derived mathematically. – joriki May 8 '13 at 10:28
Thank you very much ! – hibc May 8 '13 at 10:30
Yes, joriki. I agree if you call it an assumption. When we make a mathematical model of some situation then assumptions are inevitable. Mathematics comes in when the model is there and leaves again after translation of the conclusions back to the situation. – drhab May 8 '13 at 10:55