An eccentric billionaire with OCD..

The Game Room
1

An eccentric billionaire* with OCD installs a perfectly unbiased no-zero roulette wheel in his home (the sections of the roulette wheel are 50% red and 50% black. No greens, no zeros, etc).

He points a digital video recorder at the wheel and spins it, recording the result of where the ball landed. He then spins it a second time and again records the result, showing where the ball landed. He then burns the two spins to a DVD as separate tracks (spin 1 = track 1, spin 2 = track 2), and labels the DVD “Trial #1”.

He then repeats the exact same process again, except he labels the next DVD “Trial #2”.

He then repeats again, labelling the next DVD “Trial #3”, so on and so forth.

He does this many, many, many times over, to the point that he has tens of thousands of DVDs labelled “Trial #1 right through to Trial #99,999”.

This eccentric billionaire then invites you to his home, where he explains his eccentric OCD hobby to you, about how he has filmed many, many two-spin trials of his roulette wheel, storing the recorded footage as separate tracks on DVDs. You gaze at his vast library of roulette DVDs. Then…

[ol]
[li]You are invited to select a DVD at random, which you do.[/li][li]The billionaire then places the DVD in a player, and selects the “Play Random Track” button on his DVD remote.[/li][li]The Random Track plays on a screen (you know that each DVD contains two tracks, but you have no idea whether you are now watching the first or second track).[/li][li]You watch the track; you see a ball whizzing around a spinning roulette wheel.[/li][li]The ball lands on RED, and the track finishes.[/li][/ol]
The eccentric billionaire then says to you: “I will now play the other track on this DVD. But before I do, can you tell me, what is the probability that the other track will also show the ball landing on red?”

  • Acknowledging there’s no reason the protagonist in this story has to be a full on billionaire…
2

50%. One spin does not affect the outcome of any other spin. If the wheel is half black and half red, no greens or zeroes, and perfectly unbiased, every spin has a 50% chance of landing on red.

3

Are we missing something? It’s 50 percent and all his DVD track recording is pointless to the exercise. It’s exactly the same as if he asked if any random spin on the actual wheel is also red.

4

50%, since the two spins are independent. If you’re trying to set up something analogous to the Monty Hall problem, the issue there is that the host would know (in this setup) which tracks are red and black, and can (and must!) eliminate a bad choice. Here everything is chosen completely at random.

5

Actually, no, scratch that (sorry, it’s almost 6am). There are three possibilities, all equally likely:

  1. The DVD has two red tracks;
  2. The DVD has two black tracks;
  3. The DVD has one red and one black track.

Given that you already showed one track that’s red, you’re in case 1 with probability 2/3 and case 3 with probability 1/3. Thus the probability that the other track is red is 2/3, not 1/2.

6

I’m sorry, I simply don’t follow your logic here. The spins are independent. It has to be 50%.

7

Okay, so you know it’s either case (1) or case (3)

If it’s case (1), the second track is red.
If it’s case (3), the second track is black.

It’s still 50:50.

8

No, because you’re choosing a track uniformly at random. Let E0, E1, and E2 denote the events that the DVD chosen contains exactly 0, 1, and 2 red tracks, and let X denote the event that the first track— chosen at randomly on the DVD— is red. Clearly P(E0), P(E1), P(E2) = 1/3. Then P(X|E0)=0, P(X|E1)=1/2, and P(X|E2) = 1. Now P(X) = (1/3)(P(X|E0) + P(X|E1) + P(X|E2)) = 1/2, as expected. By Bayes’ theorem:
P(E0|X) = 0,
P(E1|X) = P(X|E1)P(E1)/P(X) = 1/3
P(E2|X) = P(X|E2)P(E2)/P(X) = 2/3.

Consider the event Y that the other track is red. We have
P(Y|X) = P(Y|X and E1)P(E1|X) + P(Y|X and E2)P(E2|X) = 01/3 + 12/3 = 2/3,
not 1/2.

Intuitively, the point is that since you picked a red track, you’re twice as likely to have picked a DVD that has two red tracks (P(E2|X) = 2/3) as one with one black and one red track (P(E1|X) = 1/3).

9

And, of course, the track that a uniformly chosen DVD contains exactly 1 of each color of track is 1/2, not 1/3. It’s clearly waaaay past my bedtime, so I’m going to crash now and deal with this mess later. Anyway, the OP is referring to this, which comes down to a matter of delicately parsing the problem and then applying straightforward Bayesian analysis.

10

The 3 cases are not equally likely. P(E0) = 1/4, P(E1) = 1/2, P(E2) = 1/4. There are two ways of getting 1 red and 1 black, either the red can come first, or the black can come first. There are a total of 4 possibilities, RR, RB, BR, and BB, all equally likely.

P(X) = (1/4) *P(X|E0) + (1/2)*P(X|E1) + (1/4)*P(X|E2) =
(1/4)0 +(1/2)(1/2)+(1/4)*1 = 1/4 + 1/4 = 1/2 as expected.

P(E0|X) = 0,
P(E1|X) = P(X|E1)P(E1)/P(X) = (1/2)(1/2)/(1/2) = 1/2
P(E2|X) = P(X|E2)P(E2)/P(X) = 1(1/4)/(1/2) = 1/2.

P(Y|X) = P(Y|X and E1)P(E1|X) + P(Y|X and E2)P(E2|X) = 01/2 + 11/2 = 1/2.

You can imagine the possibilities as RR, RB, BR, and BB, equally likely. There are then 4 possible places for the red you picked to have come from, from either of the reds from RR, or from one red each from RB and BR. That means the probability of RR given a red is 2/4 = 1/2, as there are 2 ways of the combination being RR, out of 4.

11

So, going through the formal computation in detail, but more correctly:

Let E0, E1, and E2 denote the events that the DVD chosen contains exactly 0, 1, and 2 red tracks, and let X denote the event that the first track— chosen at randomly on the DVD— is red. Clearly P(E0) = 1/4, P(E1) = 1/2, P(E2) = 1/4. Then P(X|E0)=0, P(X|E1)=1/2, and P(X|E2) = 1. Now P(X) = 1/4(P(X|E0) + 2P(X|E1) + P(X|E2)) = 1/2, as expected. By Bayes’ theorem:
P(E0|X) = 0,
P(E1|X) = P(X|E1)P(E1)/P(X) = 1/2
P(E2|X) = P(X|E2)P(E2)/P(X) = 1/2.

Consider the event Y that the other track is red. We have
P(Y|X) = P(Y|X and E1)P(E1|X) + P(Y|X and E2)P(E2|X) = 01/2 +11/2 = 1/2.
Hooray for math.

Anyway, having embarrassed myself for the night (or morning, or whatever), I’m off to bed for real this time.

12

For the record I missed Itself’s retraction. It looks like Itself missed my correction.

13

Doesn’t the eccentric billionaire know that the odds aren’t actually 50%, though? Unless the spins have resulted in exactly 99,999 red results and 99,999 black results, the odds of a random video showing red or black will be slightly to one side or the other, right?

14

This is similar to, but crucially different from, a classic puzzle involving three coins - one double-headed, one double-tailed, and one normal. There the idea is that if you pull a coin at random and see a head, what is the probability that the other side of the coin is also a head? (2/3)

The crucial difference is that here you might have seen one of the following eight cases, all equally likely (disregarding Johnny Bravo’s hair-splitting):

Red, then Black, and you see track 1 *
Red, then Black, and you see track 2
Black, then Black, and you see track 1
Black, then Black, and you see track 2
Black, then Red, and you see track 1
Black, then Red, and you see track 2 *
Red, then Red, and you see track 1 *
Red, then Red, and you see track 2 *

Plainly only the asterisked cases can have applied to this instance, since you have seen Red, and we can see that in two of those cases there is a Black on the other track, and in the other two, Red. Hence the chance is even to within Johnny’s extent of error.

NB: Since there are 99,999 discs, and this is not divisible by four, obviously one of the categories must be under-represented. Mais n’enculons pas des mouches.

15

That would depend on the severity of his OCD. Maybe he only put the best 99,999 DVDs on the shelf for his guests. Maybe that is his angle with this whole exercise: release a Two Roulette Wheel Spins: Greatest Hits DVD Compilation, Volumes 1 through 99,999.

Does he have a hoarding disorder too? In which case, he probably has roomfuls of DVDs that didn’t burn or are filled with discarded takes where his cats wandered into the shot.

16

At no point did you state that it was a fair wheel only that it was divided into two even sections. So I’d say the probability that it will land on red a second time is ever so slightly greater than 50%. If there is a bias, however slight, the chance that it would have landed on the advantaged side during the first role is > 50% by whatever that bias is, and thus is there is a bias it’s slightly more probably to be towards red than towards black. For all we know there’s such a great bias, because there’s a magnet that attracts the ball or something, that it actually lands on red every time.

So, if I had to put money on the second spin I’d go for red.

17

I applaud your dedication to nitpickery (SDMB motto: ‘technically correct, the best kind of correct!’) but maybe you should have read the OP more carefully:
(underlining added by me)

18

It seems to me that the OP is a rewording of the Two Boys problem (i.e. a family has two children, at least one of which is a boy) which, if interpreted one way, leads to a probability of 1/3, but, interpreted another way, yields an answer of 1/2.
The OP has carefully stated the problem to eliminate the former interpretation, so the answer is 1/2. (I think - you never really know with probability.)

19

Hmph! Disregard my hair-splitting, will you? I’m sorry, I thought I was on the SDMB!

20

It’s 50%. There is a specific track whose outcome you do not know, and it is independent from anything you know.