Do you remember the movie clue? It has three different endings. With the home release, you can select to play all endings at once or have one of them randomly selected. I was trying to watch the clue dvd but I didn’t have the remote, so I could only choose the random option. As I was restarting the film repeatedly, I got to thinking, how can you derive the probability that all 3 endings will have been seen after the nth showing. Also, can you figure out the number of showings most likely needed to see all 3 endings?
For positive n, the probability that you’ve seen all 3 endings by the nth showing is 1 - (the probability that you’ve missed one) = 1 - (the probability that you’ve missed ending X + the probability that you’ve missed Y + prob you’ve missed Z - (prob you’ve missed X & Y + prob you’ve missed X & Z + prob you’ve missed Y & Z)) = 1 - (3*(2/3)^n - (3 * (1/3)^n)). This comes out to 1 - 3 * [(2/3)^n - (1/3)^n].
God i hate combinations so someone will probably come along and correct me. There are 3^n possible permutations of endings in three showings. Of these 2^n are all endings 1 and 2, 2^n are all endings 2 and 3 and 2^n are all endings 1 and 3. so the probability is
1 - 3x[(2^n)/(3^n)] or [3^(n-1) - 2^n]/[3^(n-1)]
Note that this implies that n must be greater or equal to 3 (makes sense). As a check, with three showings, the probability is 3! / 3^3 = 2/9 but the above formula gives 1/9 so forget everything I wrote.
You’ve engaged in (and not cancelled out) some double counting of sequences that are all of a single ending.
Yep. I see it now. The 2^n permutations of 1 and 2 includes 111111…1 as does the 2^n permutations of 1 and 3.
Please remember that random in electronic devices is not truly random, so a skewed response from that calculated is normal.
You need to specify this more. Obviously the more showings, the higher the probability of seeing all 3. I had to keep myself from snarkily answering infinity. Give us a probability, and we can calculate how many showings it takes such that the probability of seeing all 3 is as high or higher than that number.
I love that you guys can show your work, but what about an answer that I can understand?
If the player is truly random, how many times would I need to watch the movie to see all three endings? I know that it could in theory be a gazillion million times because of monkeys and typewriters and all before you see them all, but on average?
Without actually trying to do the math myself, I can reiterate what others have tried to show above.
There is no number of viewings * that will result in a 100% probability of seeing all three endings. The probability of seeing all three in three viewings is something like 22.2%, and increases with the number of viewings. You instead have to tell us what is the percentage that you would consider “good enough,” say, 99.999%.
You can’t say “on average.” to wave away this analysis. “On average” you will see all three endings in three viewings 22.2% of the time.
(The monkey and typewriter thing is a whole nother discussion that doesn’t apply here. That simply says that if you iterate through all combinations and permutations of the keys on a keyboard up to, say, book length, it will include any published work ever written (or that will be written). Nothing at all to do with probability, although combinatorics is a necessary foundation for probability.)
_________________________________________________-
*I did this by brute force to get 6 permutations of 1-2-3 out of 27 combinations with replacement. The generalized formula for more than 3 viewings is more complicated because you have to figure out how many resulting combinations contain all three endings. That is trivial for the case of three viewings.
I believe by “on average” Mr. Goob is trying to say how many showings do I need to have a 50% chance of seeing all 3 endings?
I don’t see what the problem with Mr. Goob’s question is. What’s the expected number of viewings necessary to see all 3 endings?
The only answer I can say with 100% certainty is “at least three.” Beyond that, you’d need to give us a probability you’d be willing to accept as “on average.”
Using Indistinguishable’s formula:
1 0
2 0
3 0.222222222
4 0.444444444
5 0.617283951
6 0.740740741
7 0.825788752
8 0.88340192
9 0.922115531
10 0.948026216
11 0.965333875
12 0.976883605
13 0.984587188
14 0.989724165
15 0.993149234
16 0.995432753
17 0.996955146
18 0.997970089
19 0.998646724
20 0.999097815
ETA: So an answer that would satisfy the OP would probably be 10. Unless they needed more than 95% confidence* that they would get to watch all 3. 14 if they wanted 99% surety*.
*Am I using those words correctly in the statistical sense?
I verified Indistinguishable’s formula is correct using a different method; to aid in understanding that formula, here are some sample values:
n=3 => A(n) = .222
n=4 => A(n) = .444
n=5 => A(n) = .617
n=6 => A(n) = .741
…
n=10 => A(n) = .948
n=14 => A(n) = .990
After the fifth viewing (the first which makes it more likely than not you’ve seen all three), it is 62% likely you have seen all three endings. After 10 viewings, it’s close to 95% likely, and after 14 viewings it’s nearly 99%.
Not if you’re thinking of confidence intervals.
Try to define in a precise mathematical way what you mean by “expected number of viewings” and you will see the problem with the question.
I think there is some confusion with “expected value.” “Expected value” is a term that means, roughly, the outcome aggregated from all possible outcomes. If I bet you $1 on a fair coin flip, and win back $1 on heads, the expected value of my winnings is zero. I have a 50% chance of winning $1 and 50% of losing 1. That averages to zero. If I am doing this in a casino, and the house only pays .90 back on my 1 bet, then my expected value becomes negative .05. (Do you wonder why anyone would take a bet like that? This is basically what happens when you play the roulette wheel, or the lottery.)
But the OP question is *not * an expected value problem. A simple but very similar problem to the OP is, “How many times to I have to flip a coin to guarantee that I will see at least one head?” There is no “expected number” of flips. You can never guarantee you will see at least one head. With a given number of flips, there is a specific *probability * of including at least one head. You can only say, “After a certain number of flips, I will *probably * see at least one head.” But what do you mean by “probably”? Do you mean half the time? Ninety percent of the time?
In fact, after 2 flips, there is only a 75% chance of getting at least one head. After 5 flips, there is a 96.875% chance of getting at least one head. After 10 flips it’s 99.902%. How high do you want to get that probability before you’re happy?
Other posters have put the numbers up for the three-endings problem, which works exactly the same way.
CookingWithGas, I don’t think you’re quite right. To take the flipping coin example, there is a 1/2 chance that the first flip will yield the first heads. There is just a 1/4 chance that the second toss will reveal the first heads, and so on. I was asking for the number of viewings most likely to result in the desired result (all 3 endings). Surely this can be thought of in a similar way.
Yeah, guys… I don’t see what’s so problematic about the OP’s second question. If a person sits down and starts watching, there is a certain probability that it will take them three views to catch every ending, a certain probability that it will take them four views to catch every ending, etc. The OP wants to know which of these is the highest. As it happens, there’s a tie; there’s a 2/9 probability that it will take 3 viewings, and a 2/9 probability that it will take precisely 4 viewings, and these are the highest ones. (See Santo Rugger’s chart)
Furthermore, there would be nothing in principle problematic about “On average, how many views will it take to see all three endings?”. Add up (probability that it takes n views * n) for each n, and you get the expected (i.e., arithmetic mean) number of views to catch all three ending.
In this case, where n is at least 3, there are 3 * (2^(n-1) - 2) ways to take exactly n viewings to catch all three endings, each with probability 1/3^n. Therefore, summing up 3 * n * (2^(n-1) - 2)/3^n over all n >= 3, we obtain a mean of 5.5 viewings.
Putting the OP’s question this way gives a restatement of the Coupon Collector’s Problem. The Wikipedia page gives the same answer as Indistinguishable.
Another way to get the same result: the number of views it takes to catch all three endings is 1 + the sum of S_n over all positive n, where S_n is 1 if one still hasn’t caught all three endings after the nth viewing, and 0 otherwise. By linearity of expectations, this means the expected number of views is 1 + the sum of the probability of not catching all three endings by the nth viewing over all positive n = 1 + the sum over all n of 3 * [(2/3)^n - (1/3)^n] = 1 + 3 * [1/(1-2/3) - 1/(1 - 1/3)] = 1 + 3 * (3 - 3/2) = 5.5
ETA: I didn’t see ubergoober’s post when I made this one, but it gives a nice explanation as well (and a nicer ultimate formula; 3 * (1/1 + 1/2 + 1/3) = 5.5)