# Clue probability stumper **URL:** \ **Category:** Factual Questions\ **Created:** [August 19, 2008, 5:53am UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528 "2008-08-19T05:53:33Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![pandNH4](https://avatars.discourse-cdn.com/v4/letter/p/eb8c5e/32.png) [@pandNH4](https://boards.straightdope.com/u/pandNH4)\ **Post date:** [August 19, 2008, 5:53am UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/1 "2008-08-19T05:53:33Z")
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? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [August 19, 2008, 6:01am UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/2 "2008-08-19T06:01:16Z")
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]. ---
**Author:** ![Saint\_Cad](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saint_cad/32/18907_2.png) [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)\ **Post date:** [August 19, 2008, 6:18am UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/3 "2008-08-19T06:18:15Z")
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. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [August 19, 2008, 6:23am UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/4 "2008-08-19T06:23:29Z")
You’ve engaged in (and not cancelled out) some double counting of sequences that are all of a single ending. ---
**Author:** ![Saint\_Cad](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saint_cad/32/18907_2.png) [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)\ **Post date:** [August 19, 2008, 6:55am UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/5 "2008-08-19T06:55:02Z")
[QUOTE=Indistinguishable] You’ve engaged in (and not cancelled out) some double counting of sequences that are all of a single ending. [/QUOTE] 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. ---
**Author:** ![Harmonious\_Discord](https://avatars.discourse-cdn.com/v4/letter/h/74df32/32.png) [@Harmonious\_Discord](https://boards.straightdope.com/u/Harmonious_Discord)\ **Post date:** [August 19, 2008, 11:39am UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/6 "2008-08-19T11:39:57Z")
Please remember that random in electronic devices is not truly random, so a skewed response from that calculated is normal. ---
**Author:** ![muttrox](https://avatars.discourse-cdn.com/v4/letter/m/a8b319/32.png) [@muttrox](https://boards.straightdope.com/u/muttrox)\ **Post date:** [August 19, 2008, 2:30pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/7 "2008-08-19T14:30:48Z")
[QUOTE=pandNH4] Also, can you figure out the number of showings most likely needed to see all 3 endings? [/QUOTE] 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. ---
**Author:** ![Mr.Goob](https://avatars.discourse-cdn.com/v4/letter/m/c77e96/32.png) [@Mr.Goob](https://boards.straightdope.com/u/Mr.Goob)\ **Post date:** [August 19, 2008, 2:31pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/8 "2008-08-19T14:31:02Z")
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? ---
**Author:** ![CookingWithGas](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cookingwithgas/32/485_2.png) [@CookingWithGas](https://boards.straightdope.com/u/CookingWithGas)\ **Post date:** [August 19, 2008, 3:17pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/9 "2008-08-19T15:17:17Z")
[QUOTE=Mr. Goob] 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? [/QUOTE] 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. ---
**Author:** ![muttrox](https://avatars.discourse-cdn.com/v4/letter/m/a8b319/32.png) [@muttrox](https://boards.straightdope.com/u/muttrox)\ **Post date:** [August 19, 2008, 3:19pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/10 "2008-08-19T15:19:14Z")
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? ---
**Author:** ![Rysto](https://avatars.discourse-cdn.com/v4/letter/r/ecccb3/32.png) [@Rysto](https://boards.straightdope.com/u/Rysto)\ **Post date:** [August 19, 2008, 3:20pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/11 "2008-08-19T15:20:27Z")
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? ---
**Author:** ![JerseyFrank](https://avatars.discourse-cdn.com/v4/letter/j/bcef8e/32.png) [@JerseyFrank](https://boards.straightdope.com/u/JerseyFrank)\ **Post date:** [August 19, 2008, 3:33pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/12 "2008-08-19T15:33:25Z")
[QUOTE=Mr. Goob] 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? [/QUOTE] 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.” ---
**Author:** ![Santo\_Rugger](https://avatars.discourse-cdn.com/v4/letter/s/e95f7d/32.png) [@Santo\_Rugger](https://boards.straightdope.com/u/Santo_Rugger)\ **Post date:** [August 19, 2008, 3:58pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/13 "2008-08-19T15:58:37Z")
Using **Indistinguishable** ’s formula: ```auto 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? ---
**Author:** ![CJJ](https://avatars.discourse-cdn.com/v4/letter/c/ecc23a/32.png) [@CJJ](https://boards.straightdope.com/u/CJJ)\ **Post date:** [August 19, 2008, 3:58pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/14 "2008-08-19T15:58:55Z")
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%. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 19, 2008, 4:40pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/15 "2008-08-19T16:40:27Z")
[QUOTE=Santo Rugger] \*Am I using those words correctly in the statistical sense? [/QUOTE] Not if you’re thinking of confidence intervals. ---
**Author:** ![CookingWithGas](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cookingwithgas/32/485_2.png) [@CookingWithGas](https://boards.straightdope.com/u/CookingWithGas)\ **Post date:** [August 19, 2008, 5:23pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/16 "2008-08-19T17:23:42Z")
[QUOTE=Rysto] 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? [/QUOTE] 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. ---
**Author:** ![pandNH4](https://avatars.discourse-cdn.com/v4/letter/p/eb8c5e/32.png) [@pandNH4](https://boards.straightdope.com/u/pandNH4)\ **Post date:** [August 19, 2008, 6:41pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/17 "2008-08-19T18:41:50Z")
**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. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [August 19, 2008, 6:52pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/18 "2008-08-19T18:52:11Z")
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. ---
**Author:** ![ubergoober](https://avatars.discourse-cdn.com/v4/letter/u/ee7513/32.png) [@ubergoober](https://boards.straightdope.com/u/ubergoober)\ **Post date:** [August 19, 2008, 7:09pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/19 "2008-08-19T19:09:34Z")
[QUOTE=Indistinguishable] 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. [/QUOTE] Putting the OP’s question this way gives a restatement of the [Coupon Collector’s Problem](http://en.wikipedia.org/wiki/Coupon_collector%27s_problem). The Wikipedia page gives the same answer as **Indistinguishable**. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [August 19, 2008, 7:12pm UTC](https://boards.straightdope.com/t/clue-probability-stumper/460528/20 "2008-08-19T19:12:14Z")
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) [Next page](https://boards.straightdope.com/t/clue-probability-stumper/460528.md?page=2)