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smartyman
Yes, I'm afraid empirical evidence upholds the accepted solve. I wrote a program to run through 10,000 tests of how long it would take to generate all numbers from 1 to 140 using a random number generator. It came up with
 Spoiler (Rollover to View): 140.038.

Of course this depends on the quality of the random number generator. The first time I created the program on a platform with a not-so-good random number generator because after 10,000 tests it came up with an answer of
 Spoiler (Rollover to View): 141.101!

So, I copied it to a different platform and came up with the right answer.

Posted: Fri Apr 27, 2007 4:30 pm
makdabo
 Spoiler (Rollover to View): One of the not-so-bright academy fellows says "it's different every minute" which is obviously based only on the so-far experience of the last 10 minutes (or perhaps a bit more). But so what? It's just a convenient way of expressing what had been observed thus far. The chance that no two consecutive shifts would resulting in the same picture in that time are pretty good. An application of Occams razor would lead us to the simpler conclusion that the pictures are probably random with replacement and not the special case of random with replacement except that the current picture is held out of the next selection.

Posted: Fri Nov 17, 2006 8:18 pm
X9Tim
fitzyfitz wrote:
 Spoiler (Rollover to View): probability of observer *not* seeing their portrait after X minutes is 139/140 ^ X (139/140)^96 = 0.502 (139/140)^97 = 0.499 so it takes 97 minutes to reach a >50% chance of seeing your own portrait

I did this calculation the same way.

Now I've read the comments I still can't believe this isn't right.

Just to check it, I did a quick (probably too short statistically) test using excel and random numbers chosen from 1-6. After 30 trials, the average time waited by the 6 numbers ranged from 4 to 11.5 averaging 5.9944444. This is probably closer to 6 than it is to 4.8 as the maths suggests....

I'll try to keep an open mind

--

Still I've got the points now and THAT'S what matters!!

Posted: Sat Oct 28, 2006 8:53 am
volvox

Not that it probably matters anymore, but my first answer was totally different:

 Spoiler (Rollover to View): 175

My reasoning:

 Spoiler (Rollover to View): Johnstone appeared three times in 11 minutes. Out of 140 possibilities every minute, this seemed astonomically unlikely. So I figured the distribution was rigged to make Johnstone appear more frequently, thereby suckering people into waiting around to see if they would appear. I thought this was the cleverness Sente was referring to. I assumed that he appeared in the first 10 minute block twice, and since he appeared at the 11th minute, he would probably appear again within the next 10 minute block. So, if he appears 2 times out of 10, that leaves 8 times out of 10 for everyone else. 8/10 = 4/5, so the average time of any non-Johnstone would be 5/4 as long = 140 * 5/4 = 175 minutes.

Posted: Wed Oct 18, 2006 8:54 am
torne
I think Mind Candy's answer is wrong :)

Although the system does accept the solve posted earlier in this thread, I think that's actually the wrong answer for the problem as stated on the card (and that my first solve attempt was correct)

Explanation:
 Spoiler (Rollover to View): The distribution is geometric, mean is 1/p as other posters have stated. However, p should not be 1/140 as everyone has assumed. The text explicitly states 'the picture changes to a different one every minute', which would mean that p is 1/139, since each time it will become one of the 139 pictures that it isn't currently displaying. This results in a answer of 139 minutes rather than 140, which the system doesn't accept.

Posted: Sat Aug 26, 2006 2:58 pm
Hawkeblu
The shakespearean dilemma

This reminds me of the Rosencrantz and Guildenstern thing where they keep tossing a coin and it always comes up heads. One of them rationalizes that they shouldn't be surprized at all because since there's always a 50% chance of it coming up heads, there's a 50% chance of it always coming up heads

Hawkeblu

Posted: Mon Aug 21, 2006 8:05 am
Carma1313
personally I think...

 Spoiler (Rollover to View): ...it should be 139, due to the fact one of the 140 has already been shown.. However... When thinking back to my Statistics A Level (HATE STATISTICS!!!) every time a new picture shows it has exactly a 1/140 chance of being said person, not 1/139... and the next one after that will be 1/140 not 1/138 because the previous portraits have no influence on the next portrait. I think both 140 and 139 should have been accepted because of the two different ways you could look at probability.

Posted: Tue Aug 15, 2006 7:31 am
Lysithea
mikew03 wrote:
 Spoiler (Rollover to View): I strongly disagree that the answer is 140. Assuming it really is a random distribution on average a person will wait half that time to see his picture. I don't see anything tricky about the wording. The answer is simply wrong. If I walk into a room where 140 pictures are being randomly displayed (one of which is a picture of me) on average I will wait 70 minutes to see mine. It might be one minute, it might be 140. But the average IS 70.

I strongly disagree with you mikew03. There are different types of random distributions. The fact that the same portrait (Johnstone) has shown up 2-3 times in the last 10 minutes indicates that
 Spoiler (Rollover to View): this distribution has the lack of memory property. That means that no matter which portraits have already been shown, the probability of any particular portrait being the next one is always 1/140. (This is just like rolling a die, no matter which numbers have been rolled recently, the probability of getting a 3 on the next dice roll is always 1/6.) If the 140 portraits were in a pre-determined sequence, and that sequence repeated every 140 minutes, then your answer of 70 minutes would be correct. (for example, if you deal cards from the top of a shuffled deck and put each card on the bottom of the deck then the average number of cards you will deal until reaching the 3 of clubs will be 26.)

To me, the wording on the card is clear, it is only the concept involved that can be hard to grasp. The accepted answer is the correct one. In statistical terms
 Spoiler (Rollover to View): As noted above, this is a geometric distribution. The average of a geometric distribution is 1/p where p is the probability of the event occuring, in this case p=1/140 (and 1/p=140). Therefore the average waiting time will be 140 x 1 minute = 140 minutes.

Posted: Fri Aug 11, 2006 11:07 pm
situp
There's 140 pictures, and I am 1 of those 140 pictures, so on average I will appear 1/140 times. There's no complication here. It really is as simple as it seems.

Posted: Fri Aug 11, 2006 8:41 pm
Agent Lex
mikew03 wrote:
 Spoiler (Rollover to View): If I walk into a room where 140 pictures are being randomly displayed (one of which is a picture of me) on average I will wait 70 minutes to see mine. It might be one minute, it might be 140. But the average IS 70.

 Spoiler (Rollover to View): It might be 1 minute, it might be 280, it might be 5 weeks until you see it. The average is (close enough to) 140.

Quote:
 Spoiler (Rollover to View): Just ask the people working on the 13th labor, they are at 60% now and are getting nervous they don't have the right code to solve the problem, because on average they should have solved it by now.

 Spoiler (Rollover to View): That's a different problem, because with each code processed it's done and will never need to be processed again, while this picture can obviously show the same person 3 times in 10 minutes or so, as seen on the card with Johnstone.

Posted: Thu Aug 10, 2006 4:50 am
mikew03
 Spoiler (Rollover to View): I strongly disagree that the answer is 140. Assuming it really is a random distribution on average a person will wait half that time to see his picture. I don't see anything tricky about the wording. The answer is simply wrong. If I walk into a room where 140 pictures are being randomly displayed (one of which is a picture of me) on average I will wait 70 minutes to see mine. It might be one minute, it might be 140. But the average IS 70. Just ask the people working on the 13th labor, they are at 60% now and are getting nervous they don't have the right code to solve the problem, because on average they should have solved it by now.

Posted: Thu Aug 10, 2006 12:47 am
European Chris
 Rifflesby wrote: I object to the wording of this card. The answer is only [what it is] if you're looking for the average of all the lengths of time it took everyone to see their own portraits. But the question *as given* is asking how long this particular guy should expect to wait, "on average", and for that question fitzyfitz's math is correct.

The maths is right, but doesn't answer the right question. It's not asking how long someone should wait to have a 50% chance of seeing themselves but how long on average time someone should wait-and the answer to that is related to the expected value

It's simlar to 'infinty series' (with the train and the bee), it just gives you more information than you need to try and confuse you. So I'm sure you do object to the wording of the card, but that's the bit that makes the puzzle.

Posted: Sat Aug 05, 2006 9:34 am
Rifflesby
I object to the wording of this card. The answer is only [what it is] if you're looking for the average of all the lengths of time it took everyone to see their own portraits.

But the question *as given* is asking how long this particular guy should expect to wait, "on average", and for that question fitzyfitz's math is correct.

Unless, y'know, I'm just totally reading it wrong. Always a possibility :)

Posted: Sat Aug 05, 2006 1:53 am
Guin
it would have been funnier if it had been

 Spoiler (Rollover to View): a mirror

Posted: Wed Aug 02, 2006 1:15 pm
The V.G.P.S.
Answer is def. as above. Remember its not asking for any guarantee,its the average that counts

Posted: Wed Aug 02, 2006 6:19 am
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