Ah, but “insightful,” that alternative answer has its problems as well. 99.9% isn’t acceptable as a Puzzler Solution. It’s just a hopeful sort of reliance on random numbers when in reality, the Warden makes no promise to pick prisoners in that fashion. Is this why you proclaimed this thread exhibits my argument is bogus? Sheesh. Do you want to re-think that at all?
I am really enjoying the input by @insightful and @GoldEye, because the first individual embodies the confusion engendered in thinking people by this Puzzler (and the subsequent desire to gloss over the facts and just move on), and the second has given his utmost to come up with Anything that could possibly have been done to solve the silly thing – including going to the lengths of using algorithmic tools which of course would not be available to these prisoners. @GoldEye, in his efforts, finally put to rest any lingering faith in Ray’s answer (or any answer).
Just to be clear on that last point, what @GoldEye came up with is a form of “likely”-ness. In other words, IF the Warden’s choices are made along a line of strict statistical probability, we Probably have a solution, mostly. Is that a solution to a mathematical problem? Nope. What’s our sample size? Uh… one guy making the choices – the Warden. Yeah, it’s bogus.
This past week, the show featured no mention of all the traffic concerning this puzzler on this forum. Instead, they read out ‘an email’ which was easily answered and chuckled over. Wisely, we were then offered a puzzler in the classic “move the matchstick around” series.
I’m cross-posting this from the clone thread which was started after this one. SO:
Let’s go back to the Puzzler itself. We’re given a quirky real-life situation set in a prison, and told to figure out how to free the prisoners early without risking loss of life. So, how do we (1) ELIMINATE the chance of the alligators getting a banquet, and (2) free all the prisoners before release or natural death has occurred?
Ray’s solution was, choose a Counter who will visit the room 44 times to find that ALL other prisoners have been there at least once.
Does anyone still think this answer is not bogus?
@stormybluemerc - While I agree with you on the bogusness of their answer, do you not know that the shows are all reruns? The guys couldn’t have mentioned “all the traffic concerning this puzzler on this forum”.
Sure enough! ;-D
@StormyBlueMerc…Let’s all openly acknowledge that the puzzler in question is not realistic, but rather an entertaining backstory to a mathematical puzzler. If I were a real prisoner in a real jail, my interpretation of this “challenge” would be that the Warden is indulging a sadistic kink of his, building everyone’s hopes up, only to have them dashed, and refuse to participate in his charade.
Once we accept that the puzzler is a “McGuffin” of sorts–an object to drive the plot forward–we can discount pesky real-world factors like:
- The Warden not keeping his word
- Prisoners dying, being killed, being in infirmary when their number’s called
- A particularly nasty prisoner sabatoging the game…etc.
Also, I would interpret the Warden’s statements about the choosing protocol to mean “The choosing process is random, and trends toward proportionally equal reults,” instead of what was actually said: “The choosing process is arbitrary, and trends toward numerically equal results.” (I would have preferred it if the puzzler just said, “I’m gonna put everybody’s name on a poker chip, shake 'em up in a pickle jar…”)
Now, what is the “best” solution, from the prisoner’s view, to this problem, using tools that prisoners would have? Well, I would “play spades” in the common area of the prison…only I’d deal out one card for each prisoner…one joker in this “deck.” I’d tally up each time a designated prsioner got the joker, and do this until all priosoners have had jokers, thus concluding this “game” of “spades.” Once the game ended thusly, I’d make a note of how many times the “most chosen” prisoner got the joker.
I figure I could play this game daily in the rec area without arousing too much suspicion. After a month (or 30 repetitions of the “game,”) I could be pretty confident regarding the random distrubution.
So, as the counter, I’d just call “time” once my number of visits to the switch equal the largest number I encountered in the 30 trial runs. Even in the worst-case scenario where I’m the most chosen prisoner (I’m probably not), I’m around two S.D.s of certainty regarding my answer…at which point the increase in certainty of my answer is probably not worth the continued loss of freedom.
As for this solution being “hopeful sort of reliance on random numbers,” well, plenty of SAFETY CRITICAL manufacturing is handled this way–destructively test every 1000th widget or so, and apply statistical proceessing to determine the probablility of a variance of normal distribution exceeding some hard limit only once ever six sigmas (or whatever the acceptable risk is). Mathematical proofs are largely irrelevant solutions outside the field of mathematics.
[WEB LACKEY–WHERE ARE MY PARAGRAPHS DISAPPEARING TO???]
MeanJoe, +1 to your statement: “Let’s all openly acknowledge that the puzzler in question is not realistic, but rather an entertaining backstory to a mathematical puzzler.” However, I don’t think the prisoners have the opportunity to play your spades games. GoldEye, they also don’t have access to your computer analysis. So I’m back to: Tom and Ray’s answer is 100% correct for the reasons previously stated.
@insightful: playing spades in jail is kind of a religion. When a friend of mine was looking at some time, my advice was “Don’t associate with the pervs, and learn how to play spades.”
He doesn’t actually need to interact with the other inmates to run the simulation; he just needs access to 23 cards. (Or 23 scraps of paper, one of which is marked.) I just thought “playing spades” was a good cover if anyone asked questions.
I think think doing this on a “statistics and probability” basis would be smarter, in actual application, than trusting 22 felons not to sabotage things, or screw up the instructions.
@meanjoe: You take me back 40 years to a former job where playing hearts at lunch was a religion. It got kinda rowdy at times. Regarding your suggestion, I guess I’ll just stick to defending Tom and Ray’s solution as “valid enough” to avoid the “bogus” label without getting into alternatives (head hurting).
yup, state the OP (me) presented a bogus analysis, cuz figuring it out hurts yer head. bye again, unsightly.
I wish all the puzzlers were automotive in nature
SBM, do you agree that the original puzzler solution methodology would work if there were only 3 prisoners?
@insightful, don t waste your time.
@wesw, point taken. Given that the puzzler states: “I will select one prisoner at random and escort him to the switch room.” Yet SBM holds that: “Warden doesn’t promise mechanically-driven randomness.” It’s hard to have a conversation when one party ignores that part of the statement of the puzzle upon which the solution is based. Random is random!
OK…Let me have a try…
SBM, you stated:
"Ray’s solution was, choose a Counter who will visit the room 44 times to find that ALL other prisoners have been there at least once. "
No, that is NOT Ray’s solution. The Counter needs to COUNT to 44 (not visit 44 times) to find that ALL other prisoners have been there at least once. But since he doesn’t increment his count every time he visits, it is likely that he will need to visit (many) more times than 44.
The other 22 prisoners are instructed to flip switch A on ONLY TWICE. The Counter increments his count ONLY when he finds switch A in the ON position. Therefore, allowing for an initial switch A position of ON, the HIGHEST number that the Counter could EVER count to is 45. And when he reaches 44, he KNOWS that one of these two cases must be true:
1) Switch A was initially off, and each of the 22 prisoners has visited the room at least** twice
or
2) Switch A was initially on, and 21 prisoners have visited at least** twice and 1 prisoner has visited at least** once.
In either of these cases, it is safe for him to announce to the warden that all prisoners have visited the room.
(** Why “at least”? Because any of the other 22 prisoners might visit the room numerous times and find switch A is already ON. Or they might visit and find switch A OFF, but having already flipped it ON twice in previous visits, will leave it alone. In these cases, their visits are insignificant.)
So, yes, it will probably take A LOT visits by all concerned before the Counter can make his announcement. The insignificant visits will likely far outnumber the significant ones, because a visit can only be significant for the purpose of incrementing the count IF:
1) It takes place when switch A is OFF,
AND
2) it is made by a prisoner who has not yet met his quota of 2.
Consider all the insignificant visits, when:
– switch A is on because the Counter has not yet come in to count a previous significant visit and reset the switch to off;
or
– the visitor has already met his quota of 2;
or
– the Counter visits and finds A off because no significant visits have occurred since his last visit.
But a fundamental premise of this puzzle, as has been pointed out by others repeatedly, is that EVENTUALLY, the warden will send ALL prisoners to the room enough times to allow the Counter to know that either 43 or 44 significant (i.e. countable) visits have taken place.
SBM, did this help you understand the solution?
–CBKiteflyer
CBKiteflyer:
Agreed. No one ever said it’s an efficient solution, but it does work.
Well, it works IFF you assume 100% compliance on the part of the prisoners. Even though it would be in their best interest to do so, ask any M.D. if he get 100% compliance from his patients, and he’ll about swallow his stethoscope.
I stand my my/insightful’s probability-based approach as being better: no need to count on “the better angels” of convicted felons, and a hell of a lot quicker to boot. I happen to be particularly parital to my simulation-based approach because I’m more experimental than theoretical (as I used to tell my dates…
Sheldon’s…uh, I mean StormyBlueMerc’s…objection arises from a flatly pedantic reading of (one part of) the puzzler’s instructions, coupled with overlooking another part of them. At some point, one needs to step back and do a plausibility test: does it make for a solvable/fun puzzle? Given that an arbitrary, malicious warden makes for a worthless puzzler, it is much, much more likely that the instructions re: the choosing process was a layman slightly misconstruing the subtle differences between random and arbitrary, and between proportionally equal and numerically equal.
At any event, the “random and proportionally equal” is the only solvable/fun interpretation, so it’s the one I’d solve, regardless…
I f you have a die handy, you can run Ray’s solution for 6 prisoners. I chose the counter as “1” and 2 to 6 as the rest. Start with the A switch [ON] or [OFF] and keep track of it as you go. (Obviously, no need to keep track of the B switch.) I entered each roll in a column on piece of notebook paper divided into 4 columns and kept track of when each non-counter reached 2 on-moves, etc., etc. Each “run” took about 15 minutes and it’s fascinating to watch the solution unfold. Starting with switch A [ON], it took 109 total visits to the switch room; if [OFF], 86 visits before the counter reached 10. I’m sure these totals can vary widely from these numbers. It’s also apparent that it is quite inefficient. In both cases, all 6 prisoners had visited the switch room well before the counter declared it to be so.