The three box conundrum

Ely. Think about the part that is not water, it used to occupy one percent of the potatoes but now occupies two percent of the potatoes. It hasn't changed at all so everything else (the water) must have halved.

I'm surprised no one has mentioned Sod's Law (if they have, I apologise for having a low boredom threshold too and therefore, haven't read every post above) which applies to most people/situations and states that 'it doesn't matter which box you pick, you'll pick the wrong one.' [%sig%]

I'm not even going to pretend I'm clever...pasted from Google Question 49088 Submitted on 4/20/2004 Mr. Man boutht 100 lbs. of potatoes that were 99% water. After the potatoes were left outdoors for a day, they were 98% water. What was the weight of the slightly dehydrated potatoes? Solution 49088 Let the loss weight be x lb. (99%)(100) - (98%)(100 - x) = x (0.99)(100) - (0.98)(100 - x) = x 99 - (98 - 0.98x) = x 99 - 98 + 0.98x = x 1 + 0.98x = x 1 + 0.98x - 0.98x = x - 0.98x 1 = 0.02x 1 / 0.02 = 0.02x / 0.02 whatever 50 = x 100 - x = 100 - 50 = 50 The weight of the dehydrated potatoes was 50 lb.

I doubt if Jon has 30 Aston Martins to give away. >> At the start of the story you pick Box A. I think you will agree there is a one in three chance that the keys are in Box A. That must mean there is a two in three chance that the keys are in Box B or Box C. Again, you can't quarrel with that. << That is utter bollocks. There is a 1 in 3 chance the keys are in box A. There is a 1 in 3 chance they are in box B, and a 1 in 3 chance they are in box C. What's all this 2 in 3 crap about? That's as bad as saying if you buy 2 lottery tickets you halve the odds against winning, when in fact the odds are just the same for both tickets. I think Jon's explanantion is mathematically far sounder, if for no other reason that it doesn't depend on 'probability' but mathematical odds. ie It isn't 'probable' that the keys are in B or C, it's a fact.

Yet another way of seeing this problem. Imagine there were 100 boxes and only one contained the keys. You pick Box 1. Then your uncle eliminates 98 boxes because they are empty and leaves just box 100. He now says you can either stick with your first choice Box 1, or switch to box 100. Common sense tells you that the chances of your getting it right first time with a 100 boxes is one in a hundred - very low odds. Now your uncle is telling you the keys are either in box 1 (one in a hundred chance) or box 100 which he has selected. Which would you choose. Obviously you would choose box 100. Why? Because your uncle has helped you by eliminating all the other boxes that did not contain the keys, and he now guarantees that either your first choice (one in a hundred odds) or his choice (99 in a hundred odds) contains the keys. Now all you have to do is imagine this problem starting with 99 boxes and then 98 and then 97 and each time you can see why it makes sense to go for the box your uncle leaves after he has eliminated all the empty ones. Keep going and eventually, of course, you will get down to just three boxes. And lo and behold, the same logic that works when there were 100 boxes, applies when there are just three. It's just harder to get your head round it.

LOOK!!!!!! If the stupid fucking uncle opens either B or C and it is empty, then the odds are that the keys are in either A or which ever he DIDN'T open. That's TWO stupid sodding boxes, so the odds are 50/50. >> if you select A, and stick with A, you are going to be correct one in three times << What a load of bollocks! You could pick 'A' a thousand times but if the keys are in B or C each time you will NEVER be right. jesus christ, you bloody maths freaks are the bloody limit.

BOING BOING BOING

hehehe - how are you Choose???

Mississippi it's not often you are wrong mate, but you are here. Or at least you appear to have misunderstood what I am saying. All I am saying is if Uncle says pick one box (and you pick A) your odds are one in three. If Uncle says pick two boxes (and let's you pick both B and C) your odds are two in three. You have to agree with that. You absolutely have to agree with that. Come on mate, tihs is your chance to be on the winning side against Hen.

Dublin, you said "Mississippi it's not often you are wrong mate." - I have to laugh - give me an example of him being right! I have a long list of things he has said that have proved to be the exact opposite of reality but I can't think of anything specific he has said - about anything other than music - that has turned out to be correct.

Folks, You probably chose the wrong box to start with, so the other unopened box probably contains the prize. Simple as that. If you want to get technical there are two important points to remember. Monty's actions are not predictable, he is reacting to your initial choice. After you have made your inital choice you are then given extra information. If anyone is in any doubt I seriously suggest they get a dice and piece of paper and try it for themselves, or if they can't be bothered I will happily write a compilable computer simulation, put the executable up in my own webspace, and they can run it and see it work experimentally. Unless there's anything good on telly tonight that is.

Ah, I see, i DID infact misunderstand what you were sying Dubber. Mykle why don't you just crawl back under you rock, you low-life?

OK, I didn't want to pull this out, but I got 100% in my stats exams, and probability problems were my strongpoint. A Level maths was a goddamn chore, and I wouldn't have got that mark if I couldn't spot the errors in a problem like this. With respect, it doesn't matter which way you phrase it, Dublindian - your answer is wrong, so I'm always going to be able to find the flaw in your working. In the first case: putting boxes C and B in a forth box called D is manifestly *not* the same as opening one of them. If you open one of them, it eliminates it from the problem. There are now only two boxes to consider. If you put them in box D, there are still three boxes. Even if you put the contents of boxes B and C into box D, it still isn't the same as eliminating a box. How to demonstrate it? Start the problem from the final decision. In the original problem, the keys are still randomly distributed between the two boxes that are left. 50/50. Heads or tails. In your adaptation, the distribution is not random - the dice has been loaded in favour of box D. Your second adaption is just silly, and defeats you on its own. Do you honestly believe that, when faced with two closed boxes, there is a 99% chance that one of them is right? I said I'd find the specific fault though, and it lies here: "and he now guarantees that either your first choice (one in a hundred odds) or his choice (99 in a hundred odds) contains the keys." No. You have to reset the probability. The box you originally chose does not remain at one in a hundred. Again, think of racehorses. Think of two racehorses. If you pick a slowcoach that has a 1% chance of winning, and then the other one dies, your original choice no longer has a 1% chance of winning! It's guranteed to! Hang on - there's a much easier way to explain it to you: Your uncle offers you two boxes, A & B. One definitely has the keys in. You pick A. He opens B, revealing there to be nothing inside. According to your methods, there is still only a 50% chance of the keys being in A! Because you base the probability of it being in A entirely on what the probability was at the start (or is there another reason you wrongly transfer B's 1/3 to C every time?) You must recalculate the odds based on what you have left. B has been eliminated. A is all that is left. There is 100% chance of the keys being inside.

Your fabled command of the English language coupled with your immense vocabulary never ceases to impress me George - who could we point to as an example of elegant, mature posts were it not for you?

You? I doubt it.

Vazzzzm.. *new Aerodynamic model*

Dan: if you could write a BASIC programme to simulate the problem and give me a link to your web space I would be very grateful! I was going to write one myself but realised I’d forgotten how to generate random integers and I wasn’t sure how to implement the opening of an empty box. I was thinking it would be easiest to assume that box A always contained the keys and therefore only the choice would need to be random - but I will be interested to see how you tackle the problem. Thanks.

yep yep yep.... thanks Dan and Jude (mainly Dan because Jude's explanation was algebra and no words) I've got it now. put simply you only ever have a pound of potato the rest is water. So, if it exists as one percent then you must have 100 lb overall, if that same pound exists as two percent then you must have 50lb overall. This one I like.

I was gonna write it in C, but I could do one in VB if you like. Right now I got bigger problems, I can't find a corkscrew!!!!!!!!!!

Actually, I favour the racehorse analogy. Three racehorses of absolutely equal ability - A, B and C. Clones, no less. They race - whoever wins is totally down to chance. You put your money on horse A. Horse B is taken ill! The bookie is a friend, and offers you the chance to change your bet. Now, who here seriously thinks that Horse C now has twice the chance of winning as Horse A? Or do the 100 horses one. 100 cloned, equal ability horses. You pick Number 1. Numbers 2 to 99 die. Has horse 100 suddenly become 99 times as fast as horse 1?

Right, firstly forget all of this bollocks about horses. When you Pick Box A, you have a one in three chance of being right. If your uncle said, "You can switch and open Box B and C" you would ALWAYS switch, because you'd have a 2 in 3 chance of being right. Your uncle telling you that Box B is empty doesn't make any difference to the critical point, which is that you are offered the chance to choose to 'open Box B and C' and that the keys are twice as likely to be in that subset of boxes as they are in Box A. If it didn't seem counter-intuitive, it wouldn't fool people. It fools a great many statisticians, but it is absolutely categorically right - it just 'feels' wrong. Dan's approach is right - run the test 200 times and you will find that the switchers do better than the stickers, at a ratio of around 2 to 1. If you do it 10,000 times it will be nearly exactly 2 to 1. The horses just confuse the issue. Imagine instead if your uncle said to you, "I know for a FACT that one of Box B or Box C is empty - do you want to open both Box B and C instead of just box A?" - you would ALWAYS take that chance because if you double the amount of boxes you open you double your chance of winning. And that's effectively what he does, except he shows you which of the boxes is empty, so you don't have to open it. IF AND ONLY IF he puts the keys in A or C AFTER opening Box B to show you it is empty are the odds for switch and stick the same... Sheesh... This is one of the most famous probability 'paradoxes' (not really a paradox, because it is not unsolveable, just counter-intuitive) around. If the maths didn't work, do you think we wouldn't know by now?

Race horse not good analogy cos race run AFTER bet key placed in box BEFORE bet still no corkscrew!!!

OK Hen, let me ask you a simple question, that will hopefully lead you down a path that will nail this once and for all. You mentioned horses so let's use that analogy. Imagine there was a race involving, say, 25 horses in which each one had an equal chance of winning. The race has already been run. Your uncle already knows the result but you don't. He has the tape of the race and he is going to play it for you, but first he asks you to pick one winner. So you pick horse A. OK, you have a one in 25 chance of being right. (We will assume you are not going to search for the result on Google). OK, your uncle now says, "Hen me old nephew, I'm going to give you some help here. I'm going to eliminate 23 of the horses that definitely did not win. That leaves just two horses: the one you picked and another one. "All I'm going to say is that one of these two horses wins the race, and if you like you can stick with your first choice or switch to this other horse. OK which horse will you pick. Bearing in mind your uncle will give you £100,000 if you pick the winning horse. Come on Hen, don't give me a load of spiel, just answer the question. To increase your odds of winning the cash, which horse do you pick: the one you selected first, or the one that is left after your uncle's process of elimination?

So you DO get the option between opening one box on the one hand and two boxes on the other?

I meant 'option of'. My grammar is crap, these days.

*shoots everybody and STILL has the car* *stick yer spuds up ya ....*

They have equal chances, so it doesn't matter. The rerun doesn't change anything, Dublindian. According to you, eliminating those 23 horses suddenly makes the 'other horse' 24 times as likely to beat the original. This only holds true if it was already 24 times better than the original, which it wasn't. So you're wrong again. If it comes down to two horses with equal abilities, they each have an equal chance of winning. You might as well make them run again, against each other. Give it up, for goodness' sake! It really doesn't matter which way you pose it. I've proved you wrong in numerous different ways, all of which would be obvious to you if you'd relinquish this obsession with *not* resetting the odds! You absolutely *have* to do this! If you didn't, as I've already demonstrated, then in the case of two boxes, one of which is eliminated, you would still have only a 50% chance of winning, even though you've only got one box left! You *have* to - repeat, *have* to - reset the odds when anything is eliminated, otherwise you are NOT using maths.

Only in the sense that, upon removing box B from the equation, your uncle is basically giving you that choice. "But Uncle," you might have said, "I'd love to open both those two but I can only open one box" "Don't worry young George" he might have replied, his benevolence glowing from him like a meltdown in a Werthers Original factory. "I shall remove one of the other two so you don't have to open it." As your ever widening smile lets forth a wish-filled gasp, he opens box B to show it is empty. "There, now the other two are just one!" Quick as a flash and seeing the mathematical advantage in this option you dive onto box C and rip open the lid! "Uncle!" you cry, lifting a tiny but perfectly designed model of an Aston Martin out with trembling hands. He chuckles to himself, "There you go Georgie boy, and when you're old enough perhaps we can get you the real thing. Now lets see what your Auntie has for our supper." And with that he ruffles your hair and turns to poke the log that cackles on the hearth while you marvel at your new toy and can't help thinking, you tight old bastard. I've got the main dealer driving up from Surrey to offer me a price on this and four girls from Spearmint Rhino on standby!

Since Hen is happily math literate we can do an exhaustive anaylsis of the problem. We will remove the arbitrary box letters since it complicates the issue and simply enumerate the as RIGHT, WRONG0, and WRONG1 (with the keys in the RIGHT box, obviously). I will now create a table that should be easy to understand. Box Chosen | Box Opened | Switch | Stick ---------------------------------------------------------------------------- | | | WRONG0 | WRONG1 | WIN | LOSE | | | WRONG1 | WRONG0 | WIN | LOSE | | | RIGHT | WRONG0 | LOSE | WIN I think this makes it blindingly obvious that you win twice as ofter if you switch. (They had a corkscrew nextdoor)

Bollocks. My beautiful text layout didn't work.

Actually Andrew when it comes to winning over the doubters it turns out your "1-in-3-versus-2-in-3" line of reasoning is a load of bollocks. I know I've already used it on this thread about five flaming times, and got nowhere. The argument that works for most people is the one where you imagine there are several times more boxes or horses than the three in the original problem. Once people understand why it works for a large number, they can then apply the same logic to a gradually reducing quantity, till eventually they reach the magic three. Like you, I favour the other approach. But it just seems to generate more confusion than enlightenment.

Hen, you seem to persistently ignore the fact that the result is know to one party and that is the crux of the altered odds. You've had a hard day - don't post again until tomorrow. If you're right you'll still be right then - and if you're wrong you'll have time to realise that you are and see why.

As I've said twice now, the way it read to me was a 1 in 2 choice v a 1 in 3 choice.

Mykle, I'm a little late getting back to you. What I was trying to get at in my post is that most of the time in the discussions and debates Mississippi has on this forum they tend to be about matters of opinion, where no one can legitimately claim to be right or wrong. So in a manner of speaking he cannot be "wrong". This is different, however. There is only one correct view and if he sides with Hen then sadly he is wrong.

I know George, and I totally see why you feel that way, and to be honest I too felt that way when I first came across this problem, but in fact it isn't like that. What appears to be a 50 50 choice is actually skewed because your uncle has revealed something significant about one of the unpicked boxes. That's why I have come to the conclusion that there is little to be gained from pursuing the 1 in 2, 1 in 3, 2 in 3 approach. It just seems to get people going round and round in circles. Much better to consider the larger number approach (ie start with 25 boxes or 25 horses) and work downwards from there.

YES YES, but if the bastard has opened one of the 2 unpicked boxes and shown tyou it is empty, that leaves the box you riginally picked and the other unopened box. As the keys are in one or the other the choice IS 1 from 2, therefore at that stage of the game the chances are 50/50. The third box having been removed from the equation. Whether or not the sadistic git knows which one contains the keys is immaterial to me as I DON'T know.

I agree the choice at that stage is one from two (ie one set of keys, two boxes), but the odds are not evenly distributed between the two boxes. The first box you picked was chosen when there was a choice of three. Those one-in-three odds remain attached to that choice throughout the life of the problem. That leaves a two-in-three chance for whatever remains (it must be so because the total odds must always add up to one). Strange though it may sound your uncle revealing that Box B is empty concentrates those remaining two-in-three odds all into Box C. It must be so, because the odds of the keys being in B are now revealed as zero, and the odds of the keys being in A are one in three.. Tell you what. However tough this might sound, it's nowhere near as tough as trying to figure out how to use Nero!

Back to Hen. Your answer to my 25 horse question shows you have missed an important point about all of this. What you have to hold in your mind is that your uncle already knows the result, but you don't. When you pick your horse, you have a one in 25 chance of being right. You simply cannot argue with that. That is a fact of mathematics. That means there is a 24 in 25 chance that it will be one of the other horses who wins. If you are a probability genius you will certainly have to accept that. Now along comes your uncle and eliminates 23 of those remaining horses. He tells you they definitely did not win. So ask yourself, which is more likely? That you picked the winning horse when you went for a one-in-25 stab in the dark, or that your uncle has provided you with the winning horse, when he eliminates all the losers? Remember he knows the result of the race and you don't! Come on mate, there can only be one answer to this.

Just for you're entertainment..hehe..snighr..sniger..'bad man John! Baced on Ms. vos Savant's original column, read it carefully, saw a loophole and then suggested more trials. On the first, the contestant picked Door 1. "That's too bad," Mr. Hall said, opening Door 1. "You've won a goat." "But you didn't open another door yet or give me a chance to switch." "Where does it say I have to let you switch every time? I'm the master of the show. Here, try it again." On the second trial, the contestant again picked Door 1. Mr. Hall opened Door 3, revealing a goat. The contestant was about to switch to Door 2 when Mr. Hall pulled out a roll of bills. "You're sure you want Door No. 2?" he asked. "Before I show you what's behind that door, I will give you $3,000 in cash not to switch to it." "I'll switch to it." "Three thousand dollars," Mr. Hall repeated, shifting into his famous cadence. "Cash. Cash money. It could be a car, but it could be a goat. Four thousand." "I'll try the door." "Forty-five hundred. Forty-seven. Forty-eight. My last offer: Five thousand dollars." "Let's open the door." "You just ended up with a goat," he said, opening the door. The Problem With the Problem Mr. Hall continued: "Now do you see what happened there? The higher I got, the more you thought the car was behind Door 2. I wanted to con you into switching there, because I knew the car was behind 1. That's the kind of thing I can do when I'm in control of the game. You may think you have probability going for you when you follow the answer in her column, but there's the pyschological factor to consider." He proceeded to prove his case by winning the next eight rounds. Whenever the contestant began with the wrong door, Mr. Hall promptly opened it and awarded the goat; whenever the contestant started out with the right door, Mr. Hall allowed him to switch doors and get another goat. The only way to win a car would have been to disregard Ms. vos Savant's advice and stick with the original door. Was Mr. Hall cheating? Not according to the rules of the show, because he did have the option of not offering the switch, and he usually did not offer it. And although Mr. Hall might have been violating the spirit of Ms. vos Savant's problem, he was not violating its letter. Dr. Diaconis and Mr. Gardner both noticed the same loophole when they compared Ms. vos Savant's wording of the problem with the versions they had analyzed in their articles. "The problem is not well-formed," Mr. Gardner said, "unless it makes clear that the host must always open an empty door and offer the switch. Otherwise, if the host is malevolent, he may open another door only when it's to his advantage to let the player switch, and the probability of being right by switching could be as low as zero." Mr. Gardner said the ambiguity could be eliminated if the host promised ahead of time to open another door and then offer a switch. Ms. vos Savant acknowledged that the ambiguity did exist in her original statement. She said it was a minor assumption that should have been made obvious by her subsequent analyses, and that did not excuse her professorial critics. "I wouldn't have minded if they had raised that objection," she said Friday, "because it would mean they really understood the problem. But they never got beyond their first mistaken impression. That's what dismayed me." Still, because of the ambiguity in the wording, it is impossible to solve the problem as stated through mathematical reasoning. "The strict argument," Dr. Diaconis said, "would be that the question cannot be answered without knowing the motivation of the host." Which means, of course, that the only person who can answer this version of the Monty Hall Problem is Monty Hall himself. Here is what should be the last word on the subject: "If the host is required to open a door all the time and offer you a switch, then you should take the switch," he said. "But if he has the choice whether to allow a switch or not, beware. Caveat emptor. It all depends on his mood. "My only advice is, if you can get me to offer you $5,000 not to open the door, take the money and go home."

I propose a simplification of the plot. After you choose a box your uncle turns his back and you sneak a look in one of the other boxes, and see it is empty. Once you start messing around with two people trying to outwit each other you enter the realm of game theory, which has never got anyone anywhere.

Even I can understand that, and I'm a self confessed number dunce. You should be a Maths teacher... you seen the golden hello they are offering these days?

That's one of the reasons I changed the puzzle and inserted the following statements into my opening post: >>Your uncle is totally honest... Your uncle now offers you a further choice. You can stick with your first selection, Box A, or you can switch to Box C. He adds that you should read nothing into this offer. He was going to give you this choice whether or not you selected the right box in the first place.>>

Let's play then George - bring a pile of cash though, you'll need it.

I know Liana, there's a big demand for maths teachers, but since I dropped out of university after just one term, I doubt I've got what it takes.

'Probably' best to keep out of it the..?

Oh I dont know.. it takes a lot to get me to understand these dratted probability thingys.. I loathe maths with a passion. You should have another stab... open uni maybe?

You all going about this the wrong way - puzzling in deed What you do is - you seduce the uncle and get him to tell you but there again that would only work if your female so you males would have to torture him some other way - play him some horried washing up music - or get him to watch none stop Emerdale farm for a few days and nights that should do it ...

OK, I think I see where you're coming from. Instinct would suggest that he'd eliminated the right one. Mathematically, your problem is this: if you select the 1 in 25, your uncle has 24 possible other candidates to leave in. So there are actually 24 ways in which you could be right, just as there are 24 ways you could be wrong. Likewise, Dan - the problem with your diagram is that there is a split path from the second point. If you pick the right box, your uncle has the choice over which of the other two boxes to eliminate. This means there are two paths that result in you having chosen the right one initially.

OK here it is, the explanation that completely lays this one to rest. Imagine the three box conundrum is acted out 3,000 times by uncle and his two nephews: Mac and Zak (ie 6,000 times in all). Mac always sticks. Zak always switches. We know uncle places the keys randomly every time. We know there are three boxes. We know, therefore, there is a one-in-three chance the keys will turn up in any particular box. What's more, because the exercise is being carried out such a large number of times (3,000 in each case), the chances are that the results will even out neatly. In other words the keys will show up: - close to 1,000 times in Box A - close to 1,000 times in Box B - close to 1,000 times in Box C. Let's assume the results are exactly 1,000 in each case. Mac goes first. He always picks A and always sticks. Out of his 3,000 attempts he wins 1,000 times. No one can challenge that? Now it's Zak's turn. He too picks A to begin with, but then always switches to one of the other boxes. Here's how his 3,000 attempts break down: We know on 1,000 occasions the keys will be in A. So on those occasions Zak fails to get the keys. Why? Because Zak never selects A. We know on 1,000 occasions the keys will be in B. So on those occasions uncle will reveal Box C to be empty. Because Zak always switches, he will choose B and he will be right - 1,000 times. We know on 1,000 occasions the keys will be in C. So on those occasions uncle will reveal Box B to be empty. Because Zak always switches, he will choose C and he will be right - 1,000 times. So to sum up: Zak is wrong 1,000 times and he is right 1,000 plus 1,000 times. Which in my day used to add up to 2,000 times. So there you have it. Mac's sticking strategy gives him success 1,000 times. Zak's switching strategy gives him success 2,000 times. Now which bit of that does anybody not understand?

>>>If your uncle said, "You can switch and open Box B and C" you would ALWAYS switch, because you'd have a 2 in 3 chance of being right. Your uncle telling you that Box B is empty doesn't make any difference to the critical point, which is that you are offered the chance to choose to 'open Box B and C' and that the keys are twice as likely to be in that subset of boxes as they are in Box A.<<< Yes but the point, Andrew, is that your uncle has an influence on the result. If he did not allow you to open both boxes and you had to choose between B and C the odds are still 1/3. So your uncle saying you can open BOTH boxes sways the result in the same way his removal of box B sways the result. That is the critical point: The Uncle's influence. All you are doing is repeating points made by others (and me) in a more complex way. John, the ambiguity problem (and this thread) is best summed up by John Cleese (writing at his intellectual best) - roughly 30 years ago. From Monty Python: Neville Shunt's latest West End Success - It All Happened on the 11.20 from Hainault to Redhill via Horsham and Reigate, calling at Carshalton Beeches, Malmesbury, Tooting Bec and Croydon West is currently appearing at the Limp Theatre, Piccadilly. What Shunt is doing in this, as in his earlier nine plays, is to express the human condition in terms of British Rail. Some people have made the mistake of seeing Shunt's work as a load of rubbish about railway timetables, but clever people like me who talk loudly in restaurants see this as a deliberate ambiguity, a plea for understanding in a mechanised mansion. The points are frozen, the beast is dead. What is the difference? What indeed is the point? The point is frozen, the beast is late out of Paddington. The point is taken. If La Fontaine's elk would spurn Tom Jones the engine must be our head, the dining car our oesophagus, the guards van our left lung, the cattle truck our shins, the first class compartment the piece of skin at the nape of the neck and the level crossing an electric elk called Simon. The clarity is devastating. But where is the ambiguity? Over there in a box. Shunt is saying the 8.15 from Gillingham when in reality he means the 8.13 from Gillingham. The train is the same, only the time is altered. Ecce homo, ergo elk. La Fontaine knew its sister and knew her bloody well. The point is taken, the beast is moulting, the fluff gets up your nose. The illusion is complete; it is reality, the reality is illusion and the ambiguity is the only truth. But is the truth, as Hitchcock observes, in the box? No, there isn't room, the ambiguity has put on weight. The point is taken, the elk is dead, the beast stops at Swindon, Chabrol stops at nothing, I'm having treatment and La Fontaine can get knotted.