lavoisier

Very simple explanation of this problem, I think it was given on a BBC programme about mathematics: - when you make your initial choice, you have 2 possible outcomes: right casket (1/3) or wrong casket (2/3) - next step: the barker reveals one empty casket, and you're asked if you want to switch Case A - you don't switch --> then your chances of having chosen the right one remain 1/3 Case B - you switch --> then if you had originally chosen the right one (1/3), you lose; if you had chosen the wrong one (2/3), you win, because both empty caskets are now excluded, and you can't pick but the right one by switching. In practice you swap your winning chances with your losing chances. So: - if you don't switch, the chances are 1/3 win, 2/3 lose as in the initial choice. - if you switch, the chances are 2/3 win, 1/3 lose, which means you've doubled your chances of winning. The presenter on the programme said there's no shame in not getting this the first time round - apparently the problem puzzled even some professional mathematicians.

OK, I got it now. I hadn't considered that the two containers were not identical, because they had flags attached to them. This automatically invalidates the assumption of equal probability of the two outcomes, thus ruling out any comparison with coin flipping. If someone had good reasons to believe that Germany was likely to win most matches, and, as you suggested, had trained Paul to go for the German flag, the game was already biased from the start. So no miracle here. That's a relief.

Hi CharonY, there's one thing I don't understand about your reply. Even if we assume the octopus was 'steered' or influenced in some way to choose the right box, this still leaves an important question: who knew the outcomes of the matches before they had been played? Unless this octopus is actually a ruthless businessman, who fixed the matches in advance to secure himself a career in show-business.

Thank you both for your replies. I seem to understand that there is quite strong evidence that the octopus was not acting randomly. I also found this on Wikipedia, which seems to be very close to my example: http://en.wikipedia.org/wiki/Statistical_hypothesis_testing#Example_2_-_Clairvoyant_Card_Game This octopus remains a mystery to me.

Hi everybody, you might have heard of an octopus who correctly predicted the winners of some matches in the Football World Cup. I think it was 7 matches he guessed right. I was discussing this with a friend last Sunday, and I was rather amazed when he told me there was nothing special in the octopus' achievement: he could have just guessed them right by chance. And his explanation was that if you toss a coin 7 times, any sequence of heads/tails has the same probability of happening: we only pay special attention to the one with all tails or all heads. I later examined the problem in detail. As you all know, there are 128 possible sequences, and in fact it's true that any individual sequence (where order matters) has exactly 1/128 chances to happen. On the other hand, if you believe the octopus acted completely at random, you should explain how he picked the right sequence out of a set of 128 possible sequences, most of which contain a mix of wrong/right. He only had about 0.8% chances to pick this one, as opposed to 99.2% chances of failing at least one prediction, and still he did pick it! So now I'm left with this fundamental doubt about probability. Say your experiment is 'toss a coin 7 times'. If you repeat the experiment N times, on average you expect to find 7 heads (or 7 tails) N/128 times. But if you only do the experiment once, can you reach any conclusion about the coin being fair (or in our case, the octopus being psychic) based on the likelihood of the sequence you observe? Or do you always need to repeat the experiment at least a few times? Is there any statistic test one can use to determine that? I'm sure this is a very simple problem for mathematicians, but I do find probability concepts puzzling at times. Thanks L.

Hi am24, sometimes the boundary between SN1 and SN2 is a bit blurred. These exercises are OK for learning the concept, but in practice my advice is don't be too dogmatic about it. The basic message I think you need to grasp is that all the conditions that make the carbocation more stable favour the SN1 mechanism, and all those that make it more unstable favour SN2. It is known that carbocations can be stabilized by field effect (e.g. alkyl substituents - that's why tert-butyl cation is more stable than isopropyl cation) and by resonance (e.g. benzyl cation is more stable than cyclopropylmethyl cation). The better the leaving group, the easier the formation of the carbocation. In general, I->Br->Cl- for alkyl halides. Adding Ag+ favours the formation of the carbocation by making the halogen a better leaving group (because it can form a strong bond with the metal). In addition to that, the solvent can form interactions that stabilize the carbocations. And temperature plays a role too. Therefore, just writing down any two reagents and asking for a definite prediction of the mechanism is a bit pretentious. But there you go, I'm afraid you'll have to live with it as long as you're a student.

Interesting debate... Personally, I think that very little or no culture in general is strictly 'necessary' (i.e. you can't do without). At the end of the day we are just animals, and our fundamental needs are just preserving ourselves and our species (therefore getting food, shelter, etc.). However, as our society is much more complex today than it was in prehistoric times, some knowledge (and in particular mathematics) helps you deal with other people and with some simple tasks most of us need to perform (e.g. using money). And in fact culture, both scientific and artistic, enriches our lives with something more than just satisfying our basic needs.

That would be true if you did this reaction in presence of at least 2 Eq of a suitable base; but when both OH and COOH aren't deprotonated, does the same explanation still hold? Sometimes these chemoselective reactions are difficult to rationalise. o-aminophenol, for instance, reacts with some electrophiles at the oxygen in presence of Et3N (because you have O- and NH2 competing), and at the nitrogen in presence of pyridine (because you have OH and NH2 competing). Here, with both groups still in the acid form, I think it's more a matter of electron density.

Hi everybody, I am trying to solve a mathematical problem that came out of a TV programme I was watching; I will explain how this happened, and I hope someone can help (I believe the problem has long been solved). I think the programme was called 'psychic challenge', a sort of game where they wanted to check whether some people were 'real' mediums or not by putting them through some tests. One of the games consisted in showing the 'medium' 5 men and 5 women, and knowing that they were 5 couples, match each wife to her husband. One of the mediums guessed just one out of five, and everybody was happy with that, but then it occurred to me that if you matched these people at random, guessing only one couple right would be more probable than guessing all of them wrong, i.e. you would be more of a 'medium' if you didn't guess any of them right! Then I tried to write a formula to calculate the probability (I'm not a mathematician, so please excuse any blunder). I will not go through all the details of my reasoning; I will just say that first I realised that the usual conditional probability formula didn't work here, so I went back to basics (set theory), expressed the probability of a sequence of dependent events in terms of independent ones, and ended up with this formula: [math]P(k,N) = \frac{1}{k!} \sum_{i=k}^{N}\frac{(-1)^{i-k}}{(i-k)!}[/math] This would be the probability that, matching at random the N wives to the N husbands one gets k (and only k) of them right. Applying it to N = 5 yields: [P(0,5) = 11/30, P(1,5) = 3/8, P(2,5) = 1/6, P(3,5) = 1/12, P(4,5) = 0, P(5,5) = 1/120] which is indeed correct if you enumerate the 120 possible dispositions and count them. My question is: is there a simplified version of the formula, like a closed form of the alternating factorial summation?

OK, I got your point. In fact this raises an interesting question, i.e. in what cases would the energy of the meteor be transformed into different forms of energy which wouldn't directly affect our ecosystem (i.e. setting aside the effects of the suspended dusts and stuff)? I'm trying to reason about that - I beg the physicists here to accept my apologies, I'm obviously not in the field. The total mechanical energy of the meteor is the sum of its potential and kinetic energy. While it falls towards the Earth, the latter becomes the main term. I don't know if the kinetic energy of the meteor when it's very far from the Earth is a significant part of the total, but anyway... If the kinetic energy was entirely transformed into heat, we'd have the consequences you calculated. But what other forms of energy can we get? If, say, the meteor was 'unbreakable', it would hit the surface, and then what? Would it dissipate its energy by breaking the chemical bonds in the rocks it hits and/or as potential energy by changing the position of the rocks (i.e. forming a crater)? And if the rocks on the surface were too hard to break, what would happen? Where would the energy go? My impression is that it's quite difficult to predict exactly what would actually happen. However, it's probably true that any impact which converts most of the energy of the meteor into heat or its equivalents (e.g. burning in the atmosphere or vaporizing a portion of the oceans), rather than into potential energy, is more dangerous for our ecosystem. Unless most of the heat and dusts are eventually expelled into space. And it's also probably true that disintegrating the meteor before the impact wouldn't be particularly helpful, especially if all of its parts would hit us anyway.

You're right, it would be a catastrophic event anyway. I was not trying to minimise it. We can all stay assured that if something like this happens, it won't be fun for anyone. However, your starting point was the energy of the hypothetical meteor which is supposed to have killed the dinosaurs. So we're talking about an event that, if hasn't actually happened, at least has been simulated by someone (on the basis of a given set of parameters) in order to justify the extinction of these animals. This means that its effects (according to the model used) are already known, nobody needs to reinvent them. My worry is that sometimes we need to over-simplify models, and when we deal with VERY complex systems (let alone those subject to chaotic behaviour), at the end of the day our prediction may be largely unreliable.

OK, I have the proof. It's in Planetmath, both for the mean and the variance. Thank you anyway for your help.

Sorry, I'm new in this forum. I didn't mean to confuse anybody. If and when I post new threads I will use LaTex. However, your guess was correct. I've found a partial answer to my question, i.e. the function PR(i) is the hypergeometric distribution, just expressed in a different format. Mathworld explains how the mean and the variance are calculated, starting from the Bernoulli distribution. It looks very easy in that way. But it also gives for granted that some awful sums are easy to prove, which was my initial doubt.

Hi, I'm unsuccessfully looking for the proof of some sums. I'm sure they are very well known, so I wonder if some mathematician could redirect me to useful resources on the web, or just give me a hint about where to start for solving them. The first is the famous binomial. If: PP(i) = p^i * (1-p)^(s-i) * s! / ( i! * (s-i)! ) with 0<p<1, i and s positive integers, then I know the proof for: Sum(PP(i), i, 0, s) = 1 But I can't work out the proof for: Sum(i * PP(i), i, 0, s) = s*p and for: Sum((i-s*p)^2 * PP(i), i, 0 , s) = p*(1-p)*s A related problem leads to the definition of: PR(i) = Product((n*p-j)*(s-j)/((n-j)*(j+1)), j, 0, i-1) * Product((n-n*p-j)/(n-i-j), j, 0, s-i-1) where 0<p<1, 1<=s<=n, s and n positive integers. Here I don't have a clue for any of the three: Sum(PR(i), i, 0, s) = 1 Sum(i * PR(i), i, 0, s) = s*p Sum((i-s*p)^2 * PR(i), i, 0 , s) = p*(1-p)*s*(n-s)/(n-1) Any help will be greatly appreciated.

Thanks for your reply. However, it doesn't justify either the fact that the uncertainty decreases as more and more voting papers are opened, or the fact that opened voting papers are never recounted. I.e. your sample s of the population n is such that s<<n. I had some further reading about this problem, and in fact I found the answer in a statistics book. msd^2 = p*(1-p)/s * (n-s)/(n-1) Here, as s approaches n the uncertainty tends to 0, as it should be. This raises another question, which I have posted in a new thread.

Hi, I'm afraid this problem will sound ridiculously elementary to mathematicians, but I'm dumb enough to be unable to get around it. I've searched the web for answers, but I get either too complicated stuff, or stuff I already know but which looks unrelated to my actual question (e.g. the normal distribution). After an election, N (closed) voting papers are found when opening the total of the ballot boxes. Then people start opening them, counting the votes and hence calculating the provisional results. So far, no problems. However, if at a given moment we've opened S (randomly chosen) voting papers out of the total N, is there a formula, allowing us to calculate the 'confidence interval' of the results? I'm not sure whether that's the right name for it, what I mean is the analogue of the mean standard deviation for physical measures. I guess the formula exists, because in fact in 'real' elections, before all the papers have been opened they DO say something like 'the candidate got 30% of the votes +/- 5%'. But I can't figure out if this 5% is simply a function of N, S (and possibly an arbitrary parameter like the confidence level used in statistical tests), approaching 0% for S-->N, or if it also depends on the provisional results. What puzzles me most is the fact that in physical measures your SD comes from errors in measuring a single, 'continuous' variable, whereas here the variables are necessarily more than one (the %s of the single candidates plus blank, void...), and they are discrete (rational with denominator N). Can anyone help me with that? Thanks

Hi Sulfinator, 2,6-lutidine (pKa of BH+ = 6.65) is a much weaker base than TEA (pKa of BH+ = 10.75), and it's more lipophilic, too, that's why the bicarbonate wash can't remove it. If your acid-labile group isn't exceedingly acid-sensitive (check in Greene's P.G.) you may want to try an acidic wash with something milder than H2SO4, maybe a buffered solution of some organic or weak inorganic acid. It only needs to have a pH lower than 6.65 (in practice say <5.65), so you have a huge choice. I would also do a bicarbonate wash after that, just to make sure that no acid stays there and cleaves your group when you concentrate the organic extract (it happened to me!). It's probably worth trying the simple option before turning to metals and the like. Hope it helps.