# Higgs boson chasers: Now only 1-in-300 MILLION chance we're wrong

CERN boffins are growing in confidence that the particle they spotted in the latest data from their Large Hadron Collider is indeed a Higgs boson. The Atlas experiment team has upped its level of certainty for Higgs-ness in a paper [PDF] for Physics Letters B, putting the sigma level at 5.9, which translates into a one-in-300- …

> Nevertheless, upping the significance of the particle find brings us a step closer to knowing that this is a new and Higgs-like elementary particle that has never been seen before. ®

HARRRRUMPH!

"Nevertheless, upping the significance of the particle find brings us a step closer to knowing that our mathematical descriptions of nature (based on Lagrangians, Gauge Symmetry, Lie Groups and yadda yadda) are *unreasonably* correct even though they cannot in the end be *totally* correct."

> Nevertheless, upping the significance of the particle find brings us a step closer to knowing that this is a new and Higgs-like elementary particle that has never been seen before. ®

HARRRRUMPH!

And you haven't SEEN it now either. But you can prove it or something like it exist.

Pedantic or what?

How can you prove that what you see is correct? Your optic sensors are fallible. The transmission architecture to your processing centers is fallible. The processing apparatus that post-processes the imagery is fallible.

Worse, that processing apparatus does not even have the capacity to process the imagery in real time; it substitutes imagery from previously stored data to compensate for the extremely low resolution imagery available from the sensors anywhere excepting the very center of their scope.

Thus the image you “see” is actually a composite of what truly exists. It is a mishmash of sensor distortions, transmission errors, filter bugs, memory retrieval errors and recompositing glitches that you choose to accept as reality. There is no scientific evidence to back up the claim that “what you can see” in fact represents reality at all.

All things being equal, there’s a reasonable chance that “what you can see” is a closeish approximation of “what is.” But it is not now and never will be a completely accurate representation of reality.

So if your standard for “what is real” is “what you can see,” then you have abjectly rejected science (and the fundamental principals that it is based upon) in its entirety.

[i]You[/i] are an entirely fallible piece of equipment.

#### Re: Trevor, would a new pair of

Adding a little bit of hardware to compensate for defects in the sensory apparatus can make the system more accurate. Unfortunately both the sensor design and the underlying system can only be corrected for so much.

#### The chance of being killed by a shark...

...is one in 300 million, yet 12 people died from shark attacks last year. It ain't over 'til the fat lady s...

(http://sharkfacts.org)

(http://www.telegraph.co.uk/earth/wildlife/9067719/Shark-deaths-hit-two-decade-high-worldwide-in-2011.html)

#### Re: The chance of being killed by a shark...

I think one of the problems there is that the 'probability of being killed by a shark' is such a staggeringly ill-defined notion as to make the 1-in-300-million figure completely risible.

#### Re: The chance of being killed by a shark...

But you're working with a sample size of 7 billion. What you're saying is that if we repeated the experiment 7 billion times, "just" one in 300 million is insufficient to describe certainty. And you'd be right. But that isn't what this says.

This says that, for any one person, observed on any one particular day, the odds of being killed by a shark / being the Higgs boson are 300 million to one. So it's the chances of you, in particular, personally, now, getting killed by a shark. Which is vanishingly small, even if you're swimming at the moment.

Don't try to interpret probabilities without (at minimum) a maths degree, or some sort of probability / statistics speciality. Because one you have a maths degree, you will realise just how dangerous it is to put your toe into probability without understanding tiny, minor differences that the human mind is built to see as equivalent when they can produce WILDLY varying results.

I refer you to the birthday problem, and the Monty Hall Problem. Even if you understand them, or have been shown them, would you really have been able to spot them on your own and get the answer right first time without any help or hints among a sea of mathematics? Chances are that you wouldn't have. And the tiniest change to circumstances makes every probabilistic calculation just as fraught with danger.

#### Re: The chance of being killed by a shark...

In case anyone here isn't familiar with those two problems, let me throw up the quick-and-dirty version of them.

Birthday Problem: In a room full of people, what are the odds of ANY TWO having the same birthday?

Monty Hall Problem: 3 curtains: 1 prize, 2 Zonks. You pick one. Monty reveals one of the ones you didn't pick (a Zonk) and gives you a chance to switch over to the other unopened curtain. Are you better off KEEPING or SWITCHING?

#### Re: The chance of being killed by a shark...

Birthday problem: You didn't say how big the room was. We would need to know this to figure out how many people are in the room. Obviously, you would need an infinite number of people to be certain two had the same birthday. The interesting bit (to ignorant, non-degree toting people with an interest in mathematics, or even statistics - like me) is that you only need 23 people to reach a more than 50% probability that any two people have the same birthday.

Monty Hall: Best to change (providing Monty knew what was behind the curtains), I believe.

#### Re: The chance of being killed by a shark...

Actually, the Birthday problem is more like (and the wording of the question makes a VAST difference to the answer here!):

How many people would you need in a room to ensure that there was a 50% chance of two of them sharing a birthday?

The answer is surprising, even when you know it.

#### Re: The chance of being killed by a shark...

How dare you criticize science. Don't you know it's the atheist's religion?

#### Re: The chance of being killed by a shark...

The problem with the Monty Hall problem is that it's SO counter-intuitive, it's almost unfair to ask it if you know the answer.

There was an American genius who published the question and answer in her newspaper column a few years back and invoked outrage from mathematics professors and all manner of "professionals" who should have known better - literally 100's of them wrote in to complain that the answer was wrong. It was embarrassing, especially because she was RIGHT, and it's one of the classic problems in probability and has been published, proven and known about for hundreds of years. And any statistician or other probability specialist should really not only be aware of it, but be able to prove why or - at the very least - run a simple simulation to provide themselves with anecdotal evidence which will corroborate the answer before they even LOOK at an equation.

Everyone wants to just say "It's the same chance - you still only have two doors so it's is still only 50-50 so switching or not makes no difference" and it can be quite hard or even impossible to explain to people why that's not true (the answer lies in the fact that your host has "prior" knowledge of the doors and aids you by opening a particular door that they had to KNOW which one had the booby prize behind).

I tell you now, I would never have got it right first time even if someone had thoroughly grounded me in probability and told me it was a "tricky" answer. And that's exactly what I'm talking about. Don't mess with the affairs of probabiliticians, for they are subtle, and stand a better-than-average chance of winning.

#### Re: The chance of being killed by a shark...

Yeah, but what if the shark has a FRIKKIN' LASER?

#### Re: The chance of being killed by a shark...

The Monty Hall Problem is makes much more sense once you've properly run through it, assuming that's not your speciality... there is a 1/3 chance you'll have the box from the off, and there is a 2/3 chance it'll be in the other two boxes... treated as a set, once one confirmed incorrect answer is removed from the set, there is still a remaining 2/3 chance that it's in that set, but now with just a single box as the option.

Course, you'd still be damn annoyed if you used your Monty Hall powers on a game show and it cost you that Yahama Jetski.

#### Re: Birthday

"How many people would you need in a room to ensure that there was a 50% chance of two of them sharing a birthday?"

Does the result change if it's a leap year?

#### Re: The chance of being killed by a shark...

Us pastafarian know that science is a dangerous trap. FSM Akbar.

#### Re: The chance of being killed by a shark...

>How dare you criticize science. Don't you know it's the atheist's religion?

Yes the socially acceptable thing to worship is a 2000 yo Jewish zombie.

#### Re: The chance of being killed by a shark...

"Yes the socially acceptable thing to worship is a 2000 yo Jewish zombie."

So if he wasn't a Jew, you wouldn't have a problem with it, right?

Maybe you want to go back and resume your attempts to equate homosexuality and pedophilia?

#### @Lee Dowling: Re: The chance of being killed by a shark...

The columnist was not necessarily a genius, but she was a savant. ; )

The story is here: http://en.wikipedia.org/wiki/Monty_Hall_problem: (rest of this post is an excerpt from the Wikipedia article. Also see good explanation of the Birthday Problem here: http://en.wikipedia.org/wiki/Birthday_problem )

"The problem was originally posed in a letter by Steve Selvin to the American Statistician in 1975 (Selvin 1975a) (Selvin 1975b). A well-known statement of the problem was published in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990 (vos Savant 1990):

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Vos Savant's response was that the contestant should always switch to the other door. If the car is initially equally likely to be behind each door, a player who picks Door 1 and doesn't switch has a 1 in 3 chance of winning the car while a player who picks Door 1 and does switch has a 2 in 3 chance. The host has removed an incorrect option from the unchosen doors, so contestants who switch double their chances of winning the car.

Many readers refused to believe that switching is beneficial. After the Monty Hall problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming that vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy."

#### Re: The chance of being killed by a shark...

Interestingly, there is a much much bigger danger...

Interpreting statements with a beautiful degree in 'n' and yet still managing to miss the ironic nature of a statement, especially when the original statement ended with the phrase:

"It ain't over 'til the fat lady s..."

The "s..." was there to imply that the fat lady's singing had indeed been cut short, because it was over.

But then perhaps that is just too subtle for someone with a maths degree...

(http://en.wikipedia.org/wiki/Irony)

#### Re: The chance of being killed by a shark...

no I don't add the jewish part due to racism but to accent just how ridiculous the whole thing is. Faith is for people with high EQ but a basic lack of science reasoning ability. Yes their are some but the majority of Nobel Prize winners were never dunked in a river.

#### Monty hall problem

The problem is too often posed without explicitly mentioning that the host has prior knowledge; that is the crux of the answer. The more priors we have the more the probability changes; we should consider that in then Higgs case.

Serious question, not being sarky: why is the LHC being shut down later this year? It seems to be a bloody useful piece of kit.

#### Like a DeLorean with quantum go-faster stripes

They are shutting it down to upgrade the hardware so that collisions can go from (I think) 7 Tev/c² in the center-of-mass to 12 TeV/c² in the center-of-mass without blowing the tunnel sky-high. This is still lower than the hoped-for 14 TeV/c² but that value is no longer considered safe (for the copper interconnects, that is). It will be back up in 2 years or so. I suspect the software will have been upgraded too, new experiments and filters suggested and the existing data hadooped something fierce.

#### Re: why is the LHC being shut down later this year?

Ah, thank you chaps. I've obviously misunderstood something somewhere along the line. Glad to hear it, too.

Stylee said: "Birthday problem: You didn't say how big the room was. We would need to know this to figure out how many people are in the room. Obviously, you would need an infinite number of people to be certain two had the same birthday."

No, you only need 367 people, and it really doesn't matter how big the room is.

Well, it's a hypothetical room but to be fair... Where you got 367 from is beyond me. You can fit 23 people in plenty of rooms.

#### Re: To Infinity and...

Well, to be fair it's more for effect... I can see how he got 367, but the amount of coincidence and a fundamental misunderstanding of the problem aside, I really can't see why he got 367.

#### Re: To Infinity and...

Yes, 367 would guarantee two the same, accounting for leap years, and as long as we are ignoring year of birth - infinity was obviously a daft thing to say, innit. However, the size of the room (and people) does matter if we are talking about 'filling a room' with people.

#### Re: To Infinity and...

*I can see how he got 367, but the amount of coincidence and a fundamental misunderstanding of the problem aside, I really can't see why he got 367.*

That's because you didn't read carefully enough. He wasn't answering the Birthday Problem question (number of people required to have a birthday match with probability 0.5); he was responding to a statement about how many people were required to have a matching birthday "with certainty" (which he understood as "with probability 1.0" - and I agree that's what the person he was quoting seems to have meant).

So the figure 367 is correct, and he got it by reading the statement he was responding to, instead of seeing what he expected to see there.

#### Long odds

One in three hundred million? That's never going to happen.

Now if you'll excuse me, I have to buy some Euromillions tickets for a shot at Friday's massive rollover.

"Where you got 367 from is beyond me"

Its obvious to anyone who thinks about the problem and how to solve it. There are 366 possible birthdays. Therefore, in any group of 367 people there must be at least two with the same birthday no matter how those birthdays are distributed. Where you got the idea that you need an infinite number of people from to be certain two have the same birthday is beyond me, but you said it was "obvious".And I repeat, the size of the room is not important. Its how many people who are in the room, no matter how big it is, that matters.

Now someone is goign to say the answer isn't 367, its only 23,.....

Wouldn't you only need 2 people.

Either they have the same birthday or they don't, it's a 1 in 2 chance, therefore a 50% chance.

#### Oh dear...

What's the chance of your birthday being the 5th of March?

That's the chance that your birthday is the same as mine.

(I hope you're trolling...)

@AC: It'll be interesting to see how many posts in we get before that!

Actually, with 367 people the odds are not all the days in a year would be someone's birthday. There would more than likely be several sets of people with the same birthday in that group. Thus leaving some days vacant.

That's because the two of you are solving two different problems. You're trying to figure out how many people would it take to make it at least even money that any two people in the room have the same birthday. It can't be two--the odds there are ~365:1 (pray one of them doesn't have February 29th).

The person going for 367 is trying to solve the problem of "How many people would you need to be CERTAIN at least two people have the same birthday?"

Khaptain, so by your logic, you either win the lottery or you don't so you think you have a 50% chance of winning? In fact, you can win the main prize or one of several other lesser prizes so maybe you think there are 10 possibilities, nine of which are wins, so you have a 90% chance of winning.

Ah! So **that's** how it works!

This week, I'll be buying **two** lottery tickets then. That should pretty much guarentee me a prize :-)

#### Re: just like rolling a six on a dice.

Good to see that at least one person is keeping up :-)

#### Depends on the kind of day you're having...

Obviously the answer to the birthday puzzle makes sense. But as far as the Monty Hall puzzle or the likelihood of rolling a 6, if anyone is having the kind of day I had Tuesday, it's likely they'd never win. You could still roll anything but 6 in anything but an infinite number of tries. It's extremely unlikely beyond pondering, but possible. Just as there is a very, very small but finite possibility that an elementary particle will not appear within the statistical expectations, but light years away, or that the particles that make up your body could line up perfectly, allowing you to walk through a wall. It might take longer than a googol times the lifetime of the universe to occur, but it still has some possibility of happening at any time. Reality is what we make it... or what it makes us.

#### The God Particle (haiku by James Ph. Kotsbar)

It’s the right "footprint."

With 5-sigma certainty,

there it was; we’re sure.

#### Collapsing reality

I'm willing to be wrong, but my understanding of the (uncle?) Monty problem is that he opens all the bad doors.

Therefore, to exaggerate for effect, you buy one lottery ticket and just before you see the winning numbers, but after Monty has seen them, he gives you the choice of keeping your ticket or a ticket he hands you - one of them is the winning ticket.

I think under those conditions you can see it makes a LOT of sense to switch.

However, if the conditions were different and you were presented with two tickets and told that one was the winning ticket, then, yes the odds of picking the right ticket is 50-50 (50% dice comment above?)

Well, that's my understanding of the frame of the question - I don't know if Monty always opened a bad door or if he would open all of the bad doors in an exaggerated example.