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Pedro Pablo Pérez Velasco

Attends U.A.M.

Lives in cobeña

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### Pedro Pablo Pérez Velasco

Shared publicly -On this day, exactly one week ago, New Horizons made it to Pluto.

To commemorate this day...

http://www.fromquarkstoquasars.com/that-pluto-controversy/

Image credit: Eduardo Salles http://cinismoilustrado.com/

To commemorate this day...

http://www.fromquarkstoquasars.com/that-pluto-controversy/

Image credit: Eduardo Salles http://cinismoilustrado.com/

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### Pedro Pablo Pérez Velasco

Shared publicly -A brief history of Pluto.

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### Pedro Pablo Pérez Velasco

Shared publicly -This is just too amazing for words.

Thanks to New Horizons' Pluto flyby, we finally have a complete family portrait.

Friends, say hello to the relatives...

http://www.fromquarkstoquasars.com/nasas-live-coverage-of-pluto-flyby-begins-now-see-it-here/

Image made by @bhgross144

Thanks to New Horizons' Pluto flyby, we finally have a complete family portrait.

Friends, say hello to the relatives...

http://www.fromquarkstoquasars.com/nasas-live-coverage-of-pluto-flyby-begins-now-see-it-here/

Image made by @bhgross144

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### Pedro Pablo Pérez Velasco

Shared publicly -**Proton collision in the Large Hadron Collider**

The Large Hadron Collider (LHC) is the world's largest and most powerful particle collider and the largest single machine on the planet. Built by CERN between 1998 and 2008, it resides within a 27km (17mi) circular tunnel 175m (574ft) beneath the Franco-Swiss border near Geneva, Switzerland. Over 10,000 scientists and engineers from 100+ countries were involved in the design, construction, and use of the collider.

The LHC allows physicists to test the predictions of different particle physics theories and to study high-energy physics. This animation shows an actual proton collision event in 2010 detected by ATLAS, one of seven particle detectors built into the LHC. Protons are accelerated up to 99.9999991% the speed of light in the main LHC ring. By smashing together protons moving in opposite directions at extremely high speeds, it becomes possible to study massive particles previously not observable with lower-energy accelerators.

Source: https://youtu.be/NhXMXiXOWAA

#ScienceGIF #Science #GIF #LHC #LargeHadronCollider #Physics #StandardModel #CERN #Animation #ATLAS #Collision #ParticleAccelerator

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### Pedro Pablo Pérez Velasco

Shared publicly -Vortex Bladeless has designed a tower that dispenses the blades and uses what is commonly known as vorticity to generate electricity.

Vortex Bladeless has designed a tower that dispenses the blades and uses what is commonly known as vorticity to generate electricity.

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### Pedro Pablo Pérez Velasco

Shared publicly -Most of us have been exposed to the concepts of astrophysics, the Big Bang theory, and evolution. Some may be very well versed, others may have a rudimentary understanding Either way - I can promise you will never have heard them explained so eloquently, clearly and simplistically as you will in the following video. One of the worlds favourite science communicators takes us on a journey from before the beginning of the universe, all the way throu...

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### Pedro Pablo Pérez Velasco

Shared publicly -**The power of randomness**

Here's a puzzle:

**I write down two different numbers that are completely unknown to you, and hold one in my left hand and one in my right. You have absolutely no idea how I generated these two numbers. Which is larger?**

**You can point to one of my hands, and I will show you the number in it. Then you can decide to either select the number you have seen or switch to the number you have not seen, held in the other hand. Is there a strategy that will give you a greater than 50% chance of choosing the larger number, no matter which two numbers I write down?**

At first it seems the answer is

*no*. Whatever number you see, the other number could be larger or smaller. There's no way to tell. So obviously you can't get a better than 50% chance of picking the hand with the largest number - even if you've seen one of those numbers!

But "obviously" is not a proof. Sometimes "obvious" things are wrong!

It turns out that, amazingly, the answer to the puzzle is

*yes*. You

*can*find a strategy to do better than 50%. But the strategy uses randomness.

I'd seen this puzzle before - do you know who invented it?

If you want to solve it yourself, stop now or read

*Quanta*magazine for some clues - they offered a small prize for the best answer.

Otherwise, you can read Greg Egan's answer, which seems like the best answer to me.

I'll paraphrase it here:

Pick some function f(x) defined for all real numbers, such that:

the limit as x → -∞ of f(x) is 0,

the limit as x → +∞ of f(x) is 1,

whenever x > y, f(x) > f(y).

(There are lots of functions like this; choose any one.)

Next, pick one hand at random. If the number you are shown is x, compute f(x). Then generate a uniform random number z between 0 and 1. If z is less than or equal to f(x) guess that x is the larger number, but if z is greater than f(x) guess that the larger number is in the other hand.

The probability of guessing correctly can be calculated as the probability of seeing the larger number initially and then, correctly, sticking with it, plus the probability of seeing the smaller number initially and then, correctly, choosing the other hand.

This is

0.5 f(x) + 0.5 (1 - f(y)) = 0.5 + 0.5(f(x) – f(y))

This is strictly greater than 0.5, since x > y so f(x) - f(y) > 0.

So, you have a more than 50% chance of winning! But as you play the game, there's no way to tell

*how much*more than 50%. If the numbers on the other players hands are very large, or very small, your chance will be just slightly more than 50%.

**Puzzle 1:**Prove that no deterministic strategy can guarantee you have a more than 50% chance of choosing the larger number.

**Puzzle 2:**There are perfectly specific but 'algorithmically random' sequences of bits, which can't predicted well by any program. If we use these to generate a uniform

*algorithmically random*number between 0 and 1, and use the strategy Egan describes, will our chance of choosing the larger number be more than 50%, or not?

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### Pedro Pablo Pérez Velasco

Shared publicly -A brief history of Pluto.

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### Pedro Pablo Pérez Velasco

Shared publicly -Why do tennis players serve differently when it's their first serve? Here are two arguments that go in opposite directions.

Argument 1. The first time round, they can afford to take a bigger risk on their serve, because they'll have a second chance if they miss.

Argument 2. If the first-serve probability is p, the probability of winning the point given that the first serve goes in is q, the second-serve probability is r, and the probability of winning the point given that the second serve goes in is s, then to maximize your chances of winning, you should do a "first serve" if pq>rs and a "second serve" if rs>pq.

The second argument, though it has a certain plausibility to it, is wrong, for the following reason: it ignores the fact that if it's your first serve, then you don't have to win the point,

So what is the right strategy, for given p,q,r,s? If it's your second serve, then you've got just one attempt, so Argument 2 is correct in that situation. That is, you do a second serve if and only if rs>pq, which presumably it is for most tennis players (though I'd be very interested to see whether there are in fact some players who in some matches would have been better off doing first serves all the time).

As for the first serve, let's assume that rs>pq. If you actually do a first serve and then follow up with a second serve if you miss it, then your chances of winning the point are pq + (1-p)rs. If you do second serves both times, then your chances are rs + (1-r)rs. So it's better to do a first serve on your first serve provided that p(q-rs)>rs(1-r), or equivalently rs-pq>(r-p)rs. I can't quite get my head round what the right-hand side means here, but let's take an example or two. Suppose that p=1/2, q=1, r=1, s=2/3. (That is, your first serve is devastating if it goes in, but does so with only a 50% chance, and your second serve always goes in but leads to only a 2/3 chance of winning the point.) Then rs-pq = 1/6, and (r-p)rs = 1/3. In that case it would be better to do a second serve on your first serve.

Suppose now that p=1/2, q=1, r=1, s=1/2. That is, a first serve with a 50% chance of going in guarantees the point, whereas a second serve that's guaranteed to go in gives you a 50% chance of the point. Here it's rather easy to see why doing a first serve is better: you have a 50% chance of winning the point there and then, but if lose it you get a second chance, whereas for a second serve it's the same but without the second chance. Doing the arithmetic, we find that rs-pq = 0 and (r-p)rs = 1/4. Since there's a non-zero difference, we can perturb these numbers a little bit and get an example where pq<rs but it's still better to start with a first serve.

Now let's see whether Djokovic and Federer got their tactics right in the Wimbledon final this afternoon. (Before doing the calculation, I strongly expect the answer to be yes, since being good at judging the percentages must surely give you the edge as a professional tennis player.)

For Djokovic we had p = 0.66, q = 0.74, r = 1 (approximately) and s = 0.6. So pq<rs, which is a good start, as it shows that Djokovic was right not to do first serves for his second serves. But also, rs - pq was approximately 0.11 and (r-p)rs was approximately 0.204, which was indeed bigger.

For Federer the figures were p=0.67, q=0.74, r=1, s=0.49. Interestingly, pq = 0.4958, so it looks as though Federer would have done minutely better by doing first serves all the time. (Of course, there may well be other factors to take into account, but still I find this interesting.) Not surprisingly in the light of this, given that pq and rs are at least approximately equal whereas (r-p)rs is distinctly positive, he was definitely right to do first serves for his first serves.

Based on this little experiment, I'd be willing to bet that many tennis players play suboptimally in many matches in this simple respect. Of course, I've oversimplified things quite a bit since there are different levels of risk one can take even with a first serve. So let's do the very slightly more complicated analysis. Suppose that if your serve goes in with probability p, then your probability of winning the point is f(p). Here f(p) decreases as p increases: given that it goes in, a safer serve is less likely to lead to your winning the point.

Then when it's your second serve, you should choose p to maximize the product pf(p). That's straightforward. Suppose that that maximum value is t. Then if your first serve goes in with probability p and you do the right thing for your second serve, then your probability of winning the point is pf(p) + (1-p)t, so you want to maximize p(f(p)-t). That is equivalent to maximizing log(p) + log(f(p)-t), so you want 1/p + f'(p)/(f(p)-t) to be zero. (Note that f'(p) is negative.) Equivalently, you want p to equal (f(p)-t)/(-f'(p)). Or -pf'(p) = f(p) - t. Here the right-hand side measures the advantage you get, if your first serve goes in, over your winning probability if it doesn't go in. The left-hand side seems harder to interpret: certainly I don't see an intuitive explanation for why the maximum should occur at a p for which this equation is satisfied, or an intuitive explanation of how you should choose the risk level of your first serve.

Can we at least show that a first serve should be riskier than a second serve? Suppose we take p = r+delta, which results in q = s - eta, with delta and eta small. Then to first order rs-pq = eta p - delta q, while (r-p)rs = delta rs. It's possible for eta p - delta q to be pretty small but positive, so it looks possible that for some players it's best to do second serves all the time. I think I may be such a player -- my second serve is risibly bad but its chances of going in drop off very quickly if I try to make it harder to return.

Argument 1. The first time round, they can afford to take a bigger risk on their serve, because they'll have a second chance if they miss.

Argument 2. If the first-serve probability is p, the probability of winning the point given that the first serve goes in is q, the second-serve probability is r, and the probability of winning the point given that the second serve goes in is s, then to maximize your chances of winning, you should do a "first serve" if pq>rs and a "second serve" if rs>pq.

The second argument, though it has a certain plausibility to it, is wrong, for the following reason: it ignores the fact that if it's your first serve, then you don't have to win the point,

*as long as you lose it by missing your serve*. So it's wrong to say that you are aiming to maximize the probability of winning the point during that rally.So what is the right strategy, for given p,q,r,s? If it's your second serve, then you've got just one attempt, so Argument 2 is correct in that situation. That is, you do a second serve if and only if rs>pq, which presumably it is for most tennis players (though I'd be very interested to see whether there are in fact some players who in some matches would have been better off doing first serves all the time).

As for the first serve, let's assume that rs>pq. If you actually do a first serve and then follow up with a second serve if you miss it, then your chances of winning the point are pq + (1-p)rs. If you do second serves both times, then your chances are rs + (1-r)rs. So it's better to do a first serve on your first serve provided that p(q-rs)>rs(1-r), or equivalently rs-pq>(r-p)rs. I can't quite get my head round what the right-hand side means here, but let's take an example or two. Suppose that p=1/2, q=1, r=1, s=2/3. (That is, your first serve is devastating if it goes in, but does so with only a 50% chance, and your second serve always goes in but leads to only a 2/3 chance of winning the point.) Then rs-pq = 1/6, and (r-p)rs = 1/3. In that case it would be better to do a second serve on your first serve.

Suppose now that p=1/2, q=1, r=1, s=1/2. That is, a first serve with a 50% chance of going in guarantees the point, whereas a second serve that's guaranteed to go in gives you a 50% chance of the point. Here it's rather easy to see why doing a first serve is better: you have a 50% chance of winning the point there and then, but if lose it you get a second chance, whereas for a second serve it's the same but without the second chance. Doing the arithmetic, we find that rs-pq = 0 and (r-p)rs = 1/4. Since there's a non-zero difference, we can perturb these numbers a little bit and get an example where pq<rs but it's still better to start with a first serve.

Now let's see whether Djokovic and Federer got their tactics right in the Wimbledon final this afternoon. (Before doing the calculation, I strongly expect the answer to be yes, since being good at judging the percentages must surely give you the edge as a professional tennis player.)

For Djokovic we had p = 0.66, q = 0.74, r = 1 (approximately) and s = 0.6. So pq<rs, which is a good start, as it shows that Djokovic was right not to do first serves for his second serves. But also, rs - pq was approximately 0.11 and (r-p)rs was approximately 0.204, which was indeed bigger.

For Federer the figures were p=0.67, q=0.74, r=1, s=0.49. Interestingly, pq = 0.4958, so it looks as though Federer would have done minutely better by doing first serves all the time. (Of course, there may well be other factors to take into account, but still I find this interesting.) Not surprisingly in the light of this, given that pq and rs are at least approximately equal whereas (r-p)rs is distinctly positive, he was definitely right to do first serves for his first serves.

Based on this little experiment, I'd be willing to bet that many tennis players play suboptimally in many matches in this simple respect. Of course, I've oversimplified things quite a bit since there are different levels of risk one can take even with a first serve. So let's do the very slightly more complicated analysis. Suppose that if your serve goes in with probability p, then your probability of winning the point is f(p). Here f(p) decreases as p increases: given that it goes in, a safer serve is less likely to lead to your winning the point.

Then when it's your second serve, you should choose p to maximize the product pf(p). That's straightforward. Suppose that that maximum value is t. Then if your first serve goes in with probability p and you do the right thing for your second serve, then your probability of winning the point is pf(p) + (1-p)t, so you want to maximize p(f(p)-t). That is equivalent to maximizing log(p) + log(f(p)-t), so you want 1/p + f'(p)/(f(p)-t) to be zero. (Note that f'(p) is negative.) Equivalently, you want p to equal (f(p)-t)/(-f'(p)). Or -pf'(p) = f(p) - t. Here the right-hand side measures the advantage you get, if your first serve goes in, over your winning probability if it doesn't go in. The left-hand side seems harder to interpret: certainly I don't see an intuitive explanation for why the maximum should occur at a p for which this equation is satisfied, or an intuitive explanation of how you should choose the risk level of your first serve.

Can we at least show that a first serve should be riskier than a second serve? Suppose we take p = r+delta, which results in q = s - eta, with delta and eta small. Then to first order rs-pq = eta p - delta q, while (r-p)rs = delta rs. It's possible for eta p - delta q to be pretty small but positive, so it looks possible that for some players it's best to do second serves all the time. I think I may be such a player -- my second serve is risibly bad but its chances of going in drop off very quickly if I try to make it harder to return.

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### Pedro Pablo Pérez Velasco

Shared publicly -What do you think?

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Story

Introduction

born, grown, reproduced and, eventually, dead

Education

- U.A.M.present

Basic Information

Gender

Male

Work

Occupation

quantitative analysis (Risk)

Places

Currently

cobeña

Previously

madrid

Links

YouTube

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