## ABC not in Svalbard [this time!]

Posted in Mountains, pictures, Statistics, Travel, University life with tags , , , , , , , , , , on January 28, 2021 by xi'an

Alas, thrice alas!, there will be no one attending the ABC in Svalbard in Svalbard  next April. As the travel conditions to and around Norway are getting tougher, it is just too unrealistic to expect traveling to the Far North even from Oslo. Too bad, but hopefully there will be another opportunity in a near enough future…

However, don’t give up the fight!, the mirror meetings in Brisbane and Grenoble are still planned to take place, along with an on-line version accommodating most of the speakers invited so far. Anyone interested in holding another mirror meeting?! Please contact me.

## ABC in Svalbard [update]

Posted in Mountains, Statistics, Travel, University life with tags , , , , , , , , , , , , , , on December 16, 2020 by xi'an

Even though no one can tell at this stage who will be allowed to travel to Svalbard mid April 2021, we are keeping the workshop to physically take place as planned in Longyearbyen. With at least a group of volunteers made of researchers from Oslo (since at the current time, travel between mainland Norway and Svalbard is authorised). The conference room reservation has been confirmed yesterday and there are a few hotel rooms pre-booked through Hurtigrutensvlabard.com. Anyone planning to attend just need to (i) register on the workshop webpage, (ii) book an hotel room for the duration of the workshop (or more)., and (iii) reserve a plane ticket as there are not that many flights planned.

Obviously this option should only attract a few brave souls (from nearby countries). We are thus running at the same time three mirror workshops in Brisbane (QUT), Coventry (University of Warwick), and Grenoble (IMAG & INRIA). Except for Warwick, where the current pandemic restrictions do not allow for a workshop to take place, the mirror workshops will take place in university buildings and be face-to-face (with video connections as well). Julyan Arbel has set-up a mirror webpage as well. With a (free) registration deadline of 31 March, the workshop being open to all who can attend. Hopefully enough of us will gather here or there to keep up with the spirit of the earlier ABC workshops. (To make the mirror places truly ABCesque, it should have been set in A as Autrans rather than Grenoble!)

## Bernoulli factory in the Riddler

Posted in Books, Kids, R, Statistics with tags , , , , , , , , , , on December 1, 2020 by xi'an

“Mathematician John von Neumann is credited with figuring out how to take a p biased coin and “simulate” a fair coin. Simply flip the coin twice. If it comes up heads both times or tails both times, then flip it twice again. Eventually, you’ll get two different flips — either a heads and then a tails, or a tails and then a heads, with each of these two cases equally likely. Once you get two different flips, you can call the second of those flips the outcome of your “simulation.” For any value of p between zero and one, this procedure will always return heads half the time and tails half the time. This is pretty remarkable! But there’s a downside to von Neumann’s approach — you don’t know how long the simulation will last.” The Riddler

The associated riddle (first one of the post-T era!) is to figure out what are the values of p for which an algorithm can be derived for simulating a fair coin in at most three flips. In one single flip, p=½ sounds like the unique solution. For two flips, p²,(1-p)^2,2p(1-p)=½ work, but so do p+(1-p)p,(1-p)+p(1-p)=½, and the number of cases grows for three flips at most. However, since we can have 2³=8 different sequences, there are 2⁸ ways to aggregate these events and thus at most 2⁸ resulting probabilities (including 0 and 1). Running a quick R code and checking for proximity to ½ of any of these sums leads to

```[1] 0.2062997 0.7937005 #p^3
[1] 0.2113249 0.7886753 #p^3+(1-p)^3
[1] 0.2281555 0.7718448 #p^3+p(1-p)^2
[1] 0.2372862 0.7627143 #p^3+(1-p)^3+p(1-p)^2
[1] 0.2653019 0.7346988 #p^3+2p(1-p)^2
[1] 0.2928933 0.7071078 #p^2
[1] 0.3154489 0.6845518 #p^3+2p^2(1-p)
[1] 0.352201  0.6477993 #p^3+p(1-p)^2+p^2(1-p)
[1] 0.4030316 0.5969686 #p^3+p(1-p)^2+3(1-p)p^2
[1] 0.5
```

which correspond to

1-p³=½, p³+(1-p)³=½,(1-p)³+(1-p)p²=½,p³+(1-p)³+p²(1-p),(1-p)³+2(1-p)p²=½,1-p²=½, p³+(1-p)³+p²(1-p)=½,(1-p)³+p(1-p)²+p²(1-p)=½,(1-p)³+p²(1-p)+3p(1-p)²=½,p³+p(1-p)²+3(p²(1-p)=½,p³+2p(1-p)²+3(1-p)p²=½,p=½,

(plus the symmetric ones), leading to 19 different values of p producing a “fair coin”. Missing any other combination?!

Another way to look at the problem is to find all roots of the $2^{2^n}$ equations

$a_0p^n+a_1p^{n-1}(1-p)+\cdots+a_{n-1}p(1-p)^{n-1}+a_n(1-p)^n=1/2$

where

$0\le a_i\le{n \choose i}$

(None of these solutions is rational, by the way, except p=½.) I also tried this route with a slightly longer R code, calling polyroot, and finding the same 19 roots for three flips, [at least] 271 for four, and [at least] 8641 for five (The Riddler says 8635!). With an imprecision in the exact number of roots due to rather poor numerical rounding by polyroot. (Since the coefficients of the above are not directly providing those of the polynomial, I went through an alternate representation as a polynomial in (1-p)/p, with a straightforward derivation of the coefficients.)

## visitors allowed in Svalbard

Posted in Statistics with tags , , , , , , , , , , , , , , , on November 8, 2020 by xi'an

## from Svalbard [with snow]

Posted in Statistics with tags , , , , , , , , , , , , on April 25, 2020 by xi'an