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Will Critchlow
This is my old gmail account. Find me here:
This is my old gmail account. Find me here:


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I'm no longer using this account. I'll migrate it with Google finally lets me (a couple of weeks? Yeah, right). In the meantime, you should encircle me(bleurgh) over here:

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I'm moving over to using my Google Apps G+ account. You can find me over here:

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It's been interesting to see the emergence of "Black Friday" in the UK lexicon as well this year (mainly Amazon, but also the press)...
Originally shared by ****

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I feel like the answer has to be "no" but that would be uninteresting so I await the real(badaboom!) answer tomorrow. cc +Tom Critchlow +Tom Critchlow (which Tom should I be encircling?)
I've been away from the internet in general for a while, but I wanted to share a math puzzle I heard recently. There are several problems that involve two envelopes, but this one is my favorite. As with all my favorite puzzles, it doesn't rely on tricky or intentionally deceptive wording—I'm stating the premise as clearly as I can:

I pick two real numbers through some process unknown to you. It might be random and it might not. Maybe I always pick "3" and "100". Maybe I roll two dice. Maybe I write C code by mashing a keyboard until it compiles and prints two numbers (or produces Windows ME). Maybe I always use 0 for the first number, and for the second I call my aunt and ask her for a negative real number, which I multiply by the estimated number of protons in the universe. (At this point, my aunt is used to that kind of call from me.)

I put these two numbers on slips of paper and put them in two envelopes. I thoroughly shuffle the envelopes, and then you choose one via a fair coin toss. You open it and look at the number. You are now given the option (as in the infamous but very different Monty Hall problem) of switching to the other envelope.

Your goal is to pick the envelope with the higher number. Can you come up with a strategy that guarantees you a better-than-even chance of winning?

It has to do better than 50% no matter how I picked the numbers--if your strategy includes the rule "switch if the number is a 2", it's wrong, because I could always be picking 2 and negative 100, and in that scenario your strategy will fail at least 50% of the time. This means your strategy must work even if I have guessed what your strategy is and am cherry-picking numbers specifically to defeat it.

I cannot emphasize enough that you do not know anything about my process. Not only do you not know the numbers, you don't know which random distribution I am picking them from—or whether it's even a random distribution at all—and there is no deductive basis for estimating the odds that I'm using any particular method. (This gets at what is sometimes called the difference between risk and uncertainty.)

I'll post again tomorrow with the correct answer and a bit more about the puzzle.

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So I read this article after someone shared it on the twitters: - I was intrigued by this claim: "hardcore pornography’s effectiveness in achieving rapid desensitization in subjects has led to its frequent use in training doctors and military teams to deal with very shocking or sensitive situations." [citation needed].

It's a hard thing to search for information about - for example:

returns a bajillion copies of the original article.

Anyone know if this kind of thing is actually done? Or is Ms Wolf making it up?

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I'm doing Movember. Until there is more evidence, I'm going with the Guinness 'stache. I'd appreciate any sponsorship here: . We do pro bono work @Distilled for a breast cancer charity and the guys @Distilled are doing Movember together...

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Doing a great job too...
I'm live tweeting the #searchlove conference in NYC. Follow the hashtag #searchlove on Twitter for updates live.
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