81–90 of 166 posts
And this, friends, is the perfect example of why the modern tech interview process is pure insanity.
Well to be fair Steve Ballmer is a terrible leader and if he had to take the tech interviews he wouldn't have passed and Microsoft wouldn't have stagnated for 10 years, before Satya Nadella took over and brought the company back on its feet.
I feel like there's an even simpler proof that you can beat adversarial-ballmer, with exactly the same expected positive outcome as binary search vs random ballmer.
I call my algorithm "randomly offset binary search". It goes like this:
1. Pick a random number between 0-100, call this 'offset'
2. Perform the binary search algorithm, except at each step add 'offset' to the value and mod by 100.
That's it. Now, even if Ballmer knows you're using this strategy, he can't make it perform any worse by selecting any specific number. Therefore your expected outcome is still $0.20 per game, beating the strategy proposed in this blog post.
Edit: Oops, nope, this comment is wrong, ty fgna for pointing that out! I feel like there's an even simpler proof that you can beat adversarial-ballmer, with exactly the same expected positive outcome as binary search vs random ballmer. I call my algorithm "randomly offset binary search". It goes like this: 1. Pick a random number between 0-100, call this 'offset' 2. Perform the binary search algorithm, except at eac…
Edit: Oops, nope, this comment is wrong, ty fgna for pointing that out! I feel like there's an even simpler proof that you can beat adversarial-ballmer, with exactly the same expected positive outcome as binary search vs random ballmer. I call my algorithm "randomly offset binary search". It goes like this: 1. Pick a random number between 0-100, call this 'offset' 2. Perform the binary search algorithm, except at eac…
This sort of misses the forest for the trees, although neat application. Ballmer's argument is essentially about tail risk. Expected value is absolutely not a good way to make bets if you value survival, because you only get one shot. Same reason you wouldn't go all in every time you get a poker hand that's "expected" to win. Because you'll (very probably) be bankrupt in a few hands. Sure the mean is +$0.07 or whatev…
Kelly Criterion Betting more than the Kelly fraction increases the risk of ruin, especially in the long run. https://en.m.wikipedia.org/wiki/Kelly_criterion Note: Not saying that this is applicable in the original post's situation. It's relevant to the parent comment though, and very useful in many situations, such as investing.
Earlier quoted context omitted.
I don't think this is true. Most people will not be bankrupt after losing a dollar. If this is true then Steve failed badly at communicating this context. To be honest, I think Steve just didn't grasp the mathematical deepness of the problem.
I think you've misread. The bankruptcy was in an analogy to poker. The point is if you get _one round_ to play - essentially all or nothing - should you play? No.
On the contrary, in general yes you should, because life as a whole will give you a variety of these sorts of risks. Steve Balmer's offer is just one episode in a lifelong series of risks offered you by the universe at large.
Earlier quoted context omitted.
Well to be fair Steve Ballmer is a terrible leader and if he had to take the tech interviews he wouldn't have passed and Microsoft wouldn't have stagnated for 10 years, before Satya Nadella took over and brought the company back on its feet.
Many Satyas successes started under Ballmer.
> I’m thinking of a number between 1 and 100 People are unable to think randomly. They'll avoid the obvious "not random" numbers 2 and 99, for example. I read somewhere that most people, asked to pick a number between 0 and 10, will pick 7. And the next digit would probably be odd, and not 5, because 5 is not random. That leaves you with 71, 73, 77 and 79. 77 is not random, so 71, 73 or 79. I'd pick 73 as my first gu…
If I were to take the bet with him, I'd make him write down the number first hide it (turn the paper over / put it under a book, whatever).
Edit: Oops, nope, this comment is wrong, ty fgna for pointing that out! I feel like there's an even simpler proof that you can beat adversarial-ballmer, with exactly the same expected positive outcome as binary search vs random ballmer. I call my algorithm "randomly offset binary search". It goes like this: 1. Pick a random number between 0-100, call this 'offset' 2. Perform the binary search algorithm, except at eac…
Neat. A nice way to see this is to imagine that the numbers 1-100 are arranged around a clockface; you randomly spin the clock before doing a conventional binary search starting from the top.