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The expected value of the game is positive regardless of Ballmer’s strategy

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Re: The expected value of the game is positive regardless of Ballmer’s strategy

#151

Post by glimshe »

lucianbr wrote: post #108
glimshe wrote: post #104

Earlier quoted context omitted.

I'd love to be wrong like Ballmer. The net balance of his decisions were billions of dollars.

The net balance of his decisions, his circumstances, his random events and who knows what else. Please please stop with this "if he's rich he must be smart" argument. Please?

As long as we stop with "if he's rich it must have been luck alone". Even a lottery winner has to buy a lottery ticket.

Ballmer was a top salesman and a notorious workaholic. Of course he needed luck, but I doubt most in his position would have netted the same billions.

Re: The expected value of the game is positive regardless of Ballmer’s strategy

#153

Post by RaftPeople »

muststopmyths wrote: post #132
cbanek wrote: post #59

Earlier quoted context omitted.

My favorite Ballmer practice is stack ranking. It completely screwed up the entire company. I worked on Windows Mobile at the time the iPhone came out. We were all shitting ourselves.

Stack ranking existed at MSFT before Ballmer became CEO, i.e., when billg was CEO. It was a practice that Jack Welch brought into the corporate world and Microsoft was just guilty of following what were thought to be best practices at the time. Source: worked at Microsoft before Windows Mobile was a thing. As an aside, Windows Phone was my favorite phone OS and Ballmer seemed like the one who actually cared about it…

> best practices

Tangent: I love that phrase. Anytime I hear someone make that statement I think about the 9,000 things that have been considered "best practice" until we actually understood that they weren't even "good practice". It gets thrown around as if it's been studied and confirmed when in reality it means "a bunch of people are doing it"

Re: The expected value of the game is positive regardless of Ballmer’s strategy

#154

Post by lern_too_spel »

Joker_vD wrote: post #140
contravariant wrote: post #127

Earlier quoted context omitted.

What idiot wouldn't put destruction of the world as '-infinity' of value? Equating money with value is a simple trap as well. Who cares if you can win millions when a single loss wipes out all your savings? Since anything below a certain level of money leaves you trapped with no way out it could be argued that the value of being destitute is not 0 but -infinity which makes any risk of losing all money unacceptable. T…

> What idiot wouldn't put destruction of the world as '-infinity' of value? Literally anyone, you included. Every day, there is a chance that the roof collapse on you while you're sleeping; that's -Inf of value! Yet you don't put infinity resources into preventing that since you consider the probability low enough to not obsess over it.

No, I don't put infinity resources into it because that would make me die even earlier due to not having money or time to do what is necessary to stay alive.

Re: The expected value of the game is positive regardless of Ballmer’s strategy

#155

Post by nojvek »

It would only make sense if Ballmer writes the number he is originally guessing on a piece of paper and fold it before game begins. And win/loss is checked with what is written on the paper.

Otherwise it is a hidden mutable information game where Ballmer dynamically changes higher/lower for maximum tree depth and always make you lose.

Re: The expected value of the game is positive regardless of Ballmer’s strategy

#156

Post by lern_too_spel »

TheOtherHobbes wrote: post #122
nneonneo wrote: post #118

Earlier quoted context omitted.

The St. Petersburg "paradox" is not a paradox if we consider any real-world implementation of it. The EV accumulates at the rate of $1 per flip. So, if we want to make the EV at least $1,000,000, we must find a counterparty that is willing to pay at least $2^1000000 (or at least 2^1000000 units of "utility" if we're trying to avoid the depreciating utility effect). That's plainly unrealistic. As soon as the counterpa…

It's interesting that in this context the assumption of infinite growth is obviously unrealistic - but in the context of trad econ, infinite growth is considered a bedrock assumption. Putting them together suggests it's flagrantly irrational to apply naive toy models to the real world. Even if they do have a nice mathy sheen. Engineers (mostly) know this, but for some reason gamblers and economists (mostly) act as if…

When you're this far away from saturation, infinite growth is a good approximation.

Re: The expected value of the game is positive regardless of Ballmer’s strategy

#157

Post by lern_too_spel »

nighthawk454 wrote: post #41
qsort wrote: post #35

Earlier quoted context omitted.

I don't agree, I think he was just plain wrong. Unlike most people here I actually think questions like this are a decent way to see how people think. I would expect people with math/stats/cs background to be able to at least start the conversation about this problem. However when you hide hypotheses or add your own BS constraints as a gotcha without explicitly stating them is where you lose me. If the question is "w…

> If the question is "would you play this game" the reasonable mathematical translation is "determine if the expected value is greater than zero". Not really! And that may be the point of the question. It's not testing if you can pattern match to plausible CS concepts. If you get one play, and the goal is to win, do you take the chance? The whole question is about the difference in likelihood in the limit (expected v…

If it were about winning or losing that round, there wouldn't be so many different monetary values associated with each outcome. Instead, each outcome would be win, lose, or draw. Because there are monetary values, we have to decide if it's a good bet. If somebody offers me a chance to wager $1 on the outcome of a die roll where the payoff for choosing correctly is $10, I will do it, even though it's more likely that I will lose than win.

Re: The expected value of the game is positive regardless of Ballmer’s strategy

#158

Post by Lockal »

gukoff wrote: post #150
Lockal wrote: post #80

Here is a chart for probabilities for starting value: https://docs.google.com/spreadsheets/d/e/2PACX-1vThljkK2nUIL... I find it interesting: it is definitely symmetrical, but I did not expect that in the final result 1/98 could be more important as a starting value, while 2-17/82-97 are not used at all.

This really depends on the pure strategies that you choose. The initial set of strategies wasn't very diverse and compensated for the binary search "weaknesses" on the ends of the spectrum by sometimes guessing 1 and 98. But after adding some more pure strategies to the set, we've got a far better mixed strategy that prefers the numbers between 28-70 as the first pick: https://github.com/gukoff/ballmer_puzzle#winning…

O, wow, post got update!

  > Avg win if Ballmer chooses randomly: $0.16247848000093376
  > Win if Ballmer chooses adversarially: $0.14657033010415976
So the goal is to find a set of strategies where adversarial avg win == random avg win? Or these numbers will never be equal?

Re: The expected value of the game is positive regardless of Ballmer’s strategy

#159

Post by nighthawk454 »

gukoff wrote: post #131
nighthawk454 wrote: post #19

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…

I don't view the original problem this way, but let's think about it! > the spread on that surely goes over the 0 line. Do you imagine starting with $1 or $1000? :) Let's add a condition that Ballmer has infinite money, we start with a specific budget, and we can't continue playing if we exceed budget randomly changes after each game, In the game where you start with $N, win $1 with probability p > 0.5 and lose $1 ot…

hey, thanks for the blog post and your reply! I think I follow - a generalization of a coin-flip type game. I agree that if you have more starting money, you would never lose. From the binary search idea, even if chosen adversarially, the worst case is still log2(100) ~= 6.6. So if you get 1,000 guesses or just any number of guesses >= 7 you literally can't lose. Then you should definitely play.

Setting the limit at 5 brings you to the interesting point of there being a good mix of win/loss outcomes. 4 would be too few guesses and you'd very likely lose, and 7+ you'd definitely win. So the question is only interesting _because_ the limit is chosen so that the spread puts your odds on both sides of the 0 line. Otherwise it'd be clear cut.

The standard deviation being ~$1 is interesting. To me that suggests that with a mean of $0.07 and a deviation of +/- $1, it's essentially 50/50 odds. There's technically a slight edge in your favor, probably 53/47, but barely. So given a game with essentially no edge, would you play? Framing it that way - deciding to what degree the game is winnable - it's essentially not. You should not particularly expect to win, no matter your strategy.

I think part of the trick with the Ballmer question as well is the question is not necessarily about 'can you find an optimal strategy?' - it's 'do you play the game or not?'. The paths chosen within the round don't ultimately matter to that question. It's only intermediately necessary to model the intra-round decision paths in order to get to the overall win/loss distribution for a single round.

If you do end up getting the time, do make another blog post!

Re: The expected value of the game is positive regardless of Ballmer’s strategy

#160

Post by Lockal »

draluy wrote: post #43

I dont get this. If this is true, then he found a more efficient algorithm than binary search. Why are we not using it in CS?

If a second party can submit adversarial values into a system, potentially causing a denial of service in a binary search (where comparisons are computationally expensive and data is unevenly distributed), there is a much simpler solution: avoid using sorted collections and binary search. Instead, use hashmaps. To address similar HashDoS attacks, many implementations (in Python, Rust, Java, etc.) use a randomized hash function, which vaguely resembles the idea of randomizing starting value for binary search.
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