61–70 of 88 posts
https://codegolf.stackexchange.com/questions/172178/smallest...
There must be something I'm missing. I borrowed the author's choice probabilities and wrote a script to verify the probability of winning by simulating 10 million rounds of gameplay. It seems it's true that when all three agents are following the same choice probability distribution the result is a win about 29.6% of the time. However, when one of those three agents follows an alternative choice probability distribut…
(Slightly off topic) -- I can't connect to this site from my mobile data or home connection. FF, Curl + mobile FF claim to not be able to resolve the host The DNS records seem odd: $ dig www.amolas.dev [ ... ] ;; ANSWER SECTION: www.amolas.dev. 66 IN CNAME https://alexmolas.github.io. https://alexmolas.github.io. 66 IN A 185.199.111.153 https://alexmolas.github.io. 66 IN A 185.199.109.153 https://alexmolas.github.io.…
hi! some people told me that they can access the site, but I've never knew why, and I'm not an expert on websites. Thanks for spotting it! How can I solve the problem? What does it mean to put a scheme in a cname?
Why do you have links styled to look exactly like the other plain text?
At the bottom of the article, it says "This post has been heavily inspired by this question in SO ↩" and I could not for the life of me find the link to said question until I discovered that "question" was a hidden link.
I must be missing something here. > if we follow this strategy we’ll win a little bit less than one-third of the time. That is not possible. If N players use the same strategy then by symmetry each of them must win on average 1 time in N. So against two other players we should win exactly one third of the time, not "a little bit less" than one third of the time. Also, if all players use the same strategy then it does…
But it's not very interesting to add "all players use the same strategy" as an assumption, because yes, obviously, if every player randomly picks from the set {33, 247, 17855433344} each has a 29.6% chance of winning. But that's a different game.
In fact the author doesn't assume everyone will use the same strategy; rather, the author simply asks what optimal play is, when you don't know the others' strategies. The fact that all players choose the same strategy (if they're playing optimally) is a consequence of this. Optimal play has to work regardless of the strategy your opponents pick; assuming they'll play optimally too is just a shortcut to figuring out what optimal play is.
For example, let's say you knew the other two players' strategy is to pick from this distribution. If you simply always pick 1, you win any time neither of the others picked 1, which happens with probability (1-0.456)^2 ~ 0.296 -- the same probability of winning as the optimal strategy! So why bother with the distribution? Precisely because you don't know whether your opponents are playing optimally or not. If another player gets the same bright idea as you, you counteract each other and the 3rd player wins 54% of the time (the rest are ties). An optimal strategy has to be optimal even if opponents are not using it; otherwise it's not really optimal (one may not exist).
So all players using the same strategy is not an assumption, but a consequence, and therefore much more interesting.
There must be something I'm missing. I borrowed the author's choice probabilities and wrote a script to verify the probability of winning by simulating 10 million rounds of gameplay. It seems it's true that when all three agents are following the same choice probability distribution the result is a win about 29.6% of the time. However, when one of those three agents follows an alternative choice probability distribut…
Our baseline (once again):
Agent 1 (nash distribution): .296
Agent 2 (nash distribution): .296
Agent 3 (nash distribution): .296
Another case:
Agent 1 (always chooses 1): .489
Agent 2 (even distribution -- equal chance of any number 1-10 being chosen): .411
Agent 3 (nash distribution): .054
In this situation, the nash strategy comes out far far behind either of the other two strategies.
This makes sense intuitively:
Agent 3 chooses 1 nearly half (45.6%) of the time. It will lose with that choice every time because Agent 1 chooses 1 every time.
When Agent 3 chooses 2-10 (100 - 45.6 = ) 54.4% of the time, it will lose almost every time because Agent 1 already chose 1.
The only case where Agent 3 wins is when Agent 3 chooses a value 2-10 (54.4%) AND Agent 2 chooses 1 (10%), eliminating itself.
54.4% * 10% = 5.4%, which is exactly the value discovered above.
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The specific strategies chosen by your competitors have a very large impact on your strategy's effectiveness.
I fail to understand how this can be considered an optimal strategy.
Earlier quoted context omitted.
For a short intuitive proof of that 1st sentence, imagine both players pick 1 every time. You both lose 1 point. If you're tied for score, you're still tied after the round. Now imagine you change your number to stop ties. Well now you've picked a number bigger than 1, and you lose every time. If you were tied, now you're losing.
This is only true if your goal is to have a score higher than your opponent. It's not true if your goal is to have the highest possible score. If ties are with 0, you always choose 1 in either situation. If ties are worth -1, choosing 1 always achieves a greater than or equal to score but the number will be much lower than if you some percentage of the time choose another number.
That is what “dominant” in game theory means, which was the specific claim that’s being discussed.
There must be something I'm missing. I borrowed the author's choice probabilities and wrote a script to verify the probability of winning by simulating 10 million rounds of gameplay. It seems it's true that when all three agents are following the same choice probability distribution the result is a win about 29.6% of the time. However, when one of those three agents follows an alternative choice probability distribut…
Agent 1 doesn't care if Agent 2 and 3 win less than him, he only feels a strategy is better than the Nash strategy if his expected score becomes higher than .296, which your simulation seems to show has not been achieved.
The game was completely shut down after people in a small little town won three times in a row and they started to suspect foul play. Turned out to be the local store asking people to join and the store distributed the numbers for the people to make sure they had an even distribution across a huge range.