Timeline for An infinite game possibly due to Ernst Specker
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| May 30, 2022 at 2:29 | comment | added | Joel David Hamkins | @Richard, the game with an uncountably creative Chocolatier doesn't have that defect---if there are uncountably many chocolate types, then even when only two are served each turn, there can be no winning strategy for the Glutton that doesn't depend on the history. | |
| Aug 13, 2021 at 18:54 | comment | added | Richard Stanley | If the Glutton knows in advance which chocolates will be presented (which seems reasonable to assume for a perfect information game) but not at what turn a chocolate will be presented, then he or she needs only order them in a sequence $C_1,C_2,\dots$ and choose the least chocolate that is on the platter. It is not necessary to know the previous history. Is there a different game that doesn't have this defect? | |
| Aug 7, 2021 at 9:29 | comment | added | Joel David Hamkins | @MarkS Thanks, I have now fixed that. | |
| Aug 7, 2021 at 9:29 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
deleted 4 characters in body
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| Aug 7, 2021 at 4:42 | comment | added | Joel David Hamkins | @Ingolifs The Chocolatier adds finitely many delicacies each turn, so the total on the platter is also finite at each turn, but it could be growing rapidly. Nevertheless, the Glutton can still win. Suppose the chocolate morsels are numbered like the natural numbers (but they are not necessarily placed in order)---then let the Glutton simply take the lowest-numbered one each turn. Eventually, he will eat every one of them, no matter how many accumulate on each turn. The other arguments show that if he knows the history, then we can drop the natural-number assumption. | |
| Aug 7, 2021 at 1:36 | comment | added | Ingolifs | I've been scratching my head trying to make sense of this game. I feel there is something basic missing in its description. If the chocolatier can add more than 1 chocolate to the platter each turn, how can the glutton ever win? When you talk about 'finitely many', do you mean per turn or overall? | |
| Aug 6, 2021 at 18:26 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
fixed some grammar, improved exposition.
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| Aug 6, 2021 at 17:36 | comment | added | Joel David Hamkins | I think that isn't the right way to look at it. A given strategy is winning, if all plays that conform with it win. It has nothing to do with liklihood or anything other than logical possibility. For every simply definable game, one of the players will have such a winning strategy. But with the axiom of choice, you can construct games played with natural numbers for which neither player has a winning strategy. In fact, even without the axiom of choice, it is a theorem that there are games (on some uncountable set) for which neither player has a winning strategy. | |
| Aug 6, 2021 at 17:10 | vote | accept | Richard Stanley | ||
| Aug 6, 2021 at 17:04 | comment | added | jcsahnwaldt Reinstate Monica | I see. I don't really know anything about infinite games, and I'm not a mathematician but a software engineer, so I guess I was thinking about this too "constructively", in a sense. In this case: We can prove that for every alleged winning strategy sG for G, there exists a strategy sC(sG) for C which defeats it. The probablity that C will actually play that strategy may be zero (because C doesn't know sG and thus can't choose sC(sG)), but that doesn't matter. Correct? | |
| Aug 6, 2021 at 17:01 | comment | added | Joel David Hamkins | In other words, part of what we mean by a winning strategy is that if someone truly has a winning strategy, then they can announce to the world that they intend to play according to it, and they will still win every time. (Specifically, a strategy is a function telling the player what move to make next, given the previous play, and a winning strategy is one with the property that every play of the game that accords with it is winning for that player.) | |
| Aug 6, 2021 at 16:50 | comment | added | Joel David Hamkins | In particular, I don't believe that there are any paradoxes at all in this vein. If you have a truly winning strategy, then it shouldn't matter whether anyone knows how you will play. Otherwise, it wasn't actually winning, if there is a play that defeats it. | |
| Aug 6, 2021 at 16:41 | comment | added | Joel David Hamkins | @jcsahnwaldtReinstateMonica No, that is incorrect. If you are claiming to have a winning strategy, then I can know what it is, for the purpose of proving that it doesn't work. If there is any logically possible play (including those built under the assumption that you would follow it) according to which your strategy doesn't work, then it isn't actually a winning strategy. | |
| Aug 6, 2021 at 16:38 | comment | added | jcsahnwaldt Reinstate Monica | "suppose the Glutton will follow a fixed strategy ... then the Chocolatier can present these pairs {𝑐,𝑑0}, {𝑐,𝑑1}, {𝑐,𝑑2}, in turn" – This requires that the Chocolatier knows the Glutton's strategy, doesn't it? (I think we should require that neither player knows the other's strategy. Otherwise, both could try to adapt their strategy to the other's strategy, and we'd run into paradoxes...) | |
| Aug 6, 2021 at 12:53 | comment | added | bof | Without getting into continuous games there are simple examples of games of length $\omega^*$ where both players have winning strategies. | |
| Aug 6, 2021 at 12:44 | comment | added | Joel David Hamkins | @PeterTaylor That kind of continuous strategy phenomenon can't arise for a game proceeding in well-ordered discrete steps like this, since here we can always play any two strategies against each other. In the paper you cite, however, it doesn't always make sense to play two strategies against each other, and this seems to be required for the situation you describe, of both players having winning strategies. | |
| Aug 6, 2021 at 12:36 | comment | added | Peter Taylor | There are two-person games where it's claimed by mathematicians I respect that both players have winning strategies: see Lion and Man -- Can Both Win? by Bollobás, Leader, and Walters. This may depend on a non-finiteness property which doesn't apply to the particular game your comment refers to. | |
| Aug 6, 2021 at 12:23 | comment | added | Joel David Hamkins | Another isomorphic version: the attacker adds finitely many more time bombs (to explode at infinity), but the defuser can only defuse one bomb each turn. Defuser wins if every bomb is defused. | |
| Aug 6, 2021 at 11:27 | comment | added | Joel David Hamkins | Since the Glutton has a winning strategy that uses the history, the Chocolatier cannot have any kind of winning strategy, especially not a special strategy that depends only on the current position. | |
| Aug 6, 2021 at 11:25 | comment | added | მამუკა ჯიბლაძე | So is it clear that for some version neither player has a winning strategy with each move based only on the current position? | |
| Aug 6, 2021 at 11:24 | comment | added | Joel David Hamkins | One can also present the game in the form of a bag of marbles, with one player adding finitely marbles at each turn and the other removing just one. | |
| Aug 6, 2021 at 11:19 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
added 25 characters in body
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| Aug 6, 2021 at 11:12 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |