#103348: "Better algorithm in order to get less repeats"
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细节描述
• 如果有的话,请将你在屏幕上所看到的错误信息粘贴出来.
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 请说明你当时想做什么,你做了什么,然后发生了什么
• 你的浏览器是什么?
Mozilla v5
• 请以英文复制/粘贴显示文字而非你的语言。 如果你有这个系统漏洞发生时的屏幕截图(画质不要太差),你可以使用Imgur.com来把它上传到网络,然后将链接复制/粘贴到这里来。
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 这段文本在翻译系统中吗?如果存在,它被翻译是否已超过二十四小时了?
• 你的浏览器是什么?
Mozilla v5
• 请简明而精确地解释您的建议,以便让人明白您想表达的意思。
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 你的浏览器是什么?
Mozilla v5
• 当你被封锁的时候,屏幕上出现了些什么呢?(空白的屏幕?部分游戏平台画面?错误的信息?)
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 你的浏览器是什么?
Mozilla v5
• 哪个规则没有被BGA的设计小组写进游戏里?
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 在游戏回放中,是否有不符合游戏规则的地方?如果有的话,请问是在哪一步呢?
• 你的浏览器是什么?
Mozilla v5
• 你当时是想做哪个游戏行动?
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 你在想做什么的时候,触发了这个游戏选项?
• 当你想这么做时,发生了什么事(错误信息,游戏状态信息,......)?
• 你的浏览器是什么?
Mozilla v5
• 请问这个问题发生在游戏的哪个阶段(当前的游戏说明是什么)?
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 当你想进行一个游戏行动时,发生了什么事(错误信息,游戏状态信息,......)?
• 你的浏览器是什么?
Mozilla v5
• 请描述一下显示画面上面的问题。 如果你有这个系统漏洞发生时的屏幕截图(画质不要太差),你可以使用Imgur.com来把它上传到网络,然后将链接复制/粘贴到这里来。
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 你的浏览器是什么?
Mozilla v5
• 请以英文复制/粘贴显示文字而非你的语言。 如果你有这个系统漏洞发生时的屏幕截图(画质不要太差),你可以使用Imgur.com来把它上传到网络,然后将链接复制/粘贴到这里来。
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 这段文本在翻译系统中吗?如果存在,它被翻译是否已超过二十四小时了?
• 你的浏览器是什么?
Mozilla v5
• 请简明而精确地解释您的建议,以便让人明白您想表达的意思。
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 你的浏览器是什么?
Mozilla v5
案件历史
Sorry for that but I'm not ready to invest my time in Just One anymore
So for example, if $wordID is 1, it would take the first word for every 13 card?
That would be like 5 options for the card index combinations.
I guess the fix would be to use a different word index for every of the 13 cards. that would be like 5^13 = 1.220.703.125 options.
So, you would have to recalculate the wordID for each card, instead of computing it once on game start. Seems to be a pretty trivial change, isn't it?
If you would share code I could take a look at that.
I don't have an idea why this algo would be biased towards specific words. I guess that either your card shuffling or the index selection is flawed. Are you using random_int(4) for that wordID? And how do you shuffle cards?
We are forced to use bga_rand() and I'm using it.
However for cardIds I create an array from 0 to 549, use PHP shuffle() method and take first 13 or first 550 depending on a game option.
Consider two different games.
Assume you drew [b]two[/b] equal [b]cards[/b] in both games.
The propability that [b]both[/b] cards have the same word (two repetitions) is:
- 20 % in your variant (you have to select the same index in the second game as in the first, and both words will repeat)
- 4 % in the variant with indices per card (as you select a different index for each card).
If you have 3 same cards, it's always 20% in your variant, and 0.8% in the index-per-card-variant.
I guess that would just feel more random to players. If you have one repetition, you don't notice it. If you have [b]two[/b] repetitions, you might notice it.
So, could you pretty please change it? :)
增加一些新内容到这篇报告
- 其他的游戏桌 ID / 移动 ID
- 按 F5 是否解决了这个问题?
- 问题是否发生了好几次?还是每次都发生?还是时好时坏?
- 如果你有这个系统漏洞发生时的屏幕截图(画质不要太差),你可以使用Imgur.com来把它上传到网络,然后将链接复制/粘贴到这里来。