[日/台]貓咪大戰爭 (76) 台版儲罐等Eva

1001 回覆
3 Like 0 Dislike
2023-07-31 00:13:02
4星神龍易打過3星好多
2023-07-31 00:13:08
2關都有用50本能喵魔同埋50特急
2023-07-31 03:17:18
忍唔住抽左金券
32抽,4超激(有2藍眼但重覆
2023-07-31 09:46:04
652抽
抽前rank18023 NP237
抽後rk 18317 NP4682
新超激未升 我依家就趕去打XP祭

新超激13隻 (強至弱):
天命 非命x2 莉莉 雷電jack 美女神x2 機器貓 小丑 小悟 海龍 沙紀 異能貓 伊達 武田

重複藍眼14隻(上面1隻):
黑魔女 黑傑x3 黑獸x2 白聖女 皇獸x3 白舞 黑無x2

其他重複超激:除美女神外都唔值得加,NP左47隻
2023-07-31 10:17:19
都係架 不過不要為明天憂慮啦 日日都有經打 再唔係用埋貓力達囉
2023-07-31 10:18:01
巴打儲左幾耐
2023-07-31 10:19:17
其實海龍都ok嘅
印象中dps期望值算高 又遠範
2023-07-31 10:24:36
提倫控場算ok cd極快 所以game8好似俾A佢
不過同雷精(專控異)/冰雪貓(專控紅)比就應該差減D


抽到新貓可以上https://game8.jp/battlecats/207051 望下先
然後就可以去battlecats.db自己學下分析
2023-07-31 10:24:55
一年?
2023-07-31 10:25:38
藍眼都np哂?
2023-07-31 10:26:21
10%祭金劵all in無蝕嘅
2023-07-31 10:26:47
全加啦梗係
2023-07-31 10:30:45
未all in
本身唔多,抽完得留幾十張,留番eva備用
點解巴打一年儲到6百幾張咁多
我儲左幾個月都係幾十張
2023-07-31 10:31:22
我以為你淨加美女神
2023-07-31 10:33:00

2023-07-31 10:34:06
啲合作關係咪一定要用eva角先過到
2023-07-31 10:34:52
係 絕對係
尤其係激稀 建議全上45+
2023-07-31 10:37:41
咁咪焗住留哂啲關去到10/8必中之後先打
2023-07-31 10:41:04
係架
well 不過都係13號機同強襲要姐
2023-07-31 10:59:45
抽左2個鐘
藍眼同超激隻數黎講係幾歐洲嘅

不過多左嘅重複藍眼無幫到我補洞
我原本12隻藍眼入面缺5隻
650劵正路會有7隻藍眼
而根據chatgpt 7隻藍眼就大概就得兩隻新:

There are 12 distinct prizes and we can use a ticket to draw one of them of those 12 with equal probabilities. I have 7 of those prizes already and I have 7 tickets. What is the expected number of new prizes that I can get?


Let's define a random variable X as the number of new prizes you will get after using all 7 tickets. The possible values of X are 0, 1, 2, 3, 4, 5, 6, and 7.

To calculate the expected value of X, we can use the formula:

E(X) = Σ(x * P(X=x))

where x is the possible value of X, and P(X=x) is the probability of getting x new prizes.

If you have already won 7 prizes, then there are 5 remaining prizes that you can potentially win with your 7 tickets. Since each ticket gives you an equal chance of winning any of the 5 remaining prizes, the probability of winning a particular new prize with one ticket is 1/5.

Therefore, the probability of getting x new prizes out of 7 tickets is:

P(X=x) = (5 choose x) * (1/5)^x * (4/5)^(7-x)

where (5 choose x) is the binomial coefficient that gives the number of ways to choose x prizes out of 5 remaining prizes.

Now we can calculate the expected value of X:

E(X) = Σ(x * P(X=x))
= Σ(x * (5 choose x) * (1/5)^x * (4/5)^(7-x))
≈ 1.941

So the expected number of new prizes you can get with your 7 tickets is approximately 1.941
2023-07-31 11:04:42
我200抽先得一隻新藍眼和一隻重覆藍眼
2023-07-31 11:17:56
2023-07-31 11:44:34
每個池應該aim邊隻必中就可以抽另一個
2023-07-31 11:54:26
2023-07-31 11:54:51

我覺得2號機必抽
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