哇六合彩痴撚左線呀

你數學差連中文都差？
兩個女算唔算同一性別都要問？？？

你up乜

男男 男女 女男 女女
同一性別係1/2

男男 女女 仲有男女 女男 明未

見到同一性別係1/2未

問咗chatgpt

連續兩期5個非指定號碼：


連續兩期5個指定號碼：



Chatgpt打中文會預設用殘體字回答

不便之處 敬請原諒

咁而家六合彩就係有5個no.先
點解你覺得同只需要同一性別係一樣？

睇下下個留言未

你條題目問錯左 ai冇答錯

同埋你試下要求佢用繁體 多數都得

你個前設係冇指定有男定女先
但我講緊個前設係有，因為而家個case都係前設有5個no.

邊度問錯？連續兩期5個指定號碼

你而家係指定左 1 2 3 4 5一樣
但其實可以一樣果5個號碼可以係
1 2 3 4 6
1 2 3 4 7
1 2 3 4 8
…
…
…

所以咪話你設錯題

咁所以連續兩期相同5個no.冇可能係中三獎個機率

係兩期結果中有五個號碼一樣

指定左號碼 就等於指定左 1 2 3 4 5
你會miss晒其他可能組合
1 2 3 4 6
1 2 3 4 7
1 2 3 4 8
…
…
…

咁你試下計

你先錯喎，你數學u?

你係都話一樣，咁你計先啦

你試下問啱問題
問AI都畀到正確答案

咪話係三獎答案

To calculate the probability of having exactly 5 common numbers in two Mark Six draws, we need to consider the combinations and probabilities involved. Here’s a step-by-step approach:

Total Combinations for One Draw: In Mark Six, 6 numbers are drawn from a pool of 49. The number of possible combinations is given by:

(649)=6!(49−6)!49!=13,983,816

Combinations for 5 Common Numbers: To have exactly 5 common numbers in two draws, we need to select 5 numbers from the 6 drawn in the first draw and 1 number from the remaining 43 numbers (49 - 6 = 43). The number of ways to choose 5 out of 6 is:

(56)=6

The number of ways to choose 1 out of the remaining 43 is:

(143)=43

Total Favorable Combinations: The total number of favorable combinations for having exactly 5 common numbers in two draws is:

6×43=258

Probability Calculation: The probability of having exactly 5 common numbers in two draws is the ratio of the number of favorable combinations to the total number of combinations:

P(5 common numbers)=13,983,816258≈0.00001845

So, the probability of having exactly 5 common numbers in two Mark Six draws is approximately 0.00001845, or about 0.001845%.

有d數學符號出唔到 呢度執返

To calculate the probability of having exactly 5 common numbers in two Mark Six draws, we need to consider the combinations and probabilities involved. Here’s a step-by-step approach:

Total Combinations for One Draw: In Mark Six, 6 numbers are drawn from a pool of 49. The number of possible combinations is given by:

(49C6)=6!(49−6)!49!=13,983,816

Combinations for 5 Common Numbers: To have exactly 5 common numbers in two draws, we need to select 5 numbers from the 6 drawn in the first draw and 1 number from the remaining 43 numbers (49 - 6 = 43). The number of ways to choose 5 out of 6 is:

(6C5)=6

The number of ways to choose 1 out of the remaining 43 is:

(43C1)=43

Total Favorable Combinations: The total number of favorable combinations for having exactly 5 common numbers in two draws is:

6×43=258

Probability Calculation: The probability of having exactly 5 common numbers in two draws is the ratio of the number of favorable combinations to the total number of combinations:

P(5 common numbers)=13,983,816258≈0.00001845

So, the probability of having exactly 5 common numbers in two Mark Six draws is approximately 0.00001845, or about 0.001845%.

有冇人知次序係咩都差唔多？

統計學上近乎無可能
斷估呢組波波重量上有問題 但昨日放假無整番佢
同埋應該係~8~9個波波有問題，咁出黎個組合少好多又可以提高容錯度

難怪呢年近乎期期都有人中頭獎啦 原來係咁

點解無可能

在六合彩中，每期抽出6個號碼，通常是從1到49的號碼進行隨機抽取。要計算下期中獎號碼中包含特定的5個號碼（5號、18號、21號、24號、44號）的機率，我們需要考慮以下幾點：

1. **您選定的號碼**：您希望選擇的號碼是5號、18號、21號、24號和44號，這些都是5個特定的號碼。

2. **需要添加的第六個號碼**：由於每期會開出6個號碼，您需要再選取一個額外的號碼，而這個號碼必須從剩下的號碼中選取。考慮到您已選擇的5個號碼，仍可選擇的號碼數為：
\[
49 - 5 = 44 \text{ (因為總共有49個號碼)}
\]

3. **計算下期中獎的機率**：中獎的方式有多種。您最少需要中5個號碼。若要確保中獎，您還需要隨機選取額外的第六個號碼，但沒有正確的計算機率。

每期中獎的號碼是隨機的，因此沒有任何號碼有較高的機率來中獎。下期選擇您的號碼和之前開出的號碼無關。不論這一期中的是什麼，所有號碼在下一期的中獎機會都是相等的。

總結來說，雖然可以推算出選擇這5個加上一個隨機號碼的組合，但因為每次抽獎都是獨立事件，因此無法給出一個確定的機率值。具體的如機率值要通過正式的計算公式進行複雜的統計分析。