Answer for Part 2 of Day 7 Math Problem
19 out of 180 or 10.56% 🎯
For simplicity, let us use variables.
Since C is in the bag with the less number of stubs among the two bags, there is a 100% chance to pick C in any of the 7 draws from that bag.
(If 
Hive account@jancharlest did not sneak his stub into any of the bags, then we will have gotten the answer right away, which is one out of all the stubs in the bag with 
Hive account@emafe's stub, or 1 out of 9. It does not matter how many draws happened before 
Hive account@chrisrice's stub got picked, since our perspective is from the very start of the drawings.)
Hive account@jancharlest inserted J into the bag where either C or E is, where there is an equal (50%) probability of J being in a specific bag. This leads to any the following:
If Hive account@jancharlest inserted J where C is, then the bag with C still has 8 stubs while the bag where E is has 9 stubs. In this case, the chance of E getting picked for C is 1 out of 9.
On the other hand, if Hive account@jancharlest inserted his stub where Hive account@emafe's stub is, then the bag with C still has 7 stubs while the bag with E has 10 stubs. In this case, E getting picked immediately after C is 1 out of 10.
Since there are 50% chances of having either a 1 out of 9 chance or 1 out of 10 chance, we can just average those two chances to arrive at the final answer. (1/9 + 1/10) / 2 = 19/180
There are 19 out of 180 chances or approximately 10.56% chance of Hive account@chrisrice getting partnered with Hive account@emafe.
1 HIVE has been sent to 
Hive account@jfang003's account as reward. 💰
Mentions: 
Hive account@holovision and 
Hive account@ahmadmanga (
Hive account@ahmadmangazap)
Special mentions: Hive account@chrisrice and Hive account@jancharlest 😆
RE: Part 2 of my Day 7 Problem of my Math mini-contest 😎