RE: RE: Part 2 of my Day 7 Problem of my Math mini-contest 😎
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RE: Part 2 of my Day 7 Problem of my Math mini-contest 😎

RE: Part 2 of my Day 7 Problem of my Math mini-contest 😎

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 @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 @emafe's stub, or 1 out of 9. It does not matter how many draws happened before @chrisrice's stub got picked, since our perspective is from the very start of the drawings.)

@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 @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 @jancharlest inserted his stub where @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 @chrisrice getting partnered with @emafe.

Winner: @jfang003 🏅

1 HIVE has been sent to @jfang003's account as reward. 💰

Mentions: @holovision and @ahmadmanga (@ahmadmangazap)
Special mentions: @chrisrice and @jancharlest 😆

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