Egg Roulette

I find probability to be one of the most difficult topics for students to grasp. Beyond the simple experiments of spinners, coins, and dice, students have issues operating on uncertainty. This issue is compounded when multiple events each involve such a calculation as well as the relationship between them. Soon they find themselves neck-deep in notation and lose all rationality–they forget what they are solving to begin with. 

This past week we found ourselves mired in another battle with conditional probability. The initial questions were completed at a high level:

Yawn… yet something is nice about the sharpness of the problem’s boundaries. There is something objective–mathematical–about the task at hand. The students can count the sample space and favourable outcomes. 

The question then becomes:

Breaking down the question proved difficult for the class (as it always seems to do). I gave a narrative proof, we ran experiments, developed notation, and tethered to set theory. Still, at the end of the class, the exit slip data was not good. We were a mess. I needed a way to make the problem real again. 

That night I happened upon a YouTube video in a random playlist. The next morning I picked up a dozen eggs before school and spent the noon hour in the home economics lab. I had found the hook. 

When class came, I began with the video and then asked the class:

I then asked if anyone wanted to play. When I had my volunteers, I asked one of them if they wanted to go first or second. His response–after a lengthly pause to think–was:

When I asked the other player if he was okay with that, he replied:

This couldn’t have gone more to plan. We quickly drew up the notation and explained the conditional probability involved in the two cases explained by the volunteers. 

We played the game. Eggs were smashed, videos were recorded, and snapchat stories went viral around the school. At the heart of it all was my hook–I had found the anchor problem.

NatBanting

For those of you interested, there is no advantage to going first or second. I don’t do this often, but I’ve worked through the problem below. 

H1 – Hit on the First Egg
H2 – Hit on the Second Egg
M1 – Miss on the First Egg
M2 – Miss on the Second Egg

P(H2) = P(H1 AND H2) + P(M1 AND H2)
(These are the two cases, each of which is conditional. I spent quite a bit of time discussing this)

P(H1) = 4/12
P(M1) = 8/12
P(H2 | H1) = 3/11
P(H2 | M1) = 4/11

P(H2 | H1) = P(H1 AND H2) / P(H1)
P(H2 | H1)*P(H1) = P(H1 AND H2)
(3/11)*(4/12) = P(H1 AND H2)
1/11 = P(H1 AND H2)

P(H2 | M1) = P(M1 AND H2) / P(M1)
P(H2 | M1)*P(M1) = P(M1 AND H2)
(4/11)*(8/12) = P(M1 AND H2)
8/33 = P(M1 AND H2)

P(H2) = P(H1 AND H2) + P(M1 AND H2)
P(H2) = 1/11 + 8/33
P(H2) = 3/33 + 8/33
P(H2) = 11/33
P(H2) = 1/3

A simple calculation shows that P(H1) = 1/3 as well.