On OKCupid, one of the match questions is the following:
“Once you are intimate, how often would you and your significant other have sex?
– Every day
– About every other day
– Once or twice a week
– A few times a month or less”
On OKCupid, you choose your own answer and then pick what answer you’d like potential matches to answer. It seems straight-forward – if the other person picks the same answer as you, it’ll be fine. But will it?
Let’s make the following assumptions.
- A person is either in the mood to have sex on a given day, or they are not.
- Two people only have sex if both are in the mood.
- If someone is in the mood and has sex, they are happy. If they are not in the mood and don’t have sex, they are happy.
- If someone is in the mood and does not have sex, then they are unhappy.
If both people choose “Every Day,” then it will be fine; both people will be happy every day.
If both people choose “Every other day,” let’s assume they are in the mood 4/7 days of the week. So on a given day, there is a 4/7 chance of being in the mood.
It follows, then, that on any given day the chance of both people being in the mood is 16/49, or ~32.65%. And so the probability of having sex 4/7 a week is 7C4*(0.3265)^4 * (0.6735)^3 = 12.15%.
So that’s an 88% chance of not being satisfied in a given week. Well, that didn’t work out.
(Of course, the assumptions aren’t perfect – mostly because being in the mood might carry over if the itch wasn’t scratched.)
At trivia tonight, one of the bonus round was a matching question – match the 10 movies to the character Ben Stiller plays. We (and by we I mean my teammates, as I know very little about Ben Stiller) knew 7 of the questions for sure, but had no clue for the other three.
At that point, one of my teammates asked me if it would make more sense to put the same answer for all three, rather than guessing. That way we would be guaranteed a point, as opposed to maybe getting them all wrong. It took me a minute to think it through, but I told him it didn’t matter either way and I’d rather take the chance of getting all 3 right. (We won trivia on a tie-breaker final question, so a 1 point difference would have been a big deal.)
I figured there were 6 possible ways we could write down our answer – 1 correct way (call it A B C), 3 ways that get us 1 point (A C B, C B A, and B A C) and 2 ways that get us 0 points. (B C A and C A B). Calculating that expected value gets us an EV of 1 point, the same as his suggestion – so, mathematically, they are equivalent. Then it just comes down to your willingness to take that risk, since it’s not a repeatable event.
And as we always say every week, go big or go home.
(Step one in going through a bunch of posts I’ve wanted to make.)
After reading this post on the Monty Hall problem last year, I decided to do a lesson on it. And it worked out okay. But, as Riley Lark did in that post, I did it at the end of the probability unit. So this year, I decided to go for it and do it first. And I must say it worked out quite well, because from the get-go it shows them that what they think about probability isn’t quite right.
First, to play the game itself, it’s good to have a little showboat, plus something that is easy to reset. So I built this:
(It’s a display board and I cut the doors open.) So it was much easier to play the game from the stay, and to keep my hands hidden as I do things behind the doors.
The other thing I did was, before we discussed the theoretical solution, I had them experiment. I gave each pair of students 3 playing cards, 1 red (for the car) and 2 black (for the goats), so one player played host while the other switched or stayed.
The main thing to learn is you really need to MAKE them switch, because teenagers are stubborn and are sure they were right the first time. But only staying won’t show all the necessary results.