Another Probability Question

Take a regular 5-day work week.

You are told months in advance that you’ll get two days off on a specific week. You don’t know which days it will be. You DO know that the days will follow each other. So it could be Monday and Tuesday or Thursday and Friday but NOT Tuesday and Thursday.

Now, that very week, your annoying neighbour will throw a gigantic party. Knowing him, it will be noisy, dirty and perhaps violent. What chance is there that one of your day off will coincide with his party ?

Bonus points if you can explain this as if to a 10-year-old kid.

We need to know the chance he will throw a party on any random day during the work week.

He will throw the party, that’s for sure.

You even know the day (say, Tuesday). What you don’t know yet is when you’ll have your two successive days off and thus, whether one of these will fall on the same day as his party (Tuesday).

Well, then the chance is 100%! :wink:

To figure out the chance of two events happening (two sequential weekdays off AND a party at your neighbour’s), you have to know (or estimate) the chance of both happening independently. So let’s say in all the years you’ve lived there, he has never thrown a weekday party. That would mean we could estimate the chance of it happening is about 1 in the number of weekdays you’ve lived there, which is pretty low (I’m guessing; unless you just moved in).

Now, if he throws a weekday party every week on a random day, that means the chance of the party happening on any given weekday is 1 in 5 or 20%.

So how often would you estimate your neighbour throws a weekday party?

This.

Your days off do not have a flat probability distribution function. Because your days must be contiguous, there are four options for days off:

Monday-Tuesday
Tuesday-Wednesday
Wednesday-Thursday
Thursday-Friday

Four possible outcomes, and Tuesday, Wednesday and Thursday each occur in two of those outcomes; Monday and Friday each occur in one of them. So if your block of days off is distributed totally at random, the probability of having any given day off is:

Monday: 25%
Tuesday: 50%
Wednesday: 50%
Thursday: 50%
Friday: 25%

If the party is equally likely to be on any day of the week (and you don’t know what day), then your requirement of having consecutive days off doesn’t matter: the probability of coinciding with one of your days off is 40%, since you have 2 out of 5 days off.

If you know what day of the week the party is, now you’re dealing with conditional probability; your knowledge changes the probability distribution. If you know his party will be on a Monday or Friday, your odds of coinciding with it are 25%. If you know his party will be mid-week, then your odds of coinciding with it are 50%.

Yes, that’s what I had in mind: you know the day on which his party will take place but not when your two consecutive days off will be.

Thank you for your answers, it’s all very clear.

Are Saturday and Sunday eligible party days, or does the party have to be on a week-day? Are those considered to be “your day off”? There is a 29% chance that the party would be on your day off, regardless of what week it was, and regardless whether you ever get the two extra days off…

I overlooked this when I read the OP, too, but the premises are a) he (?) will get two sequential days off on that work week and b) the neighbour will throw a party on one of the five days of that same work week.

The idea was indeed to consider just a five-day work week. Saturdays and Sundays don’t count.

Hmm… I think that we might need to know the method for determining the sequential days off. It’s quite possible that the four possibilities are equally likely, however, there could for instance be a method like the following:

First, determine one day off abitrarily/randomly from the five days available.
If necessary, flip a coin to determine if the second day off is before or after the first day selected.

This would mean a 30% chance that you have Monday-Tuesday or Thursday-Friday off, and a 20% chance of the other pairs, and the total odds of getting any specific day off would be:

Monday: 30%
Tuesday: 50%
Wednesday: 40%
Thursday: 50%
Friday: 30%

Because you mention two terms of art–probability distribution function and conditional probability–can you tell (us) what the word for that condition that “flattens” that distribution function, the one which is spelled out in the list to the list paired-day/days off results?

In fact as term of arts of art, can you throw in the words amalagous to “operand” or “arity,” but in probability, not arithmetic?

ETA: A lucky (heh) shot-in-the-dark just now in Wiki came up with “enumerative combinatorics” as the word for “listing/spelling out every damn one of them.” Does that sound right?

ETA: But not “remunerative,” which is what autocorrect gave me. For most people, I should think.

Unless there’s more information about how the days off and party day are chosen, I don’t think we can say definitively about the probability.

However, I think we can say that if we don’t know the party day yet, and it is chosen randomly with an equal (1/5) chance for any of the weekdays, then there’s a 2/5 chance it lands on one of the two days off. This holds no matter how the two days off are chosen.

However, if we already know that the party is Tuesday, but don’t know the days off, then we have to know how the days off are chosen to have a good estimate of the probability.

It’s getting more complicated than I thought or, more accurately, I hadn’t thought of all those possibilities.

Let’s say the days off are chosen purely randomly. They could be any pair as long as the days follow each other.

‘purely randomly’ isn’t a detailed enough description, since they have to be next to each other. If they could be separated, the most reasonable random way would be to choose one day using equal chances of all five days and then choose one from equal chances of the four remaining days.

But there are a lot of different ways you could choose two days next to each other, ‘purely randomly’. You could choose with an equal 1/4 chance each of the possibilities M-T, T-W, W-Th, and Th-F. Or you could choose like Chrisk suggests: choose with an equal 1/5 chance from all five days, then if you get M or F add the only adjacent day, and for T, W, Th flip a coin to choose from the two adjacent possibilities. There are probably other possibilities.

Of course, as I said, if you don’t know the party day, and it’s equally likely to be any of the five days, then the odds are 2/5 no matter how the days off are chosen. If you do know the party day, then the odds are more precise, but you need to know how the days off will be chosen.

BTW if the days off are chosen equally from M-T, T-W, W-Th, and Th-F, then if the party is on a M or F, there’s a one if four chance the days off will include the party day, but if it’s T-W-Th, there’s a 1/2 chance.

Note that if you assume the party day is chosen with equal chances from all five days, you can multiply the odds of the party being on any day by the odds of one of the days off falling on that day, add all of those days up, and you’ll get an overall chance (before the party day is chosen) of 2/5, like we said before.

well obviously if it’s an annoying neighbour and the party will be a source of great annoyance you’ve got a 100 per cent chance of it messing with your days off, then on the other hand, if it would suit you best if the party happened on a day off so you wouldn’t suffer the pain of zombie at work , then it will happen on a work night, rat bast**d neighbours have a queer feeling for what’ll f you up the most, bless their little hearts

I suspected as much as soon as hit “Submit Reply”.

I guess it shows how out of my depth I’ve become when it comes to probabilities. I used to love those classes at school but that was almost 25 years ago :frowning: .

Intro to probability includes all kinds of simplifying assumptions that they don’t bother to mention so as to keep the text simple enough for beginners. IOW, a 2 sentence problem followed with 2 pages of caveats is a bit offputting to newbies.

A bad side effect of this teaching technique is folks who don’t go much farther end up embedding all these unstated artificial assumptions into their idea of how the real world works.

We recently had a beastly thread on a dice problem where several people insisted that if there are, e.g., 4 possibilities for something and we know nothing about their probabilities, then it’s totally valid to assume they’re equally probable.

That’s actually totally wrong. But taking that assumption does give you a way to get to an answer. A totally bogus answer, but an answer nonetheless.