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September 13

Friday the 13th

Here is a question. What proportion of the 13th of a month is a Friday? Is it 1/7 as most people would expect? My gut feelings tells me that I need to find a formulae that turns (YYYY,MM,DD) into the day of the week. Is there such a formulae? 202.168.50.40 04:50, 13 September 2006 (UTC)[reply]

There's calculating the day of the week. Frencheigh 05:11, 13 September 2006 (UTC)[reply]

In the long run, of all the 13th days of the month, 1/7th will be Fridays, yes. However, when looking at shorter time periods, like a year, the ratio may be quite a bit different. StuRat 05:18, 13 September 2006 (UTC)[reply]

That's not true. I've made the calculation, and the proportion is actually 43 / 300. -- Meni Rosenfeld (talk) 05:32, 13 September 2006 (UTC)[reply]

OK, that makes it 43/300 instead of 43/301 (one seventh). A small, but significant, difference. I stand corrected. StuRat 15:09, 13 September 2006 (UTC)[reply]

That's because not only the structure of the years repeat every 400 years, but, as it it happens (I've checked), the days of the week also repeat. So you don't get a true random distribution. For reference, the number of times sunday through saturday are the 13th of the month in a cycle of 400 years is {687, 685, 685, 687, 684, 688, 684}. -- Meni Rosenfeld (talk) 05:38, 13 September 2006 (UTC)[reply]

And, as you'll notice, Friday actually has a tiny advantage over the other days ... spooky, no? Confusing Manifestation 10:30, 13 September 2006 (UTC)[reply]

I've done the calculation once (with a computer of course), and indeed, as User:Meni Rosenfeld says, I've got that Friday was the most frequent day for 13th. – b_jonas 10:51, 13 September 2006 (UTC)[reply]

Perhaps it's also worth noting that the cause of all this mess is the fact that the number of days in the 400-year cycle, 146,097, is divisble by 7. Otherwise we would get a perfect 1/7 for questions like this. -- Meni Rosenfeld (talk) 19:49, 13 September 2006 (UTC)[reply]

proability

I need some help with a problem that has been bugging me for a while: Suppose the odds of an event happening are 1/n. If I repeat the event n times, what are the odds that the given event will happen at least once? For example, if I flip a coin twice, the odds are 3/4 that at least once I get heads. If I roll six dice, what are the odds of at least one landing on 1? If I pull one card at random from 52 different decks, what are the odds that at least one will be the ace of spades? I know the answer involves calculating the odds that it WON'T happen every trial, but that's as far as I got. Thanks! Duomillia 15:32, 13 September 2006 (UTC)[reply]

You're on the right track. The odds of "getting something at least once" are the same as one minus the odds of not getting it at all, and that's an easier thing to compute. For the coins, the odds of not getting heads are 1/2 per flip, so the odds of that happening twice are (1/2)*(1/2) = 1/4, so the odds of getting heads at least once is 1 - 1/4 = 3/4 (as you said). For the dice, the odds of not getting a one are 5/6, so the odds of that happening six times are (5/6)^6 ~ 0.335, so the odds of getting at least 1 one is 1 - (5/6)^6 ~ 0.665. You should be able to see the pattern from here. -- SCZenz 15:40, 13 September 2006 (UTC)[reply]

(Edit conflicted.) Yes, you're exactly right. This is classic introductory probability theory. If the odds of something happening are p the odds of it not happening are q = 1 - p. The odds of it not happening twice are q². The odds of it not happening n times are qⁿ. So to get the odds of it happening at least once (but possibly more) it is 1 - qⁿ = 1 - (1 - p)ⁿ. The reason it is easier to calculate with the odds of it not happening is that if you calculate the odds of something happening in multiple independent trials, you have to account for when it happens (did it happen the first time or the fourteenth?) and how many times it happened. But to calculate the odds of it not happening, you just calculate the odds of the same thing (not) happening at each trial. See binomial distribution for (much) more. –Joke 15:41, 13 September 2006 (UTC)[reply]

To put it all together, the general answer is

1-\left(1-{\frac {1}{n}}\right)^{n}

. For large n, this is roughly equal to

1-1/e\approx 0.63212\dots

. -- Meni Rosenfeld (talk) 15:43, 13 September 2006 (UTC)[reply]

I think you mean 1-1/e, which would be about 0.63212. Black Carrot 19:36, 13 September 2006 (UTC)[reply]

(getting rid of evidence) Yeah, that's what I said, 1 - 1/e. (evil grin) :-) -- Meni Rosenfeld (talk) 19:44, 13 September 2006 (UTC)[reply]

So, as n approaches infinity, the odds of n in n trials approaches 63% and a little? Duomillia 21:31, 13 September 2006 (UTC)[reply]

Yeah. That's cool—and see how close it is already for n=6... -- SCZenz 22:16, 13 September 2006 (UTC)[reply]

Well, 6 is a big number in the "1, 2, 3, lots, many" system ;-) -- AJR | Talk 23:18, 13 September 2006 (UTC)[reply]

In the Discworld troll counting system, they go "1, 2, 3, many, many-one, many-two, ..., many-many-many-three, lots". Although the article does point out the alternative of "one, two, many, lots" and not bothering about the rest of the numbers. Confusing Manifestation 03:45, 14 September 2006 (UTC)[reply]

I'm confused, how do you calculate n? Klosterdev (talk) 21:39, 10 April 2008 (UTC)[reply]