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November 2

World Series outcomes

I am trying to calculate how many possible outcomes there are for a World Series (or any best‑of‑7 series). Now I know there are four values of games won possible: 4‑0, 4‑1, 4‑2, or 4‑3. But what I want to know is how many patterns of wins/losses there are. For example, a 4‑2 series could go WWLLWW or WLWLWW or LWWLWW or etc. I am actually not interested in the actual number — I could list and count them easily enough — but rather in learning the mathematical way to calculate this. There must be a formula or something that could not only be used for a best‑of‑7 but for a series of any length. Thank you. → Michael J Ⓣ Ⓒ Ⓜ 03:16, 26 October 2014 (UTC)[reply]

The total number of win/loss patterns from n games is just 2ⁿ. If you know that there are k wins out of n games then the number of patterns is Binomial(n, k) (see Binomial coefficient), which is equal to n!/(k!(n−k)!), where ! means factorial. For example, 3 wins from 7 games yields 7!/(3!4!) = 5040/(6×24) = 35 patterns. 86.160.82.218 (talk) 03:57, 26 October 2014 (UTC)[reply]

No; with a best-of-7 series, you must only count patterns where the team that wins the most games wins the last game. For example, if the series goes to 5 games (4-1) there are just 8 possible patterns: AAABA, AABAA, ABAAA, BAAAA, BBBAB, BBABB, BABBB, ABBBB. So if you know the series is 4-1 with team A winning then there are Binomial(4,1) (more commonly written C(4,1) or 4C1, and pronounced 4-choose-1) ways for the first 4 games to go and then the last game is known. For a 4-1 series with either winner it's 2 times 4C1; for a 4-2 series, 2 times 5C2. In general if the winning team wins m games and the losing team k games, it's (m+k−1)Ck ways for a specific winner and twice that for either winner. So now you have to sum that formula for all possible lengths of the series. For best-of-7, you get 3C0 + 4C1 + 5C2 + 6C3 = 1 + 4 + 10 + 20 = 35 ways for a specific winner or 70 ways for either winner. For a general best-of-n series where n is odd, the winner must win (n+1)/2 games and the loser anywhere from 0 to (n−1)/2. So you evaluate the sum of (n+1)/2 terms, from k=0 to k=(n−1)/2, with the formula (k + (n+1)/2) C k for each term. --174.88.134.249 (talk) 04:41, 26 October 2014 (UTC)[reply]

Your answer is correct, but so was 86.*'s. There's a simple bijection between ways of winning a World Series and ways of winning exactly four games of seven (⇒: pad with losses; ⇐: strip trailing losses), so the answer is

{\tbinom {7}{4}}=35

. For best-k-of-n series, the answer is

{\tbinom {n}{k}}

. -- BenRG (talk) 23:15, 26 October 2014 (UTC)[reply]

The annoying part is, I knew there was a closed-form formula like that, and still thought 86 had gotten it wrong. Thanks. --174.88.134.249 (talk) 04:48, 27 October 2014 (UTC)[reply]

Re: "For best-k-of-n series, the answer is

{\tbinom {n}{k}}

." This doesn't seem right. Consider a best-of-3 series.

{\tbinom {3}{2}}=3

, but there are 6 possible patterns: WW, WLW, WLL, LWW, LWL, and LL. —SeekingAnswers (reply) 21:19, 1 November 2014 (UTC)[reply]

The answer is

{\tbinom {n}{k}}

ways of winning a World Series,

{\tbinom {n}{k}}

ways of losing a World Series, for a total of

2{\tbinom {n}{k}}

possible outcomes for a World Series. Egnau (talk) 04:45, 2 November 2014 (UTC)[reply]

Continued square (or nth) roots of Graham's number

Take the square root of Graham's number. It's still impossibly huge. Take the square root of that. Still no good. Continue this process for as long as it takes to get a number that can be expressed using conventional notation. How many steps would be required? Or is even that answer not expressible using conventional notation? -- Jack of Oz ^{[pleasantries]} 20:38, 2 November 2014 (UTC)[reply]

Or, if the square root approach is way too slow, what would be the lowest value of n we'd need to make the question approachable using nth roots rather than square roots? -- Jack of Oz ^{[pleasantries]} 20:55, 2 November 2014 (UTC)[reply]

The number of square roots needed is very close to Graham's number itself. The same is true if we talk about nth roots, unless n is very close to Graham's number.

In other words, if you take a square root of G m times, you get

G^{1/2^{m}}

. You want this to be some reasonable number, say 10 (or a Googolplex, whatever).

G^{1/2^{m}}=10

gives you, roughly,

m=\log \log G

. What I said before is equivalent to saying

G\approx \exp(\exp(G))

. This is true in the sense that the steps used to construct G grow so much faster than mere exponentiation, that adding two more exponentiations pales in comparison.

It's that big. -- Meni Rosenfeld (talk) 22:46, 2 November 2014 (UTC)[reply]

That's really counter-intuitive, given that the larger a number, the smaller its square root is as a proportion of the number (example: 10 is 10% of 100, but 1,000 is only 0.1% of 1 million). And the effect becomes much more pronounced the greater the value of n is. So, the 1 millionth root of some huge number will usually be pretty damn small. If not, take the millionth root of that, and so on. But you're saying that Graham's Number is so fantastically huge that these sorts of extreme strategems would make virtually no dent in it at all? -- Jack of Oz ^{[pleasantries]} 23:36, 2 November 2014 (UTC)[reply]

It is counter-intuitive since we never have much occasion to deal with such massive quantities. For example, just consider k = 2^{(2^m)} with m = 1,000,000. 1,000,000 is less than 2²⁰ so taking the millionth root of k 1000 times reduces m by 20,000...barely a dent. The number k is small compared to what you can get with power towers, which, in turn, are insufficient to express Graham's Number. On our smaller scale, you can see the same phenomena by looking at how common operations don't compare with the next scale up. For example, 1,000,000 x 1,000,000 is large enough that subtracting 1,000,000 from it a 1000 times does not make a dent; as is the case with dividing 1,000,000^1,000,000 by 1,000,000 a thousand times, and so on. The difficulty with Graham's Number is that the operations to express it are so far up the scale that they are never encountered outside of mathematics, essentially, so we lack any real reference point. What I find rather mind blowing about the whole thing is that no matter how powerfully compact are notation, even if we use every bit of space with perfect efficiency, there are numbers so large that they the inversion of that operation a Graham's Number of times will not make a dent in it. --on a final point, you can phrase this is in a different direction that is, perhaps, even more counter-intuitive: if we put G for Graham's Number, then G + G, and G * G, and G^G, and etc. are all, basically, the same size since those operations are so weak relative to the one's used to construct G that the size increase is absolutely negligible.Phoenixia1177 (talk) 01:42, 3 November 2014 (UTC)[reply]

Well, there you go then. It really is super-inaccessible. Thanks all. (Next time I meet a cute guy named Graham, I won't bother asking for his number.) -- Jack of Oz ^{[pleasantries]} 02:43, 3 November 2014 (UTC)[reply]

I'd recommend the short essay "T formation" by Isaac Asimov. It appears in a few compilations, here's a giant pdf of "Asimov on numbers" that contains it [1]. The piece is a very nice and accessible pop-math introduction to thinking about very large numbers and notation systems. (It doesn't mention Graham's number because the essay predates the description of the number. The essay also predates the Conway and Knuth notations for very large numbers, so Asimov's T notation is actually somewhat anticipatory.) SemanticMantis (talk) 16:03, 3 November 2014 (UTC)[reply]

That chapter is a fascinating read. I'll read the whole publication when I have a chance. Thanks indeed. -- Jack of Oz ^{[pleasantries]} 20:45, 4 November 2014 (UTC)[reply]