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I was reading The Ascent of Money by Niall Ferguson, and he writes

"... he [John Law] wagered 10,000 to 1 that a friend could not throw a designated number with six dice at one throw. (He probably lost that one [gamble] too, since here are only 31 possible outcomes: the chance of throwing 21 is about one in ten." (the italics are mine)

How does one figure out that there are only 31 distinct outcomes (i.e. sums)?

Not sure what tags would be fruitful. So if you guys have suggestions for tags, rather than answers please inform me.

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However note that not all outcomes are equally likely. Despite there being only 31 distinct outcomes, the chance of getting a sum of 6 (resp. 36) is low, namely one in 6^6 = 46,656. So offering a 10,000 to 1 wager on that sum would be profitable. – hardmath Dec 31 '13 at 22:23
up vote 9 down vote accepted

Well the minimum is all $1$s and the maximum is all $6$s so you have $6\times 6 - 6\times 1 + 1 = 36 - 6 + 1 =31$ possibilities.

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I don't think this answer is sufficient. Can you prove that all of the integers in [6,36] can be found as sums of the dice? – Newb Dec 31 '13 at 22:00
Of course. $6$ is just $1$s. Now pick one die and keep incrementing it by $1$ until it reaches $6$. You get $7,8,9,10,11$. Pick another die, do the same thing and you get $12,13,14,15,16$. And you can go all the way up to $36$. – xavierm02 Dec 31 '13 at 22:04
Of course the probability of each of the results differs. I don't remember my prob/stat well enough but I would assume that it's pretty much a standard bell curve with minima at 6 and 36. Which would intuitively illustrate why 21 was the one picked out as most likely; it's right at the center and therefore at the top of the bell curve. – keshlam Jan 1 '14 at 2:05

We want to prove that any number from 6 to 36 can be tossed. suppose k=6n with n between 1 and 5. Then we can roll $6 n$'s to get it. now suppose we want $6n+c $ with c between $1$ and $5$. then we can roll $c(n+1)$'s and roll n on the rest to get the outcome.

The minimum is $1$ in all of them $(6)$ and the max is 6 in all $(36)$ so we can get 31 possibilities.

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