Hexagonal

[cinelli]

Hexagonal numbers are defined by
N = 3*n^2 + 3*n + 1
where n is a positive integer and they count the number of dots in the shape of a hexagon. The first few are 7, 19 and 37. Alison suggests the following even money bet to Betty. Alison would write a different digit on each of three pieces of paper and put them in a hat. Betty would pull them out one by one. If the three digit number formed by writing the digits in the order in which they were drawn were a hexagonal number, Alison would win. Otherwise Betty would win. Betty reasoned that there couldn’t be too many three digit hexagonal numbers and that there are hundreds of non-hexagonal number – she gleefully accepted the bet. Was she wise?

Cinelli

[UncleEbenezer]

There are just 12 such conventionally-speaking three-digit numbers. However, one of the digits may be a zero, so there are another 4 possible wins for Alison. That's from a problem space out of 10*9*8 = 720 candidate numbers whose digits are distinct. Very nice odds, yesno? Improved yet further by not all those hexagonal numbers having three distinct digits.

Except, Alison gets to choose the three digits, reducing the problem space to just six possible orders in which the the digits are drawn. Selecting digits 1, 2 and 7 gives her four chances out of those six: namely 127, 217, 271, and 721. Whoops!

Moral: the actual randomness is limited to what isn't under Alison's control. Look more carefully at what she's offering!

[cinelli]

Well solved, UncleEbenezer. There are just twelve 3-digit hexagonal numbers. They are

127 169 217 271 331 397 469 547 631 721 817 919

and four of those involve digits 1, 2 and 7. Quite a coincidence.

Cinelli