Lots of fun! Here are some photos…

The ceremony:

Playing with kids:

I heard this one at work today (no, it would be a pretty terrible interview question – I wouldn’t post it if it were) and thought some of the frequenters of this blog would enjoy it.

The problem is as follows:
You find yourself in a pitch black room, seated at a table. On the table are 30 dice which you have been told sum to 70. Your objective is to create two piles of dice, each with the same sum.

Details:
- The dice are smooth, you cannot tell which face is which.
- There is a temptation to find a solution that results in the piles being equal sum “in probability”, “in likelihood” or “in expectation”. I’ve not actually seen a correct such solution. Regardless, it is insufficient. There is at least one strategy that results in the piles always having exactly the same sum.
- You cannot deface or destroy or play any other tricks with the dice. Each of the 30 dice must be in one pile or the other. Dice are summed in the normal way (ie. by adding up the face parallel and closest to, the ceiling).

There is a serious solution to this problem that does not involve turning on lights, flashlights, explosives, thievery, etc. While these are “cute” and “imaginative”, they are not the solution I’m looking for.

Dear commentators and comment writers:

Vancouver likely does not have a 33% advantage, all things being equal, by being up one game and having two games left. This would mean they have 4:3 odds (57.1428…% chance of winning the series).

From high school math (we all did pay attention, didn’t we?), Vancouver can either win next game OR lose next game and win the game after that to win the series.

In “math”:

P(Vancouver winning) = P(Win Next Game) + P(Lose Next Game) x P(Win Final Game)

The trick then is picking reasonable numbers for these probabilities. Certainly , P(Vancouver winning) = 4/7 (57%) for some choices of these probabilities, but I can’t think of any motivation whatsoever to choose them.

If you consider that Vancouver has won 5/10 away games and 10/13 home games, you’re looking at P(Vancouver winning) = 88%.

If you consider Boston has won 5/11 away games and 9/12 home games, you’re looking at P(Vancouver winning) = 3/12 + 9/12 * 6/11 = 65%.

If you consider that Vancouver had no hope of winning in Boston, then they still have a 77% chance of walking away with the cup.

If both teams have two great final games (which I hope they do – they’re a hellova lot of fun to watch), the games will be decided on a lucky bounce (let’s say 50-50). Vancouver ends up with a 3:1 advantage (75% chance of winning, or 200% “advantage”).

But please, think of the poor math before you spew useless nonsense to fill time. Thanks.