I’m trying to work out the odds.

Let’s start with the first play.

There are six suspect cards and six weapon cards.

One of each goes into the envelope, leaving five of each in the deck.

Let’s make an assumption that those are evenly divided in the deal, so that I got three suspects and two weapons, and the Cub got two suspects and three weapons.

So when I looked at my hand and called a suspect I didn’t have, I had three and there were three I didn’t have. My choice of Mrs Peacock was one out of the three I didn’t have. So, a 1 in 3 chance that I got the murderer.

And when I looked at my hand and looked for a weapon I didn’t have, there were four. My choice of the candlestick was one out of the four I didn’t have. So, a 1 in 4 chance that I got the weapon.

Multiply those, and there was a 1 in 12 chance that I got the murderer and the weapon right on my first play. (And the end result would be the same if the distribution had been reversed and I got two suspects and three weapons.)

But now we go to the second play, the room, and it gets much more complicated.

There’s nine rooms. One goes into the envelope. Again, assuming even distribution, we each got four rooms. That means looking at my hand, there are five unknown rooms, so my choice of the right room is one in five. Multiply 1 in 12 by 1 in 5 and you get a 1 in 60 chance that I got all three on my second play.

Except, unlike the suspect and weapon, I don’t have a free choice of what room. I have to move Mr Green from the ballroom to another room to make the call.

We’ve got two dice, so Mr Green can move from 2 to 12 squares each move. Since he’s in the Ballroom at the start of the second turn, he’s got a choice of:

Conservatory - 4 squares

Billiard Room - 6

Kitchen - 7

Dining Room - 7

Library - 12

And, there are three rooms that Mr Green absolutely cannot reach on his second turn: the Hall, the Study and the Lounge, which are all more than 12 squares away.

That makes the odds of getting it right in two turns much more difficult, because in addition to the 1 in 5 chance of choosing the right card, I have to beat the odds that the room is within range, and that Mr Green rolls enough to get to it. And that isn’t a simple calculation, because there’s only one way to roll 12, which gives a 1 in 36 chance of getting to the Library, for instance. If I roll anything less than 12, then my choice of rooms goes down. And I have to roll as least 4 to get to even one room. There’s a 1 in 36 chance of rolling 2, and a 1 in 36 chance of rolling 3. I have to beat those odds to get to the right room. And then there’s the odds for the 6 and 7 rolls, which are different again.

**Chronos**, a little help, please? How do you calculate the odds when there are three rooms that are impossible to get on the second roll, and such a range for getting into the other five rooms?