# Won Clue in Two!

Playing Clue with the Cub at our local Starbucks.

I was Mr Green. His starting location is one of the worst, tucked around a corner of the Ballroom. I got a good throw and just made it into the Ballroom.

I had the Ballroom card, so I decided to call in a weapon and a suspect I didn’t have in my hand. Randomly picked Mrs Peacock and the Candlestick.

Cub: “I don’t have any of those.”

Me: " You’re sure?!"

Cub: “Yup.”

Next round, get a throw that just gets me to the Kitchen. Call in a weapon and a suspect that I had in my hand.

Cub: " I don’t have any of those."

Me: "Really? You’re absolutely sure?

Cub checks his hand again. “Yup.”

I declare that it was Mrs Peacock in the Kitchen with the Candlestick.

Check the solution: Mrs Peacock in the Kitchen with the Candlestick!

Boo-yah! Got it in two!

Love it when that happens! The Wife did it in one. It was embarrassing to all the rest of us because we had all just talked a bunch of smack about how she hadn’t won a game in forever.

What are the odds of that? 1 in 6 weapons times 1 in 6 characters times 1 in 12? rooms? 1 in 432?

Just the two of you playing, or were there others involved?

EDIT: lisiate, the odds are considerably better than that, since you’re holding some of the cards, and thus can eliminate them.

Oh, yeah, good point.

A few years ago, some friends of ours had some of their relatives visiting from Ireland, and invited us to come over to join them for dinner and some board games. We played the Firefly version of Clue, with six total players. One of the Irish visitors won the game on the second turn of the game, which was, obviously, just a wild-ass lucky guess.

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?

Oops - just realised that I misstated the chance of rolling a three with two dice. It’s 1 in 18, not 1 in 36, right? Two different throws to make a three; each of those throws individually is 1 in 36, but added together, 2 in 36 ways to throw 3, or 1 in 18?

You forgot the possibility that you could learn the room in your first turn.

Yes, but I’m trying to figure out the odds for what actually did happen: Mr Green discovering that it was Mrs Peacock in the Kitchen with the Candlestick.

But let’s change it. Assume I have the Ballroom and I want to go to a room I don’t have.

Again, it’s tied to the game board. If I didn’t go to the Ballroom because I had that card, there’s only one other room Mr Green can reach: the Conservatory. To get there, I have to throw a 10 or higher.

By my calculations, the chance of rolling a 10 is 3/36; chance for an 11 is 2/36; chance of a 12 is 1/36. Add those together and my chance of Mr Green making it to the Conservatory is 6/36, or 1/3.

And, assuming I’m going to the Conservatory because it’s not in my hand, and I’ve got 4 rooms in my hand, there are five unknown rooms,so the chance the Conservatory is the murder room is 1/5.

Putting all that together: 1/3 (Murderer) * 1/4 (Weapon) * 1/3 (dice roll to Conservatory) * 1/5 (Murder Room), for a final result of 1 out of 180 for that particular combination.

Have I done the maths right?