This puzzle, created by Michael Fletcher, was featured in Significance (Volume 21, Issue 5, November 2024, Page 4). It was published on October 1, 2024, by Oxford University Press on behalf of the Royal Statistical Society (RSS), the Statistical Society of Australia (SSA), and the American Statistical Association (ASA).
Dr Watson was looking forward to a quiet night in at 221b Baker Street. Dr Watson was looking forward to a quiet night in at 221b Baker Street.
“Come, Watson, grab your coat. We have an important assignment,” announced his detective friend.
“Has there been a murder?” asked Watson.
“Yes, Watson, Dr Black’s body has been found on the stairs at Tudor Mansion.”
“Good grief, Holmes! Shall I bring my revolver?”
“No need. All the weapons are supplied. Mycroft has invited us round to his house for a game of Cluedo. Let me explain.
“There are six suspects, six weapons and nine rooms. Before the game starts one ‘person’ card, one ‘weapon’ card and one ‘room’ card are placed in the murder bag. The remaining cards are then dealt out to the players.”
“So in our case,” said Watson, “each of the three players will have six cards.”
“Precisely, Watson. Then, by a process of elimination, one deduces where the crime was committed, by whom and with which weapon.”
“You mean once you have eliminated the possible, whatever remains, however improbable, will be the contents of the murder bag,” observed Holmes’s friend.
Holmes glanced quickly at Watson to check he wasn’t trying to be funny. He needn’t have worried. His companion was gazing at him in admiration.
“Exactly so. You know my methods,” said the famous detective.
When they arrived at Mycroft’s house, they found the cards had been dealt.
On looking at his cards Mycroft announced, “If I were to guess the contents of the murder bag now the probability I would be correct is 1 in a number to a power greater than 2.”
“I am saying nothing,” said Watson, “in case I give something away.”
“Commendably prudent, old chap. But in fact, the same is true for you,” said his companion.
Watson looked at his cards and realized his friend was correct. “I say, Holmes, how could you possibly know that?”
“Sherlock isn’t cheating, Dr Watson,” said Mycroft. “He doesn’t even know how many ‘person’ cards you are holding.”
How many ‘room’ cards is Mycroft holding?
Mycroft holds 0 'room' cards.
The total number of possible combinations is given by the multiplication of the options for each category: suspects, weapons, and rooms. In mathematical terms, if we let \( S \) represent the number of suspects, \( W \) represent the number of weapons, and \( R \) represent the number of rooms, then the total number of combinations is:
\[ S \times W \times R \]
Given that there are 6 suspects, 6 weapons, and 9 rooms, the total number of combinations is, without further information:
\[ 6 \times 6 \times 9 = 324 \]
And the probability of guessing correctly without any additional information is \( \frac{1}{324} \).
We begin by analyzing Mycroft's statement about his probability of guessing correctly.
“If I were to guess the contents of the murder bag now, the probability I would be correct is 1 in a number to a power greater than 2.”
Mycroft's statement can be translated into mathematics as follows: the probability of a successful guess is \( \frac{1}{n^p} \), where \( n \) is an integer and \( p > 2 \). In other words, he has \( n^p \) possible combinations remaining after seeing his cards.
To solve this, we need to find \( n \) and \( p \) such that:
\[\text{remaining suspects} \times \text{remaining weapons} \times \text{remaining rooms}= n^p \]
After evaluating potential combinations, we find two possible solutions:
This leads us to conclude that Mycroft's cards could be distributed in either of the following configurations: \( 1 \) suspect, \( 1 \) weapon, and \( 4 \) rooms or \( 3 \) suspects, \( 3 \) weapons, and \( 0 \) room.
“Commendably prudent, old chap. But in fact the same is true for you [Watson],”
Sherlock uses Mycroft’s statement along with his own cards to deduce that Mycroft’s possible card combinations must also fit one of the configurations: \( (1,1,4) \) or \( (3,3,0) \).
With simple calculations, we conclude that the three players hold the following configurations: \( (1,1,4),(1,1,4),(0,0,3)\).
Finally, the puzzle can be solved with the help of Mycroft's following statement.
“He [Sherlock] doesn’t even know how many ‘person’ cards you [Watson] are holding.”
To determine under what circumstances Sherlock would know the exact number of ‘person’ cards Watson holds, we must examine the possible distributions.
The only way Sherlock could be certain of Watson's 'person' cards is if he holds the combination \( (3,3,0) \). In this case, the only possible configurations for Watson and Mycroft would be \( (1,1,4) \) and \( (1,1,4) \), which are identical.
Thus, we conclude that Mycroft holds 3 suspects, 3 weapons, and 0 rooms, meaning he has 0 room cards.