Optimal strategies for “Guess Who?”

***Using these strategies can make the game boring. You have been warned.***

This time, I will analyze the popular guessing game “Guess Who?”. As its name suggests, the goal of the game is to be the first to guess who is the character picked by the other player. Every turn, players ask each other yes or no questions about the other’s character. The goal of the game is to be the first to guess who is the opponent’s character.

American woman

Characters from the original version

This analysis is based on the classic 1982 edition, but these strategies can be easily adapted for any version (like the crazy marvel edition). Let’s first check what would be an optimal theoretical question.

Optimal question

Better questions will remove more candidates on average. Let R be the number of removed persons for a particular question. For a total of N characters, with A characters having some chosen characteristics and N-A characters not having them, we have the following distribution for R:

\displaystyle \Pr(R=r)=\left\{  \begin{array}{ll}  \frac A N & \quad \text{when }r=A\\  \frac {(N-A)}N & \quad \text{when }r=N-A,\\  \end{array}\right.

which means that A characters will be removed with a probability of \displaystyle \frac A N, or N- A with a complementary probability.

Thus, we have that the expected number of characters removed, R, is equal to:

\displaystyle \mathbb E(R)=A \times \frac AN+(N-A)\times \frac {(N-A)}N.

We want the variable A to maximize this expectation. To find this value, we can derive this equation with respect to the variable A, in order to get the following result:

\displaystyle A=\frac N 2.

This means we need a question with a characteristics separation of 50/50.

For N=24 starting characters, this means A=12 characters should have some chosen characteristic. The next turn, we need a question which gives A=6 and so on, until we discover the opponent’s character.

Best questions to ask

The rule of 5/19

To make the game longer and more fair, the game creators separated all characteristics in the same way. For every distinct characteristic, 5 characters will have it and 19 won’t:

  • 5 with hats/19 without hats,
  • 5 with blue eyes/19 with brown eyes,
  • 5 with glasses/19 without glasses,
  • etc.

Even the gender is distributed in this fashion. This made some reactions when some young girl asked Hasbro directly why there were so few female characters.

This means “normal” questions will have more or less the same results on the long term and since the proportion is far from 50%, these questions are not optimal. In fact, the number of turns to guess the character is around 6 on average using normal questions. As seen in the next graph, an optimal strategy can reduce this number under 5 turns (and normal questions are approximated with \frac 5 {24} \approx 0.208).


The game encourages low risk strategies

Multiple characteristics at once

In the official game rules, it is written that only one question can be asked each turn:”[…] ask your opponent one question per turn. Each question must have either a ‘yes’ or a ‘no’ answer.” Yet, nothing is said about the number of characteristics. Thereby, we can combine some characteristics with an “or” or an “and”. By using these grammatical conjunctions, we can have optimal questions. For example, one opening question could be “Does your character have both brown eyes and no facial hair?” which uses the trick that both mustaches and beards can be described as “facial hair”.

Using similar questions, we can have the following flow chart that guarantees to have the answer after an average of 4.66 turns. This chart is only one example of an optimal set of questions, but it has the advantage of requiring only a few adjustments in the bottom when considering that the opposing player can’t have your card.lalilala

Other optimal strategies can be conceived by using the character names (like this person did).

What can be done against an opponent who uses the same strategy?

When excluding deceiving strategies such as bluffing, some situations are advantageous for the player to guess even if he’s not sure. If the player has it right, he wins the game, or else he automatically loses and the opponent wins. This means that this opportunity should be used carefully.

After each player has asked 4 questions, both players should be left with 3 persons (as seen in the bottom of the last chart). When the first player asks his question, there is 2 possible outcomes:

  1. He asks a “good” question and his number of persons goes down to 1. In this case, he will be assured to win in next round, and so the second player has to guess between his 3 persons, with 33% chance of winning.
  2. He doesn’t ask the right question and his number of people only goes down to 2. If this happens, it is now optimal for the second player to ask yet another question that has 2 possible outcomes for him too:
    • He asks the right question and his number of people goes down to 1. In this case, he will be assured to win next round, and so the first player has to guess between his 2 persons, with a 50% chance of winning.
    • He doesn’t ask the right question and his number of people only goes down to 2. If this happens, the first player can guess or let the second player guess. In each case, both players have a 50% chance of winning.

This means that the odds of the player 2 winning when both players are using this strategy is:

\displaystyle\Pr \left(\text{Player 2 wins}\right) = \underbrace{\frac 13\times \frac 13}_{\text{First situation}} +\underbrace{\frac 23\times \frac 12}_{\text{Second situation}}=\frac 49.

This calculation uses the fact that the first situation has a probability of \frac 1 3 of occurring and the second situation has a \frac 2 3 probability. Other strategies will lead to equal or less chances of winning, thus making this strategy the ultimate one.

Further reading

  • An analysis of simple questions.
  • Some forum guy with extensive knowledge of advanced and deceiving strategies.

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