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Someone sent me this question today:

Your friend chooses at random a card from a standard deck of 52 cards, and keeps this card concealed. You have to guess which of the 52 cards it is.

Before your guess, you can ask your friend one of the following three questions:

· is the card red?

· is the card a face card? (Jack, Queen or King)

· is the card the ace of spades?

Your friend will answer truthfully. What question would you ask that gives you the best chance of guessing the correct card?

It's relatively easy to calculate that all three questions give you an equal probability of choosing the right card - 1/26 - so it basically doubles your likelihood of guessing correctly. It's also relatively easy to generalize this to prove that any question you ask about the deck, if answered truthfully, will result in the same gain in likelihood.

Laypeople will find this surprising. But I wondered if we can generalize further to say that this is the case for any zero-sum game? For example, in Chess you could be asked to guess which piece was moved first - you could ask was it black or white, was it a pawn, was it the Knight?

I wondered if this problem was familiar to anyone and if they had seen any proofs or disproofs of the general zero-sum game proposition?

Thanks!

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  • $\begingroup$ When you say "It's relatively easy to calculate that all three questions give you an equal probability of choosing the right card . . . " what three questions are you referring to? (The three bulleted questions give you a 1/13, 1/12, and 1/4 chance of guessing correctly if the answer is "yes", respectively.) $\endgroup$ Commented Jun 18, 2021 at 4:19
  • $\begingroup$ @SelrachDunbar: In the first case your chance of guessing right is $1/26$ whether the answer is yes or no. In the third if the answer is yes your chance is $1$. The point is that after asking any of the three, your chance of guessing right weighted by the chance of a yes or no is $1/26$ as I show. $\endgroup$ Commented Jun 18, 2021 at 4:23

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Suppose there are $p$ equally probably possibilities. You ask a question that will be answered yes by $q$ of them and no by $p-q$ of them. The chance your guess is correct is then $$\frac qp \cdot \frac 1q + \frac {p-q}p \cdot \frac 1{p-q}=\frac 2p$$ which does not depend on $q$, so it doesn't matter how large $q$ is. I find this counterintuitive because I get less bits of information as $\frac pq$ deviates from $\frac 12$.

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Yes, it is counterintuitive since if you instead got to ask two questions, you could double your chance of guessing correctly by asking so that p=q=1/2. E.g. q1: is it red, q2: is it a spade etc (this gives 1/13 chance to guess correctly). In the other extreme (q1: is it ace of spades; q2; is it queen of hearts) your chance of guessing correctly is only increasing linearly in the number of questions (this only gives 3/52 chance of guessing correctly). More generally, the p=q=0.5 strategy gives chance=(1/52)*2^n where n is the number of questions (disregarding the fact that you cannot set p=q=0.5 for n>2). The "is it the queen of hearts" strategy instead gives the chance=(1+n)/52. It just happens that for n=1 we get 2^n=(1+n). So the intuition is correct -- just not for n=1. This puzzle does thus not contradict your intuition that it is a crappy strategy when playing 20 questions to go directly for "is it Donald Trump?".

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