Your friend places 100 cards face down in a row; exactly one of them is the…

2026

Your friend places 100 cards face down in a row; exactly one of them is the Jack of Clubs, and you need to guess which one it is. You pick the card at position 15. The host knows exactly where the Jack of Clubs is: he will never remove the card that holds it, and if your own pick already holds the Jack, he chooses which other card to leave uniformly at random from the remaining 99 cards. He now removes 98 of the remaining 99 cards, leaving only your chosen card at position 15 and one other card at position 87, and says, "One of these two cards is the Jack of Clubs." What is the probability that the card at position 87 is the Jack of Clubs?

  1. A.

    0.01

  2. B.

    0.5

  3. C.

    0.98

  4. D.

    0.99

Attempted by 235 students.

Show answer & explanation

Correct answer: D

Concept:

This is a generalisation of the classic Monty Hall problem: if you pick one item at random among N equally likely candidates, and an informed host - who knows exactly where the target is, never eliminates it, and breaks ties uniformly at random whenever your own pick is already correct - removes candidates down to a single survivor besides your original pick, that survivor's probability of being the target equals the probability that your original pick was wrong, which is (N-1)/N.

Application:

  1. Here N = 100 cards and exactly one holds the Jack of Clubs, so the prior probability your pick at position 15 holds the Jack is 1/100, and the prior probability it does not is 99/100.

  2. The host knows where the Jack is and never removes the card that holds it, while removing 98 of the other 99 cards, leaving only the card at position 87 besides your own pick.

  3. If your original pick at 15 was wrong (probability 99/100), the Jack is among the other 99 cards; since the host removed 98 of those 99 without ever removing the one holding the Jack, the single card he did not remove - position 87 - must be the Jack, with certainty.

  4. If your original pick at 15 was right (probability 1/100), the Jack is at 15, and the host chooses which of the 99 non-Jack cards to leave standing uniformly at random - so position 87 specifically survives with only a 1/99 chance in this branch.

  5. By Bayes' rule, P(Jack at 87 | survivors are 15 and 87) = (1/100 x 1) / [(1/100 x 1) + (1/100 x 1/99)] = 1 / (1 + 1/99) = 99/100 = 0.99 - the tiny 1/99 term from the 'you were already right' branch is why the uniform-random tie-break rule is what makes this clean result hold exactly.

Cross-check:

Scale the same puzzle down to check the logic: with only 3 cards (you pick 1, an informed host who ties uniformly at random removes 1 non-Jack card, leaving 1 other card), the identical Bayes' rule argument gives the surviving other card a 2/3 chance of holding the Jack - the standard 3-door Monty Hall result. The 100-card version follows the same logic with (N-1)/N = 99/100 in place of 2/3, confirming the answer.

Therefore, the probability that the card at position 87 is the Jack of Clubs is 0.99.

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