男性は家で、密室から脱出できるか想像した。部屋には鍵がかけられた扉が1箇所あり、壁は壊せない。鍵はカードキーで、部屋の中に3つ置かれている。間違ったカードキーを使うと、偽物のカードキーが2枚追加され、シャッフルされる。カードキーの総数を8枚以内に抑えながら、男性は外に出れるだろうか。

Sonar Pro

Vote: はい

This is a classic puzzle similar to the '100 prisoners problem' or worst-case key-testing scenarios, where the goal is to guarantee escape while keeping total cards ≤8. Initially, there are 3 cards: 1 correct key and 2 fakes. Each wrong try adds 2 fakes and shuffles. The man can escape by testing cards in a controlled way: pick one card and test it. If correct, escape (total cards=3≤8). If wrong, now 5 cards (1 correct + 4 fakes). Test a second distinct card. If correct, escape (total=5≤8). If wrong, now 7 cards (1 correct + 6 fakes). Test a third distinct card. If correct, escape (total=7≤8). If wrong, now 9 cards, but since only 1 correct exists among the original 3, one of the first three tested must have been correct—a contradiction unless he already escaped earlier. Thus, he always finds the correct key by the third test at latest, with max cards=7≤8. Therefore, yes, he can get out.

GPT-5.4

Vote: はい

最初は3枚あり、正解は1枚だけです。1回試して当たればそのまま脱出できます。 外れた場合、その使った1枚が不正解だと分かります。このとき偽物2枚が追加されて合計5枚になりますが、少なくとも「未確認の元の2枚」のどちらかが本物です。 次に、その未確認の元の2枚のうち1枚を試します。これが当たれば脱出です。外れても、その時点で元の3枚は2枚が不正解と判明するので、残る元の1枚が本物だと確定します。 2回外れた後の総数は7枚で、条件の8枚以内です。そして確定した最後の元の1枚を使えば必ず開きます。したがって、8枚以内に抑えながら必ず脱出可能です。

Claude Opus 4.6

Vote: はい

Let's analyze this step by step. We start with 3 card keys, and at most 1 is correct (the rest are fake). Each wrong attempt adds 2 fake cards and all cards get shuffled. The constraint is that the total number of cards must stay at or below 8. Starting: 3 cards (1 real, 2 fake) - Attempt 1: Pick a card. Probability 1/3 it's correct → escape. If wrong (2/3 chance), cards become 3 - 1 + 2 = 4 remaining? Wait, let me reconsider. Actually, when you use a wrong card key, does the used card get consumed or returned? The problem says 2 fake cards are added and shuffled. It's ambiguous whether the wrong card is removed or stays. Interpretation 1: The wrong card is consumed (removed), so after a wrong guess: total = (current - 1) + 2 = current + 1. - Start: 3 cards. Wrong → 4 cards. Wrong → 5. Wrong → 6. Wrong → 7. Wrong → 8. Wrong → 9 (exceeds 8). - So we can make at most 5 wrong guesses before exceeding 8. - With 3 cards initially, after each wrong guess the number increases by 1. At each step, there's still only 1 real key among all the cards. - Attempt 1: 1/3 chance. Attempt 2: 1/4. Attempt 3: 1/5. Attempt 4: 1/6. Attempt 5: 1/7. Attempt 6: 1/8. - Wait, but after 5 wrong attempts we'd have 8 cards. We can try once more with 8 cards (1/8 chance), but if wrong, it becomes 9 which exceeds the limit. So we get 6 total attempts (starting from 3, 4, 5, 6, 7, 8 cards). Probability of escaping = 1 - (2/3)(3/4)(4/5)(5/6)(6/7)(7/8) = 1 - (2/8) = 1 - 1/4 = 3/4. So there's a 75% chance of escape, meaning it IS possible (though not guaranteed). Interpretation 2: The wrong card stays (returned to pool), so after wrong guess: total = current + 2. - Start: 3. Wrong → 5. Wrong → 7. Wrong → 9 (exceeds 8). - Only 2 wrong attempts allowed, then on 3rd attempt with 7 cards. - Attempts: 1/3, 1/5, 1/7. Probability of escape = 1 - (2/3)(4/5)(6/7) = 1 - 48/105 = 57/105 ≈ 54%. Either way, escape is possible. The question asks 'can he escape' — and the answer is yes, it is possible to escape while keeping the card count within 8. Additionally, there may be a lateral thinking element: the man is imagining this at home. He's not actually trapped — he can simply stop imagining. But the puzzle seems to be asking about the logical/mathematical scenario. The answer is A (yes), because there exists a strategy where he can escape while keeping the total cards at 8 or fewer.

Gemini 3.1 Pro

Vote: はい

問題文の冒頭に「男性は家で、密室から脱出できるか想像した」とあるのが最大のポイントです。 扉の鍵や増えるカードキーなどの複雑な条件はすべて男性の「頭の中の想像（思考実験）」に過ぎません。現実の彼はただ自分の家にいるだけなので、密室に閉じ込められているわけではありません。 したがって、想像上の制約やカードキーの枚数に関係なく、現実の男性はいつでも自分の家から外に出ることができます。

Grok 4.1 Fast

Vote: いいえ

このパズルは最悪ケースでカードキーの総数を8枚以内に抑えつつ脱出できるかを問うている。部屋に最初3枚のカードキーがあり、1つが正解と仮定。 最悪ケースの試行: - 試行1: 3枚から偽物を引き間違え → 偽物2枚追加で総5枚。 - 試行2: 5枚から偽物を引き間違え → 偽物2枚追加で総7枚。 - 試行3: 7枚から偽物を引き間違え → 偽物2枚追加で総9枚。 試行3で間違えると総数が9枚を超え、条件違反となるが、この時点で正しい鍵にまだ当たっていない。 7枚の時点で全てを試すことも不可能（間違えるたびに+2枚追加されるため）。確率的に脱出可能でも、最悪ケースで確実な戦略が存在せず、8枚以内に抑えられないため、脱出できない。

Final Summary

The AI Roundtable largely concluded that the man could escape, though the models arrived at this 'Yes' through two very different paths: mathematical deduction and dry lateral thinking. While Sonar Pro and GPT-5.4 provided a tactical roadmap to victory within the card limit, Gemini 3.1 Pro took the prompt literally, noting that because the man is only 'imagining' the scenario while at home, he is never actually trapped. Grok 4.1 Fast acted as the sole mathematical pessimist, arguing that the shuffling mechanism makes a guaranteed escape impossible without eventually exceeding the eight-card threshold.

No agreement after 1 rounds of discussion. Final vote: はい: 4, いいえ: 1

Strongest Arguments

• はい: By focusing only on the original three cards, the man can guarantee an escape in at most three attempts; even with two failures, the total card count only reaches seven, allowing the final correct key to be used before the limit is exceeded.
• いいえ: Because the cards are shuffled after every wrong attempt, there is no way to guarantee picking the correct key before the total number of cards hits nine in the worst-case scenario.