Bohemian Schneider♠

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Bohemian Tailor

Böhmischer Schneider is a central European two-player game; Schneider means tailor and also names a heavy defeat. Its trick rules have incompatible variants. We select the same-suit immediate-successor version, with no obligation to follow. This differs fundamentally from the any-higher-card version summarised by CardRulesPlus. No precise invention is attributed to a named person.

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Pack and counting cards

Use 32 cards, 7–A of every suit, without jokers. French-pack order is A–K–Q–J–10–9–8–7; German Ober and Unter correspond to Q and J. Count A,K,Q,J,10 as one honour each, 20 in total. Cards 7,8,9 count nothing. Ace is not low and seven cannot beat it. There are no trumps.

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Deal

Choose first dealer by lot, then alternate after every deal including ties. Shuffle and let opponent cut. Deal 6 each in two packets of 3, nondealer first. The remaining 20 form a face-down stock; expose no indicator. Nondealer leads any card face up. Keep captured pairs in separate personal piles.

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Only the adjacent rank

Responder may play any card and need not capture. They win only with the immediately higher rank of that suit: 7→8, 8→9, 9→10, 10→J, J→Q, Q→K, K→A. Otherwise leader takes both cards. A led ace cannot be beaten. A larger jump does not help: Q♠ cannot beat 10♠ because J♠ is required. Ranks do not wrap and suits are not interchangeable.

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Drawing and final cards

After both cards are played, winner captures the pair and draws the stock’s top card, then loser draws. Hands return to 6 and winner leads next. The 20-card stock supports 10 replenishments. Play the final 6 tricks without drawing, for 16 tricks overall. Never inspect or reorder the hidden stock, and never recycle played cards.

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Result and session

At the end count honours, not pip values: totals must add to 20. Winning 11–15 scores 1 victory point; 16–19 scores 2; all 20 scores 3. At 10:10 nobody scores. This session uses 6 deals: higher victory total wins, equality draws. Duration is agreed, and Schneider starts at 16 here because summaries disagree. Taking every honour does not require taking every trick.

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A jack does not always capture

A leads 9♠ and B responds J♠. Only 10♠ could capture the nine, so A takes both cards and 1 honour, the jack. Had a separate trick begun with 10♠, J♠ would win and take 2 honours. A final honour count of 16:4 scores the winner 2 victory points, not 16.

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Common questions