Wikipedia:Reference desk/Archives/Mathematics/2017 August 13

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August 13

Odds of drawing all 4 suits, but only the highest cards in each

I've seen it stated that a "perfect hand" in contract bridge would be all 13 of the spades, but this could still be beaten by opponents' bidding 7 notrump. (I'd be tempted to open by bidding only 6 ♠, to weaken the hint that they ought to do so.) As far as I can tell, a truly perfect hand would contain all 4 suits, but only the highest cards in each (e.g. ♠ AKQJ10 ♥ A ♦ AKQJ ♣ AKQ), since then one could bid 7 notrump and be guaranteed to win all 13 tricks. (The reason for including all 4 suits is to guarantee that the first trick won't be in a void suit, which would make it impossible to control the suits for subsequent tricks. The player who's bidding first doesn't have this requirement.) What is the probability of such a hand? NeonMerlin 22:40, 13 August 2017 (UTC)

• Correction: it could be beaten by the left-hand opponent bidding 7NT. If the auction begins 7S–pass–pass–7NT, then 7NT is going down 13 tricks, presumably doubled. --69.159.60.147 (talk) 06:10, 14 August 2017 (UTC)

There are 220 hands that work here. Such a hand must contain all 4 aces. After that, you need any number of the top cards of each suit, which is equivalent to a 4-tuple of nonnegative integers which add to 9. There are ${\binom {9+3}{3}}={\binom {12}{3}}=220$ such tuples (see stars and bars (combinatorics) for more detail). There are ${\binom {52}{13}}=635,013,559,600$ total hands. So the probability you're looking for is $220/635,013,559,600,$ or about 1 in 2.886 billion. --Deacon Vorbis (talk) 23:49, 13 August 2017 (UTC)

Oh, and if you drop the requirement that you need to have all the aces, then you get ${\binom {16}{3}}=560$ such hands instead, so about 1 in 1.334 billion. --Deacon Vorbis (talk) 23:55, 13 August 2017 (UTC)

The latter figure can't be right, because there are more than 16 cards from which the top 13 of multiple suits might be drawn. For example, they could be ♠ AKQJ10987 ♥ A ♦ AK ♣ AK or ♠ AKQ ♥ AKQ ♦ AKQ ♣ AKQJ. NeonMerlin 23:59, 13 August 2017 (UTC)

It's right. You're looking for the number of 4-tuples of nonnegative integers that sum to 13 (each of which gives the number of cards of a specific suit). Again by stars and bars, this is given by ${\binom {13+3}{3}}=560.$ --Deacon Vorbis (talk) 00:07, 14 August 2017 (UTC)

It's also worth noting that if you have the top 6 cards in a suit, then any remaining cards of that suit are safe to hold (no matter how low they are), since you're still guaranteed to win with those as well after playing all the top cards. So the number of safe hands is really higher taking that into consideration. --Deacon Vorbis (talk) 01:46, 14 August 2017 (UTC)

If you take that into account, in the first case (require all suits), there are an extra 520 safe hands, for a total of 740, or about 1 in 858 million. In the second case (allow voids), there are an extra 6304 safe hands, for a total of 6864, or about 1 in 92.5 million. Still pretty remote, but not quite so bad. --Deacon Vorbis (talk) 02:10, 14 August 2017 (UTC)

There are more safe hands. If you require all suits:

• Hold the top-5 and at least 3 other cards in a suit, so no opponent can have more than 5.
• Hold the top-4 and at least 5 other cards in a suit, so no opponent can have more than 4.
• Hold the top-3 and exactly 7 other cards in a suit, so no opponent can have more than 3.

I didn't say "at least 7" because if you have all aces then you cannot have more than 10 cards in a suit. PrimeHunter (talk) 21:36, 15 August 2017 (UTC)

Oh, good point. I guess the same idea still works if you don't require all suits, for that matter. I get the sense that the OP still doesn't believe me though. --Deacon Vorbis (talk) 22:06, 15 August 2017 (UTC)