Problem No. 10, 2001


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The problem

Our new problem, the tenth, comes from Torbjörn Gustavsson of Stockholm. And considering how long it took me to solve it, I can assure you that it is tricky.

BK S:t Erik in Stockholm arranges a strong tournament for 52 pairs every year, Plattliret, and ten years ago two of the competing pairs used the Carrotti system. It is a so called strong pass system, where a pass in first or second seat shows 13-16 HCP with any distribution; 1 club is either 17+ HCP or a balanced had with 10-12 HCP; 1 diamond shows any 0-7 hand; and the remaining opening bids are natural with 8-12 HCP.
  And when the players involved were four of Sweden's best bridge players (I think they were P-O Sundelin – Tommy Gullberg and Tjolpe Flodqvist – Hans Göthe), it's easy to understand why the round where the Carrotti pairs should meet attracted many spectators. How would Carrotti handle Carrotti?
  But those waiting for action, were waiting in vain. On the first board of the round only North-South were vulnerable, and Tjolpe Flodqvist, North, was the dealer. And in these unfavorable vulnerabilities Flodqvist – Göthe used a more traditional system. So Tjolpe opened with a naural 1 heart, just like the rest of the players in the room. No fun at all.
  One board left. Now, P-O Sundelin, East, was the dealer and only East-West were vulnerable. And can you imagine: Sundelin – Gullberg also used a tradtitional system at these vulnerabilities. What an anticlimax!
  In the evening three of the players came to talk about the round where the Carrotti pairs met. This is what they said:

– A: "Carrotti versus Carrotti, and the dealer's side is vulnerable against not on both deals. The chance for that to happen has to be very small."
– B: "Yes, only one out of four."
– A: "You're kidding. Had you been a little better at mathematics, you would have known the correct answer is one out of sixteen."
– C: "Don't you mean one out of twelve?"

Which of the players are right, or is the correct answer something else. What do you think – and why?


Solution

There are four possible vulnerabilities on a board (North-South, East-West, All and None). Any of these can occur on the first board of the round, with the same frequency (25%). But for the second board of the round, things are different. In tournament bridge the boards are played in ascending order: you start with board No. 1, then play No. 2, etc; and the vulnerabilities are fixed on beforehand, so that within every interval of four boards (boards 1-4, 5-8, etc) there is one occurence of every possible vulnerability just like there is one occurence of every possible dealer.
  Who is dealer rotates one step clockwise, starting with North on board No. 1. Then it's East on board No. 2, etc, and on No. 5 we are back to North. That means you need 16 boards (4 x 4) to get every possible combination, so that North once has been dealer with North-South vulnerable, once with both sides vulnerable, etc. When you come to board No. 17, you start it all over again with board No. 1.
  In the international laws of contract bridge, you can see how the dealer and the vulnerabilities rotate on these 16 boards. If we look at every four board interval, it looks like this (N = none, A = all):

Deal No. Vulnerabilities
1-4 N – NS – EW – A
5-8 NS – EW – A –N
9-12 EV – A – N – NS
13-16 A – N – NS – EW

If you compare any interval with the next, you'll see that the everything has moved one step to left (except the first one, which now is the last). That means the relative order between the vulnerabilities will always be intact. After North-South come East-West; after East-West come All; after All come None; and after None come North-South. So if you in tournament bridge play rounds of two boards each, and know the vulnerabilities of the first board of the round – which always is an odd number, and always with North or South as dealer – you automatically know the vulnerabilities of the second (even numbered) deal. There are only four possible cases!
  Does this mean the probabilities for a round with two consecutive boards where only the dealer's side is vulnerable is 1/4?
  Yes – if we play an infinite number of boards. In Plattliret, where the story comes from, 52 pairs competed. They met every other pair, so 51 rounds were played. If you look after in which rounds North-South are vulnerable against not on the odd board and East-West vulnerable against not on the next board, you'll see that it happens in round 3 (boards 5 and 6) and round 8 (boards 15 and 16). In 51 rounds, you play six full 16-boards sets and six more boards. That means you play "round 3" seven times and "round 8" six times. So, in Plattliret, the chanse that this would happen is 13/51.
  But since it wasn't obvious that I was looking for the probabilities for this very tournament, I have accepted both 1/4 and 13/51 as correct answers.


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