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Talking about that 2♦ can be passed I like to know that partner will never pass in pass-out position.

As for leaping michaels, I like to show my hand as quick as possible but I guess as most people play 2♠ as spades (over multi), 4♠ in given auction must be spades and a minor.

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How do you act then when bidding goes 1♦-(2♣)-x-(3/4♣) or similar with strong hands including one major? I play that double suggests both majors, though not necessarily 4+4+.

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Actually I have played in irregular partnership and bid 3♦ followed by 3♠ over 3♥ (by RHO) and by double over 4♥ (aggressive, I know, but I felt like neither -100 nor +620 would be good scores) which my partner corrected to 4♠ with 3 spades.

4♠ doubled was down one against cold 4♥, but, however, bridgemate showed one-digit number for us. Most of the tables didn't go further than 3♦, 3♠ or even 1NT on our line, and my partner has 3-3 in minors which I suppose most of the pairs open with 1♣.

I definitely agree with 2♠ in IMPs but I found a hand interesting for matchpoints.

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I am not sure if I like more “solid 7+ major” or “a hand that didn't open 1NT just because of six majors (but if it was a minor it would be 1NT)”. I voted never bid it as well.

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You include fact they are from same suit when you divide it by (26 choose 6). That way you can say that in first example there is nothing in formulation that says they have to be from a specific hand, but there is that (26 choose 13).

I agree it is more intuitive to create one whole hand than one whole suit if there is one suit taken into consideration :)

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If you want, for example, to calculate chance for 4333 distribution you calculate it like this: (13 choose 4)*(13 choose 3)*(13 choose 3)*(13 choose 3)*4/(52 choose 13) and it is because from 13 cards of one suit you pick exactly 4, from 13 of other you pick exactly 3, etc. you multiply it by 4 because 4 of a suit can be in every of 4 suits and you divide it by (52 choose 13) because from 52 cards you picked 13. You can see that as dividing 13 places in groups of 4, 3, 3 and 3 and when you multiply it with 4, you reserved group of 4 for specific suit.

Now if you are interested to know just what is a chance that there are 3 cards of one suit in one hand and 3 cards of that suit in other hand you can either:

1) Pick 3 cards from 6 for one hand and 10 from 20 (of other three suits) cards for the same hand, and put remaining cards in the other hand, (multiply by 2 if needed) then divide it by number of combinations…

or

2) Pick 3 from a vacant places for one hand and pick 3 from the other hand, multiply it by 2 if switching hands doesn't cover all combinations (as for 4-2 breaks etc.) and divide it by number of combinations…

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When I first saw his comment I thought he meant the same as I do just reversed k with n-k. Now when he edited comment I see what he meant. And yes, you can get one from other:

The one Dave wrote is: (6!*14!/3!/3!/7!/7!)/(20!/10!/10!) The one I wrote is: (10!*10!/3!/3!/7!/7!)/(20!/6!/14!)

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But let's say probabilities are 36% for 3-3 break and 48% for 4-2 break. In total it is 84%. When both opponents follow to the first two rounds of the suit, we know that suit is either 3-3 or 4-2. To me it seems it is 36%/84% that suit is 3-3.

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Probably you are right, I can't find the book atm. And yes, still I have no idea how to get to 50%. Given what David said, seems like it is around 40%.

Selena Pepić

As for leaping michaels, I like to show my hand as quick as possible but I guess as most people play 2♠ as spades (over multi), 4♠ in given auction must be spades and a minor.

Selena Pepić

Selena Pepić

Selena Pepić

Selena Pepić

Selena Pepić

Selena Pepić

4♠ doubled was down one against cold 4♥, but, however, bridgemate showed one-digit number for us. Most of the tables didn't go further than 3♦, 3♠ or even 1NT on our line, and my partner has 3-3 in minors which I suppose most of the pairs open with 1♣.

I definitely agree with 2♠ in IMPs but I found a hand interesting for matchpoints.

Selena Pepić

Selena Pepić

I agree it is more intuitive to create one whole hand than one whole suit if there is one suit taken into consideration :)

Selena Pepić

(13 choose 4)*(13 choose 3)*(13 choose 3)*(13 choose 3)*4/(52 choose 13) and it is because from 13 cards of one suit you pick exactly 4, from 13 of other you pick exactly 3, etc. you multiply it by 4 because 4 of a suit can be in every of 4 suits and you divide it by (52 choose 13) because from 52 cards you picked 13. You can see that as dividing 13 places in groups of 4, 3, 3 and 3 and when you multiply it with 4, you reserved group of 4 for specific suit.

Now if you are interested to know just what is a chance that there are 3 cards of one suit in one hand and 3 cards of that suit in other hand you can either:

1) Pick 3 cards from 6 for one hand and 10 from 20 (of other three suits) cards for the same hand, and put remaining cards in the other hand, (multiply by 2 if needed) then divide it by number of combinations…

or

2) Pick 3 from a vacant places for one hand and pick 3 from the other hand, multiply it by 2 if switching hands doesn't cover all combinations (as for 4-2 breaks etc.) and divide it by number of combinations…

Selena Pepić

Selena Pepić

The one Dave wrote is: (6!*14!/3!/3!/7!/7!)/(20!/10!/10!)

The one I wrote is: (10!*10!/3!/3!/7!/7!)/(20!/6!/14!)

So it is the same compound fraction.

Selena Pepić

I was talking about Steve's second case, but I see your point and, indeed, it is easier to calculate it your way.

Selena Pepić

Selena Pepić

Selena Pepić

Selena Pepić

And yes, still I have no idea how to get to 50%. Given what David said, seems like it is around 40%.

Selena Pepić

Selena Pepić

Selena Pepić