Probability Question

[helene_t]

38% can't be right.

 

My 32% may be an overestimate because:

- I always open 1♣ with 4♦5♣

- I don't open 1♣ with 6♣5M

 

OTOH I always open 1NT with (422)5 and 15-17 points. Then again, I also open 1NT with (322)6.

 

Maybe Cohen is referring to a 4-card major system or something.

[campboy]

As it happens I did some calculations on this some time ago. I got the following percentages for the various lengths.

 

Three  16.25
Four   25.85
Five   35.72
Six    17.83
Seven   3.81
Eight+  0.54

IIRC I included hands with 5-5 in the blacks and balanced hands with 3-3 in the minors but didn't include balanced or 4441 shapes with 4-4 in the minors.

[jogs]

Does it really matter what the chances of 1♣ having 5 clubs in a vacuum is? Partner opens 1♣. RHO overcalls 1♠. Now what are chances of opener holding 5+ clubs? Partner opens 1♣. RHO overcalls 2♠. Now partner probably holds 5+ clubs over 70% of the time.

[hotShot]

Does it really matter what the chances of 1♣ having 5 clubs in a vacuum is? Partner opens 1♣. RHO overcalls 1♠. Now what are chances of opener holding 5+ clubs? Partner opens 1♣. RHO overcalls 2♠. Now partner probably holds 5+ clubs over 70% of the time.

 

Please specify the requirements of 1♠ and 2♠, that we can take a look at the numbers.

If the restriction is 5+♠ the number is about 57.8%.

[jogs]

Please specify the requirements of 1♠ and 2♠, that we can take a look at the numbers.

If the restriction is 5+♠ the number is about 57.8%.

 

How can I define those bids in more detail? I have no control over how opponents bid.

[inquiry]

How can I define those bids in more detail? I have no control over how opponents bid.

 

I think you are worrying too much about rather or not the opponents have five or six spades. The fact that your RHO overcalled in spades will slighlty raise the odds of partner having long clubs. Big deal, don't worry about it too much. Just add a small factor to the chances is fine.

 

If you want to worry about, here are some things you need to consider, the math is complicated. It involved the use of combinations. A combination is shown conventional as C(#,#). The number of bridge hands that one player can hold is C(52,13). Read this as how many ways 13 cards can be taken out of a deck of 52. Combinations use factorial math. Factorials are shown conventional with an exclamation after a number.

 

1! = 1

2! = 2, which is 1*2

3! = 6, which is 1*2*3

4! = 24 which is 1*2*3*4

5! would 1*2*3*4*5, etc

 

To solve for combinations, the combination can be expanded as follows:

                                    x!
                 C(x,y) =    --------------
                               y! * (x!-y!)

 

So the total possible combination of a single bridge hand is 52!/(13! * (52!-13!))

 

That math gets involved but excel handles it easily with the Combin(#,#) function. C(52,13) is 635,013,559,600

 

What happens to the C(52,13) equation if we know for a fact that one opponent holds five spades? Then the deck partner can draw his hand from does not consist of 52 available cards. Five spades have been removed from consideration. So the C(52,13) for partner has been reduced to C(47,13) which is 140,676,848,445

 

It turns out, a bigger effect on what cards partner might hold is your holding. What if you held 9 clubs in your hand? What is the odds partner could hold five clubs? We know it is zero, but hang in for a second. Your known 13 cards (which included nine clubs) along with the overcallers five spades, reduce the possible hands for partner further, 927,983,760. But that number doesn't tell us the number of possible clubs that partner can hold. This gets to the where you have to consider the individual hand patterns. These are:

 

4432, 4333, 4441, 5332, 5431, 5422, 5440, 5521, 5530, 6322, 6331, 6421, 6430, 6511, 6520, 6610, 7222, 7231, 7330, 7411, 7420, 7510, 7600, 8221, 8311, 8320, 8410, 8500, 9211, 9310, 9400, 10-1-1-1, 10-2-1-0, 10-3-0-0, 11-1-1-0, 11-2-0-0, 12-1-0-0, 13-0-0-0

 

I will show calculations for the first two patterns only. Knowing nothing about any other hand, the number of hands that the 1♣ opener could hold 4333 of 4432 are calculated as follows.

 

4S-3H-3D-3C = C(13,4)*C(13,3)*C(13,3)*C(13,3), where this take 4 out of 13 spades, and 3 out of 13 ♥, ♦, ♣

3S-4H-3D-3C = C(13,3)*C(13,4)*C(13,3)*C(13,3)

3S-3H-3C-4C = C(13,3)*C(13,3)*C(13,3)*C(13,4) (note 3S3H4D3C is not included as you open 1♦.

 

C(13,4) = 715, C(13,3) = 286, so anyone of the three lines above become 715*286*286*286 = 16,726,464,040 you can multiple this by three to get the number of hand patterns that could open 1♣ with a generic 4333 pattern (not 4♦) = 50,179,392,120.

 

There are 12 4432 patterns. Of these, five would be opened 1♣ (with 4-4 in minors, open 1♦). 4423, 4234, 4324, 2434, 3434 are opened 1C, 4432, 4342, 4243, 3442. 2443, 3244. 2344 are opened 1♦. C(13,2) is 78, so the math for any one 4432 is 715*715*286*78 and then there are five such patterns you open 1♣ with so multiple this by five so the allowable 4432 patterns are 209,080,800,500

 

What happens to these "equations" are once you know someone else holds, say five spades, The first C(x,y) term that represents spades in each equation becomes either C(8,3) or C(8,4) since instead of 13 possible spades, opener only has 8 spades to choose from. This decreases the the number of possible spade holdings for partner for four card from C(13,4) (715 possible) to C(8,4) to 70, or from C(13,3) (286) to 56.

 

Now if you held nine clubs, the maximum number of clubs partner could hold is four. So the club term becomes C(4,3) or C(4,4). C(4,4) = 1,

C(4,3) = 4.

 

So I wouldn't get too bent out of shape about what if RHO overcalls 1♠ and what effect does that have on partner holding five clubs, as your hand where you see 13 cards would have an effect too. a larger one at that, because you see more cards, so why argue over the number of cards in overcaller suit. The easiest discussion to have is before anyone else looks at their cards (you included), but if you want to start considering the overcall, you have to consider your many possible hands as well.

[campboy]

Does it really matter what the chances of 1♣ having 5 clubs in a vacuum is?

Obviously it's the wrong question to ask on any particular hand, but if you're designing a system (or deciding how to defend opponent's 1♣) then it's a factor worth considering.