[kgr]

Given a certain distibution of suit A, how to calculate the odds of a certain card with East including certain combinations.

Examples:
- You know that Spades are 4-2 (West-East), What are the odds that East has Club Q + odds of Club Qx with west.
- You know that Spades are 4-2 (West-East), Hearts are 3-3, What are the odds that East has Club Q + odds of Club Qx with west.

=> How to calculate this/Is there an internet site where you can calculate it?

[Elianna]

kgr, on Jun 19 2005, 11:43 PM, said:

Given a certain distibution of suit A, how to calculate the odds of a certain card with East including certain combinations.

Examples:
- You know that Spades are 4-2 (West-East), What are the odds that East has Club Q + odds of Club Qx with west.
- You know that Spades are 4-2 (West-East), Hearts are 3-3, What are the odds that East has Club Q + odds of Club Qx with west.

=> How to calculate this/Is there an internet site where you can calculate it?

You can take a course in probability (combinatorics will also greatly help) and learn how to calculate conditional probabilities.

I'm too tired to calculate these for you, except to tell you that I hope that you mean your plus as and OR not an AND. If no one answers in the morning, I'll give a more detailed explanation.

[kgr]

Elianna, on Jun 20 2005, 08:48 AM, said:

kgr, on Jun 19 2005, 11:43 PM, said:

Given a certain distibution of suit A, how to calculate the odds of a certain card with East including certain combinations.

Examples:
- You know that Spades are 4-2 (West-East), What are the odds that East has Club Q + odds of Club Qx with west.
- You know that Spades are 4-2 (West-East), Hearts are 3-3, What are the odds that East has Club Q + odds of Club Qx with west.

=> How to calculate this/Is there an internet site where you can calculate it?

You can take a course in probability (combinatorics will also greatly help) and learn how to calculate conditional probabilities.

I'm too tired to calculate these for you, except to tell you that I hope that you mean your plus as and OR not an AND. If no one answers in the morning, I'll give a more detailed explanation.

Yes, I mean plus as an OR.

You play 6NT, having:
Axxx
AQ
AQT8
JTx
-
Kxxx
KJT
KJ9
AKx

West starts with Spade Q (you let it hold) and spade J on which East discards a Heart.

This gives you 2 possible plays:
1. Cash club A and finesse for club Q later (East has club Q)
2. Cash club AK and play for a squeeze (West has club Q OR East has Club Qx OR East has Club Q singleton)

How to calculate the probabilities? ... Or is there a tool available on internet to calculate it?

[ochinko]

West has 9 empty slots (cards - spades), East has 12. Based solely on that info you have 12/21 or 57% of catching CQ in East. This makes the clubs finesse preferable than a squeeze.

A better idea though is to cash all hearts and diamonds after winning with SpK. Now your percentage estimates will be far more precise.

Petko

[kgr]

ochinko, on Jun 20 2005, 10:44 AM, said:

West has 9 empty slots (cards - spades), East has 12. Based solely on that info you have 12/21 or 57% of catching CQ in East. This makes the clubs finesse preferable than a squeeze.

Given this info:
The squeeze will be slightly better then 43% because it is possible that West has Club Qx.
How do you calculate this additional % given the info you have till now?

[Gerben42]

Have a look at Richard Pavlicek's site I think he has a tool for things like this.

[kgr]

Gerben42, on Jun 20 2005, 11:49 AM, said:

Have a look at Richard Pavlicek's site I think he has a tool for things like this.

Thanks. I can do the calculations using:

http://www.rpbridge.net/xcc1.htm

...Still interested in the mathematics if anybody has them.

Koen

[beatrix45]

You can get very close to the precise answer just by enumerating the possible combinations. For example, a specific card is twice as likely to be in the four card hand as in the two card hand. (Four chances versus two chances). There is a further consideration that comes about because it is easier (more likely) to fit two cards of a specific suit into 13 unknown cards than to fit four cards. In most cases this is not important enough to worry about. It seldom changes the odds enough to matter. A big exception is if there has been a preempt. As mentioned in earlier posts, Richard Pavlicek has a web site that can give you the tools to deal with this.

[kgr]

The probability of specific Qx with RHO if very small (around 3%).
In the above problem (squeeze or finesse):
- If available space are not equal then play hand with highest available space for the Q.
- If available space are equal then play for the squeeze.

[ochinko]

kgr, on Jun 20 2005, 02:22 PM, said:

Gerben42, on Jun 20 2005, 11:49 AM, said:

Have a look at Richard Pavlicek's site I think he has a tool for things like this.

Thanks. I can do the calculations using:

http://www.rpbridge.net/xcc1.htm

...Still interested in the mathematics if anybody has them.

Koen

Yes, I missed singleton Queen and Qx in East.

Ok, you've got the percentages from Richard Pavlicek's Card Combination Analyser. I'll try to mobilize my mathematical and English skills to show how you can arrive to these numbers without a specialised calculator.

Again, we use the slots metaphor. We imagine that we drop the cards from above, and they fall into any free slot with equal probability. We have six small clubs and the Queen of clubs.

First case: What is the percentage of East having a singleton Queen? We remember that we have 9 free slots in West, and 12 in East, so when we drop the CQ it will land in East in 12 out of 21 cases. For it to be singleton all the other clubs must fall into West's slots. The first one gets there in 9 (free slots in West) out of 20 (total empty slots) times. The next one - 8 out of 19 times, etc. We have to multilpy all those probabilities, which leads us to:

12/21 * 9/20 * 8/19 * 7/18 * 6/17 * 5/16 * 4 / 15 = 0.2%

Second case: Qx in East. This is a bit more complicated. We start like in the previous case: 12/21 (for the CQ) * 11/20 (the other small club in East), then continue with the West clubs * 9/19 * 8/18 * 7/17 * 6/16 * 5/15. Here comes the tricky part. Because we considered all small clubs, but they aren't, we have to multiply the result by 6, since the small club in East can be any of the six. This increases the chance for the squeeze with 2%. So if you like to play for the better percentage (55), you go for the finesse. You can still chose to play for the kibitzers, though, and go for the squeeze Again, these numbers will change if you cash your tricks in diamonds and hearts first.

Hope to have helped, and not messed up something.

Petko

[PriorKnowledge]

This site computes these odds for you.
http://www.automaton.../en/OddsTbl.htm

[kgr]

Petko,

Thanks a lot. I will study your reply as soon as I have a bit more time for it.
(I did also receive an explanation from Frances Hinden on rgb. Both replies give me excellent info to work with)
I just want to understand the calculations. I don't intend to use it at the table

Koen

[inquiry]

How to calculate some of this is hardly practical, but other of it is incredibly simple. Let's handle the first case first. You know that West had 4 Spades, and East 2. Nothing else about the hand.

Room in WEST hand for club queen is (13-4 = 9 spots). Room in East hand is (13-2) = 11 spots. 11+9 = 20. So there are 20 bins for the queen to fall in. 11/20 = 55% chance Queen is with RHO, 45% with LHO. We will use similar math later as a last step to calculate the chance or a doubleton queen.

If hearts are 3-3, you can modify these numbers so that of instead of 20 remaining bins there are 14, with 6 and and 8 slots in the two hands. 8/14 = 57.14%. Math so easy, anyone can do it, often in their heads.

For the other situations, like chances for singleton or doubleton queen, the math becomes much more complicated. First you have to know what a factorial is. Factorial is a number followed by an exclamation point, and it means every number from 1 to that number, multipled by each other... so that

3! = 1x2x3 = 6, and
5! = 1x2x3x4x5 = 120

If you are using excel, factorials are calculated easily with the function FACT(x) where "x" is the number to calculate.

Now that you know what a factorial is, the next thing you need to know is how to calculate a bionomial coefficient (anyone still with us? Richard could do a better job of this). A binomial coefficient (also known as a known as a combination or combinatorial math) calculates the number of ways of picking p non-ordered outcomes from n possibilities. There are two normal ways of expressing this, one is to put n directly above k inside a large set of paranthesis. Trust me I don;t know how to show that here with the limits of these forum software. But the other way is easy, this is the combinational way to showw a bionomial coefficient....

nCk

Read this as "n choose k". That is, it gives the number of k-subsets you can make out of n different items. Thus if k was 2 and n was 5, you could make 10 different grouping. Lets assume the 5 items are letters ABCDE, how many different ways can you group these? The answer is 10...

AB
AC
AD
AE
BC
BD
BE
CD
CE
DE

Of course, Excel will calculate this for you, the combinational math function is COMBIN(5,2). The equation breaks down to

nCk = n!/(n-k)!xk!

substituting in 5 for n and 2 for k

5C2 = 5!/(5-2)!*2! = 5!/(3)!*2! = 120/6*2 = 120/12 = 10

For fun, The number of possible distinct 13-card hands is (using excel combin function)...

52C13 = COMBIN(52,13) = 6.35014E+11

I guess a Bridge Encyclopedia would give you 635,013,559,600, but the rounded excel number is good enough for us.

Once dummy hits, you know 26 of the cards, so the number of hands your opponents can have is 26C13 (note, once one of these hands is "Defined" the other is know. 26C13 comes to 10,400,600 hands.

As cards are played, the possible hands decrease of course, as we saw in the very simple example above. Now comes some more complicated math. Let's take a very, very simple case first. Let's assume that you have 11 diamonds missing the king. And you want to know, what are the chance for 1-1 split versus 2-0? The answer is calculated as follows...

The total hands they can hold is 13C26 but you are going to divide this into hands where diamnds are 1-1 and where diamonds are 2-0.

CASE ONE When diamonds are 1-1, we define the equation for their hands as ...

2C1 * 24C12 = 2,496,144

CASE TWO When West has two diamonds 2-0, this is defined as

2C0 * 24C13 = 5,408,312

And CASE THREE when EAST had 2 diamonds (0-2), this is defined as

2C1 * 24C11 = 2,496,144

Note, you calculate the odds on either opponents hand (the other is the same), for each pattern. Here we show EAST hand calculation. The total number or case is (add the millions up) = 10,400,600. The chances diamonds are 1,1 is 5,408,312/10,400,600 = 52%, just as you have been taught.

The general equation is when there are "N" missing cards and you want to calculate how many "K" ways these can be in a particular opponents hand (Lets say WEST since you will be looking for doubleton club queen here) is...

Total combinations = nCk * (13-k)C(26-n).

So chances WEST has a doubleton club when 7 clubs are missing would be..

7C2 * (13-2)C(26-7) =7C2 * 11C19 = 21 * 75582 = 1587222

When you divide 158722 by 10,400,600 you get the chance of LHO having a doubleton club as 15.26%

But as you have known cards in other suits, 13 and 26 are reduced accordingly. So now we are getting close to the problem you wanted. What are the chances WEST will have a doubleton club when he holds 4 spades and 3 hearts (and EAST had 2Spades and 3 hearts)?

The total number of hands have been reduced, because we know 12 cards. So the instead of 13C26, the number of hands is now much lower. For WEST he can have only 6C14 hands. And his partner, 8C14, these are the same 3003 hands!!!

Now the chances WEST can hold exactly two clubs is...

0C7 X 6C7 = 1 x 7 = 7 hands
1C7 x (6-1)C(7-1) = 7 x 21 = 147 hands
2C7 x (6-2)C(7-2) = 21 x 35 = 735 hands
3C7 x (6-3)C(7-3) = 35 x 35 = 1225 hands
4C7 x (6-4)C(7-4) = 35 x 21 = 735 hands
5C7 x (6-5)C(7-5) = 21 x 7 = 147 hands
6C7 x (6-6)C(7-6) = 7 x 1 = 7 hands

Total hands = 3003 (as it should). So you can see where the numbers can from. West has 6 vacant slots and there are seven clubs. So he can have from 0C7 to 6C7 (he can not have 7C7 simply becasue he can only hold six cards). His remaining cards would have been (13-club)C(26-clubs), except we knew he held 3 hearts and 4 spades. We also knew his partner had 2 spades and 3 hearts. So West number of cards was 13-7 = six before we started placing clubs.. so his starting "non-club" number was 6. So if he had clubs, the second combination would be 6Cy. As we add a club we have to reduce this 6. The second number is not 26, because we first remove the known cards from both (12 majors). So 26 becomes 14. But then, we also have to remove the seven clubs are using in the calculations. That leaves seven not clubs (turns out to be the seven diamonds).

When you divide the number of hands for each distribution of clubs by 3003 (the total), you find this percentage for clubs with WEST...

0 = 0.23
1 = 4.90
2 = 24.48
3 = 40.79
4 = 24.48
5 = 4.90
6 = 0.23

Hopefully if you check one of the online things, you will see similar numbers. So the chance WEST has a doubleton club is 24.48, but this is not the chance it is doubleton queen. Here, you have to see that clubs are divided 2-5, so EAST is much more likely to have the queen. So you take the two slots out of seven (2/7) (chance WEST has the queen when holding 2 clubs) and multiply by the chance of west having doubleton club (24.48%) and you get 6.99%.

To test these numbers out go to one of the pages with online calculations. Use first 7-5 known cards and look for split of a seven card suit... then look for split of seven card suit with a queen.

Hope that helps.

Ben

[pigpenz]

this is what makes bridge so interesting you can play hand out to the best possible probablitities and that doesnt mean it will work on the particulalr given hand. but in the long run the force will be with you