[inquiry]

---

Needing 3 tricks in this suit, what are the approximate odds of 3-3 split if you cash the AK and you opponents follow with low clubs on both rounds. Vote your honest choice (don't bother calculating, if you don't want too, just pick the best answer). Of course if you want to calculate it, great.

[Cascade]

This post wouldn't let me reply yesterday. Was there something set to do that?

[inquiry]

Cascade, on Mar 9 2004, 07:37 PM, said:

This post wouldn't let me reply yesterday. Was there something set to do that?

Hi, Yes, when I set up this poll, I intentionally made it voting only so that no one would show the mathematics of the correct solution until people had sometime to think about it and vote their choice.

Once there was a wide variety of votes (there are 17 now), I took off the block so that people can explain the math, if htey like. It seems that people have a fairly good grasp of the 3-3 split situation in general, but maybe overlooking how the odds may or may not change as cards played.

Ben

[inquiry]

If you look up the odds for how a suit with 6 cards outstanding will split, you will find a table something like this....

6-0    1.49
5-1   14.59
4-2   35.53
3-3   48.45

This is what the probabilities are before any cards are played. Can played cards affect these odds? You betcha. Imagine a hand where you are now at trick 10 and no one has played a heart yet. You hold these cards

---

When you cash the SPADE ACE, both opponents follow. Now there are 6 cards remaining and the opponents only have 6 card left, so you are 100% chance that hearts are 3-3.

So how does cashing the ♣AK and have only small cards show up? First, the 6-0 distribution is no longer possible, nor are the 5-1 distributions, and of the possible 4-2 distributions, all of them with Qx doubleton are alos no longer possible. That is if the missing cards were the Q5432, the possible doubletons with doubleton Q would have been... Q2, Q3, Q4, Q5, Q6, this turns out to be 1/3 of the 4-2 splits.

Thus what is left, is the original distributions that gave you 35.53% for 3-3, but only 2/3 of the distributions that gave you a 48.45% for 4-2 split. The other distibutions are no longer available. Reducing the 48.45 by the 2/3 it becomes only 32.3%.. now you have to normilize these numbers. The 35.53 + 32.3 now equals the 100% value...

3-3  35.53
4-2  32.3
------------------ 67.83

So 3/3 is now 35.53/67.83 = 52.4
  4-2 is 32.3/67.83 = 47.6

The reason you can calculate this out to 52.38% is that the QUEEN is what is known as a "posiitve" card that gives you a clue to the distribution, all the other cards are neutral. You can watch for it... if the suit in question had been....

---

And you cash the AK and both opponents follow to both, any card they play in this suit would be neutral (they could drop Q on second round from Qx, QJx QJTx, etc) so you will not be able to rule out any of the doubletons like you could when the queen did not fall. Now the new odds for 3-3 suit is quickly calculated via....

3-3  35.53
4-2  48.45
------------------ 83.98

So,
3-3  42.3%
4-2  57.3%

So what did we learn? First odds change as cards are played, and second the concept of neutral and positive cards.

Ben

[mpefritz]

Ben, I usually agree with your analysis. But...

1) Why are you playing the original suit combination as you did?

2) You analysis of AK32 across from 654 seems off.

After 2 rounds each player has played specific cards. Given that they are all equals, there should be no RC involved. Say LHO has played AB , RHO CD with EF still outstanding. Then the question is actually comparing a priori chances of:

LHO RHO

AB CDEF
ABE CDF
ABF CDE
ABEF CD

I believe this translates to comparing 1-1 split vs a 2-0 split. The exact numbers depend on how many cards have been played in other suits, etc.

<temporary retraction of the above>

<WARNING> Do not read unless you really really want to:

there is a temporary retraction in the above logic as we need also factor in the likelihoods that AB and CD would show up in each of these cases. When those are factored in, the odds may tip away from 3-3 as AB CDEF would be played {1/C(2,2) * 1/C(4,2)} = 1/6 of the time as AB CD. ABE CDF would be played as such would {1/C(3,2) * 1/C(3,2) } = 1/9 of the time as AB CD, etc.. So those probabilites times the a priori of each would be more appropriate to compare.

Using the same analysis using the AKxx Jxx example would give each of the a priori equal chance of having the spots show up as they did. So in the Q example, it would be (essentially, not exactly) comparing the a priori of a 1-1 to a 2-0.

BUT in the AK32 654 example since the missing cards are not all known to be equals to the player playing them (say, from QTx) , the above factors (how likely AB and CD would show up from each of the mentioned combinations) be swayed back towards 3-3.... In other words maybe the Q is a "positive" card after all!

I think then, in the AK32 654 case, the real chacnes are closer to 50+ for 3-3 and 50- for 4-2.

The 52+% is close for the AKxx Jxx case and depends on how many cards have already been played. I assume there was an opening lead..

</WARNING>

Therefore, I am now in temporary partial (dis)agreement with Ben on the AK32 654 > I am in full disagreement with the original playing of the AKxx Jxx, but do agree with the chance of 3-3, 4-2 at the point he was at in that problem.

Anyone else?

fritz

[badderzboy]

I wasn't sure about any of the %'s as we now have conditional probabilty introduced so the original %'s are irrelevant as .

The only possible suit combinations that would show xx and xx into two rounds are :-

(LHO, RHO) qxx xxx ,xxx qxx, qxxx xx, xx qxxx so I would place the odds of Q dropping at 50%?

Steve

[Cascade]

badderzboy, on Mar 10 2004, 06:28 AM, said:

I wasn't sure about any of the %'s as we now have conditional probabilty introduced so the original %'s are irrelevant as .

The only possible suit combinations that would show xx and xx into two rounds are :-

(LHO, RHO) qxx xxx ,xxx qxx, qxxx xx, xx qxxx so I would place the odds of Q dropping at 50%?

Steve

Each individual 3-3 break is slightly more likely than each individual 4-2 break hence Ben's 52.4%

[Trpltrbl]

According to my calculations, there is a 68,9752% chance of getting 3 tricks. If you play the Ace and then a small towards the Jack, I think I will take my chances with that

Mike

[inquiry]

badderzboy, on Mar 10 2004, 04:28 AM, said:

I wasn't sure about any of the %'s as we now have conditional probabilty introduced so the original %'s are irrelevant as .

The only possible suit combinations that would show xx and xx into two rounds are :-

(LHO, RHO) qxx xxx ,xxx qxx, qxxx xx, xx qxxx so I would place the odds of Q dropping at 50%?

Steve

WARNING... this goes into material not that useful to anyone, and certainly not to beginners ... the best way to get percentages is to just know the basics of a few simple situations and realize how other cards affect them as shown in original problem. This is meant to address a question raised by bladderboy. I would skip reading this one myself...

Hi Steve,

The solution lies in combinational math.. .if you have excel, it has a function called combin that can this for you very quickly.

With two cards remaining in the suit, the number of possible combinations are

@COMBIN(2,0) = 1
@COMBIN(2,1) = 2
@COMBIN(2,2) = 1
Total distib = 4

(where 2,0 means one hand has 0 of 2 cards, there is only one way this can happen. 2,1 = i hand can have 1 cards..there is two way this can happen, the one can be any one of the two cards, and there is only one combination where you hold both the cards). This is how you come up with a 50% calculation.

But lets imagine this as a situation where you have 11 trumps and are missing the king. When you play a low card and the next hand plays low, do you play for the drop or the hook? Are the odds still 50-50?

The answer is the odds are not 50-50 once a low card is played. In this case, the low card was in fact a "positive" card, as it excluded the situation where WEST had a singleton King. This also allows one less vacant space in the players hand to hold the king. This affects the math, but just slightly. So the possible distributions are now Kx and x. Which one is more likely? Are they equal?

I want you to take on faith that if you can see your hand and dummy's, there are only 26 unknown cards that can be divided in @Combin(26,13) ways = 10,400,600 unique ways. (same math that says two cards split 1-1 can be divided in 2 ways).

When one opponent has one of your trumps, he can have any of the two, that is expressed as @Combin(2,1). But since he has one card in your trump suit, he has only 12 other cards, and that is expressed now, not as @Combin(26,13), but rather has @combin(24,12). First, why 24 and not 25 or 26? Of the 26 missing cards, if one hand has one, the other has one too, so we there are only 24 unknown cards not 25. And of course, if each hand has a trump, that leaves 12 non-trumps.

So the number of hands where each hand has one trump can be expressed as an equation...

@combin(2,1)*@combin(24,12) = 2 x 2,704,156 = 5,408,312

Likewise the number of hands where an opponent holds a void, or both the missing trumps can be expressed as

@combin(2,0)*@combin(24,13) = 1 x 2,496,144
@combin(2,2)*@combin(24,11) = 1 x 2,496,144

Note the second number is 13 (in 24.13) when they are void in your trumps, as they have 13 non-trumps, and 11 in the other case as when they hold two trumps, they have only 11 non-trumps.

If you add up the number of combinations here, you will see it totals back the original number of possible combination (10,400,600)

Now for some simpler math....
Suit     Cases/total possible
2,0     2,496,144 / 10,400,600 = 24%
1,1     5,408,312 / 10,400,600 = 52%
0,2     2,496,144 / 10,400,600 = 24%

So it turns out, that with two cards missing the odds favor the hook, by a little bit. This is why missing two trumps to the king, playing for 1-1 split is better odds than just banging the ACE (not a lot better, mind you.. just 4% better).

What affect does leading low and having the next hand play a low trump have on these odds? well, the 0,2 distribution is no longer possible, and half the 1,1 distributions are no longer possible. But as you can see, this still remains 52% for the 1,1 split.

Why is after cashing two top cards and both opponents follow twice low in our four-3 suit looking for the QUEEN that the odds slightly better than this 52%? (isnt the Qx missing the same as Kx in the earlier example)? The answer lies in the combinational math we used early. After the opponents followed to two rounds, the number of known cards are reduced from 26, to 22. So the combinational equations show above become...

@combin(2,0)*@combin(22,11) = 1 x 167,960 = 167,960
@combin(2,1)*@combin(22,10) = 2 x 184,756 = 369,512
@combin(2,2)*@combin(22,9) = 1 x 167,960 = 167,960

(22 because four cards are known, 11, 10, and 9 as the remaining cards after two known cards in the suit are played earlier, and if 0, 1 or 2 are held). Now do the math, total combinations here = 705,432 (add those three number above together)... Thus, the chance for a 3-3 split (illustrated as @combin(2,1) above since both opponents followed to the first round)... is 369,512/705,432 = 52.38

As for the 705,432... this is also the total all combinations after both opponents follow to two rounds of your suit... as @Combin(22,11)=705,432 possible distributions once each hand is know to have two cards in this suit (22 unknown cards, each opponent has 11 cards left).

Ben

[luke warm]

this reminds me of a riddle i once read... i only give it as preamble to my comments...

you're at a table with another person... he places a coin under one of three upside down cups... he asks you to choose which cup you think the coin is under (we'll assume you didn't *see* him place it heheh)

after you choose, he lifts one of the three cups and shows it to be empty underneath... there are now 2 cups, the one you chose and the other... he now gives you the choice between changing your pick to the last remaining cup or staying with your first pick... do you change?

the original odds were 2:1 against you choosing the correct cup... those odds did not change after you chose, simply because he showed you one of the two that was not correct... so the winning choice is to change cups... the odds were against you when you chose, therefore by changing you placed them with you

on the problem above, i think the original odds of clubs dividing 4/2 are greater than them dividing 3/3.. if so, the odds for any one card (the queen in this case) being with the 4 are 2:1... i don't see how these odds changed, all things being equal...

[inquiry]

luke warm, on Mar 10 2004, 08:03 PM, said:

this reminds me of a riddle i once read... i only give it as preamble to my comments...

you're at a table with another person... he places a coin under one of three upside down cups... he asks you to choose which cup you think the coin is under (we'll assume you didn't *see* him place it heheh)

after you choose, he lifts one of the three cups and shows it to be empty underneath... there are now 2 cups, the one you chose and the other... he now gives you the choice between changing your pick to the last remaining cup or staying with your first pick... do you change?

the original odds were 2:1 against you choosing the correct cup... those odds did not change after you chose, simply because he showed you one of the two that was not correct... so the winning choice is to change cups... the odds were against you when you chose, therefore by changing you placed them with you

on the problem above, i think the original odds of clubs dividing 4/2 are greater than them dividing 3/3.. if so, the odds for any one card (the queen in this case) being with the 4 are 2:1... i don't see how these odds changed, all things being equal...

I guess we are not going to be able to convince you Jimmy, but I will give it one more try.

But let's look at your cup game. Assume you play it 9 million times. 3 million times you will initially pick the cup with the coin under it, and your opponent will turn up either of the other two cups at random (since no coin under either of them).

In the other 6 million cases, your first choice is an empty cup. Now your opponent will SELECT to turn up the single cup without the coin under it, to keep the coin hidden.

Now, if you played this game these 9 million times, picking the "other cup" would win 6 million times and lose 3 million times. 2-to-1 in your favor. It isn't even 50/50, as you can see.

So imagine this little model as an example. The fact that you opponent here can freely choose between which cup to expose, and will never expose the coin, you go from 2.1 against with the first choice, to 2/1 for if you switch cups.

Ben

[Free]

Hehe, this reminds me of a paradox of probability:

You're in prison with 2 other guys. Tomorrow 2 of you three are going to be executed. How much chance of survival do you have?

Sollution 1: you're with 3 guys, only 1 will survive, so you have 33.33% chance of survival.

Sollution 2: you're with 3 guys, but at least one of the other will be executed. That leaves 2 guys left (one of the 2 is me), so I have 50% chance of survival.



Ofcourse, we all know that sollution 1 is the correct one...

[Cascade]

luke warm, on Mar 10 2004, 10:03 PM, said:

on the problem above, i think the original odds of clubs dividing 4/2 are greater than them dividing 3/3.. if so, the odds for any one card (the queen in this case) being with the 4 are 2:1... i don't see how these odds changed, all things being equal...

The original odds were approximately:

6-0 1.5%
5-1 14.5%
4-2 48.4% with a 2-1 chance that the Queen is with the long suit
3-3 35.5%

Lets assume the odds do not change.

This means that after everyone follows low twice the percentages are the same as they were at the beginning.

This is absurd as if everyone follows twice it is impossible for the suit to divide 6-0 or 5-1.

Therefore the odds must have changed.

It is also impossible for someone to have been dealt Qx after everyone has followed low twice. So the odds of the Queen being with the long suit if the suit turns out to be divided 4-2 have gone from 2-1 on to 100%.

We have discovered in the play that from the original combinations of cards that none of the 6-0 breaks are possible, none of the 5-1 breaks are possible, only some of the 4-2 breaks are possible and all of the 3-3 breaks are possible.

This changes the odds.

[Cascade]

Trpltrbl, on Mar 10 2004, 11:47 AM, said:

According to my calculations, there is a 68,9752% chance of getting 3 tricks. If you play the Ace and then a small towards the Jack, I think I will take my chances with that

Mike

The question was not about the best way to play this combination but about the odds of this suit dividing evenly after two rounds have seen everyone follow small.

[inquiry]

Jimmy,

I thought of a way to illustrate the point that Wayne was making using a simple card trick.

1) Get out 6 hearts, 5 spot cards and the QUEEN
2) Get out two clubs....

Shuffle these and deal two hands of 4 cards each....25 times

Now, pretend someone cashes AK of HEARTS and the hand with Q must not play it if they can avoid it. So... what we will do it NOT COUNT any hand in which the QUEEN is doubleton.....

But other than that, count each time both hands have exactly 3 hearts and one club, or two little hearts and two clubs.

The odds of clubs being split 1:1 or 2:0 are roughly equal. But what you will find as you throw out hands where the HEART QUEEN is doubleton, that the odds of the clubs being 1:1 will be twice the odds of the clubs being 2:1. It is the "positive nature" of the heart QUEEN that provides this infomation.

Ben

[luke warm]

inquiry, on Mar 11 2004, 03:34 AM, said:

So imagine this little model as an example. The fact that you opponent here can freely choose between which cup to expose, and will never expose the coin, you go from 2.1 against with the first choice, to 2/1 for if you switch cups.

Ben

that's right.. that's what i said (i think)... the point i was trying to make was, the original odds did not change... they were 2:1 against your first choice being correct, therefore changing the pick improved those odds

[Trpltrbl]

Cascade, on Mar 10 2004, 11:25 PM, said:

Trpltrbl, on Mar 10 2004, 11:47 AM, said:

According to my calculations, there is a 68,9752% chance of getting 3 tricks. If you play the Ace and then a small towards the Jack, I think I will take my chances with that 

Mike 

The question was not about the best way to play this combination but about the odds of this suit dividing evenly after two rounds have seen everyone follow small.

Then the question is just a waste of time.

Mike

[Cascade]

Trpltrbl, on Mar 11 2004, 09:31 PM, said:

Cascade, on Mar 10 2004, 11:25 PM, said:

Trpltrbl, on Mar 10 2004, 11:47 AM, said:

According to my calculations, there is a 68,9752% chance of getting 3 tricks. If you play the Ace and then a small towards the Jack, I think I will take my chances with that 

Mike 

The question was not about the best way to play this combination but about the odds of this suit dividing evenly after two rounds have seen everyone follow small.

Then the question is just a waste of time.

Mike

I do not think so.

There are often considerations that require a suit to be played other than to maximize the number of tricks taken in that suit. For example you may wish to minimize the chance of one particular defender getting on lead.

As well as that there is the obvious general benefit of understanding how odds change as the play progresses.

Wayne

[mpefritz]

I think Ben's original problem is a good one in getting newer players to think about how odds change as more information is available. Perhaps, as this is a Beginner/Intermediate discussion, an aside of how best to play the combination would have been helpful. But sometimes you are combining chances and the play of the suit as Ben played it would not be unreasonable.

In particular, in the AKxx Jxx problem once you cash the AK and the Q doesn't fall, you know the Q was either Qxx or Qxxx. This changes the possible splits and increases the chances of a 3-3 split compared to 4-2.

Let me digress to the restricted choice questions posed recently. They come in many forms:

3 cups, one with a coin underneath

3 prisoners, 2 to be shot

3 curtains, 1 car and 2 goats (The "Let's Make a Deal" problem examined in excruciating detail in American Mathematical Monthly several years back.)

AKJxxx xxx cash A and Q falls behind A. Do you hook the next round?

Let's look at the cup example and why the odds do or do not change:

In the 3 cup problem there are 3 cups called 1,2,3 and one of them has a coin under it. Once you choose a cup that you think has the coin, the other player shows you that one of the other cups is empty. He then offers you a chance to change to the remaining cup. Should you change?

Say you choose cup 1, and the other player shows you cup 2 is empty. Should you now guess that the coin is under cup 3? YES!

Why? Say the other player has a strategy of always showing you the lowest number cup that is empty. When he shows you cup 2, you have not ruled out cup 3 as being empty. The odds are now 50-50.

Now suppose he has a strategy of always showing the highest number empty cup. Because he showed you number 2, you are 100% sure that the coin is under Cup 3.

Since his strategy is somewhere in between, you should always switch cups. If his startegy is to randomly pick between 2 + 3 when they are both empty, then the odds work out to be 1/3 that the coin is under cup 1, as you would expect because the other player has given you no useful information.

What happens if YOU get to pick a cup to look under (after originally choosing 1)? If you turn over cup 2 and it is empty, the odds are now 50-50 that the coin is under cup 1. You have taken a risk of finding the coin, and when you take that risk, and it turns out in your favor, the odds change.

The prisoner problem is usually stated as:

3 prisoners are in jail. 2 will be shot in the morning. Prisoner A bribes the guard with his last pack of cigarettes to tell him which of the other 2 prisoners is going to be shot. When the guard tells him prisoner C will be shot, does this change prisoner As chances of being shot? The answer is no (nothing risked), BUT if prisoner B were secretly listening to this exchange, his chance would now be decreased (he took the risk of hearing that he would be shot).

So how does this apply to Ben's original problem?

The declarer took a chance cashing the AK. If the Q showed up on round 2, he'd be sure (less misclicks or "brilliant" unblocks) that the cards were originally divided 4-2. Zero chance of 3-3. He took a chance of finding the cards 4-2.

Since the Q did not show up, the chance for 3-3 is now increased. Thus his concept of a "positive" card.

fritz

p.s. Yes, i can calculate the numbers correctly and should include:
1) a priori chance of a given combination
2) the chance the cards would have been played through 2 rounds in such a combination -- this is key when computing the real numbers, but is washed out in this case as each would have the same factor after 2 rounds when the Q does not show up.

[inquiry]

mpefritz, on Mar 11 2004, 10:12 PM, said:

I think Ben's original problem is a good one in getting newer players to think about how odds change as more information is available.

Thanks, that was the point exactly....

BTW, how close are you to fully agreeing with my solution now?