[Michael000]

I've been trying to work out the statistical probability of a card dropping to the play of 'x' number of higher cards in an attempt to identify whether it is better to play or finesse. I note that the BBO site has a 'hand analysis' feature, can someone explain to me how I interpret the information in this feature?

[mgoetze]

You click + on all the distributions your line wins against and it adds up the odds for you.

[Michael000]

mgoetze, on 2015-January-21, 09:41, said:

You click + on all the distributions your line wins against and it adds up the odds for you.

Yes I've done that but, unless I am missing something, it simply seems to list the percentage chance of the opposition cards being split.

[Michael000]

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For example, with this hand, what is the statistical percentage probability chance of the Q dropping to the play of the A K?

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And, with this hand what is the statistical percentage probability chance of the Q dropping to the play of the A K?

[mgoetze]

Michael000, on 2015-January-22, 07:39, said:

Yes I've done that but, unless I am missing something, it simply seems to list the percentage chance of the opposition cards being split.

So you never thought to try clicking on the "More" button?

[mgoetze]

Michael000, on 2015-January-22, 07:47, said:

For example, with this hand, what is the statistical percentage probability chance of the Q dropping to the play of the A K?

[Michael000]

mgoetze, on 2015-January-22, 08:20, said:

So you never thought to try clicking on the "More" button?

I did find the 'More' button and I did recover information similar to that which you posted but my question is - does the feature have the answer to my question and if so how do I interpret or extract that answer.

The example I gave which you have re-produced with the analysis suggests that there is a 41% chance of a 2/2 split and a 25% chance of a 3/1 split and of course there is a 25% chance of the Q being a singleton in that 4/1 split . . . if my maths are correct (please do correct me) that means (or does it) that, with this hand, there is a 65% chance of the Q dropping on a straight play of the A K.

Can I know how you created/input into the analysis feature, the example hand I gave? Is there an 'Even More' button?

[barmar]

Click in the + next to the distributions that matter. At the bottom of the table it shows the total of those percentages.

In this case, the he highlighted Q singleton in West, all 2-2 distributions, and Q singleton in East. They add up to 53.13%.

[Michael000]

barmar, on 2015-January-22, 15:34, said:

Click in the + next to the distributions that matter. At the bottom of the table it shows the total of those percentages.

In this case, the he highlighted Q singleton in West, all 2-2 distributions, and Q singleton in East. They add up to 53.13%.

Thank you . . . .

If anyone can help me with the maths that gets it to 53.13% I'd appreciate it (I'm not doubting it, I'd just like to see where I went wrong)

Edit . . . I'll further explore the analysis feature, the answer may be there

Thanks again

[Mbodell]

Michael000, on 2015-January-22, 16:02, said:

Thank you . . . .

If anyone can help me with the maths that gets it to 53.13% I'd appreciate it (I'm not doubting it, I'd just like to see where I went wrong)

Edit . . . I'll further explore the analysis feature, the answer may be there

Thanks again

You asked what is the chance playing the A then the K. To know you have to count cases where this will drop the Q. That's what mgoetze's answer did. The cases where the Q drop get the plus highlighted in yellow. That is the first 1=3 split with the stiff Q which is 6.22%, all of the 2-2 splits (which didn't need to get expanded) which is another 40.70%, and then the last 3=1 split with the stiff Q which is another 6.22%

If you add up 6.22 + 40.70 + 6.22 you'll get 53.14%. That is the 53.13% answer you see at the bottom. The least significant digit is off by one because the percentages given for each case are rounded to 2 decimal places but are not an exact number. Therefore the more accurate values are used in the calculation which results in the more accurate total number, when rounded to 2 decimal places, being 53.13% instead of 53.14%. Not sure if that 0.01% was confusing you.

Note also that sometimes in the interest of space the system combines cases that don't matter. For instance it might treat Q2 73 and Q3 72 as the same in the + and list it as 2 cases, instead of 1. The percentages would still work. Basically if any strategy of plays would be the same it might get combined.

You can also use the same tool to see the percentage for cash a top spade and then finesse repeatedly for AKJT9 opposite 7532. There you win one 1=3 (stiff Q onside) for 6.22% plus half the 2-2 splits (the half with the Q onside) for 20.35% plus 24.87% for all the 3=1 splits (you cashed a high one first to pick up stiff Q off, and finesse for all the Q on), and 4.78% for the 4=0 onside. If you add those all up you get 56.22% overall. That's your chance of success at the start. Once you cash the top honor, if you see 2 small cards from the other players you know there are no 4-0 splits and no stiff Q and then the number of cases is smaller and the percentages change.

[Michael000]

Mbodell, on 2015-January-22, 23:12, said:

You can also use the same tool to see the percentage for cash a top spade and then finesse repeatedly for AKJT9 opposite 7532. There you win one 1=3 (stiff Q onside) for 6.22% plus half the 2-2 splits (the half with the Q onside) for 20.35% plus 24.87% for all the 3=1 splits (you cashed a high one first to pick up stiff Q off, and finesse for all the Q on), and 4.78% for the 4=0 onside. If you add those all up you get 56.22% overall. That's your chance of success at the start. Once you cash the top honor, if you see 2 small cards from the other players you know there are no 4-0 splits and no stiff Q and then the number of cases is smaller and the percentages change.

Thanks for the helpful reply

[Michael000]

Getting back to the first scenario with the Q 8 6 4 of spades with the opposition is it not the case that all possible splits are as follows:


. . . .WEST . . . . .EAST . .
. . . .Q x x x . . . . . . .
. . . .Q x x . . . . . .x
. . . .Q x . . . . . . .x x
. . . .Q . . . . . . . . .x x x
. . . . . . . . . . . . . .Q x x x
. . . .x . . . . . . . . Q x x
. . . .x x . . . . . . .Q x
. . . .x x x . . . . . .Q

And if that is the case, two have a 2-2 split and two have a singleton Q split . . . so on half of all possible splits the Q falls to an A K play for the probability of it falling is 50%?

[barmar]

You're forgetting that in the Qx opposite xx rows, there are 3 possible cases. Qx could be Q8, Q6, or Q4. That's why the 2-2 cases add up to 40% of the total.

[Mbodell]

barmar, on 2015-January-23, 16:33, said:

You're forgetting that in the Qx opposite xx rows, there are 3 possible cases. Qx could be Q8, Q6, or Q4. That's why the 2-2 cases add up to 40% of the total.

Right, there are:

1 ways to do 4=0
3 ways to do 3=1 with Qxx=x
3 ways to do 2=2 with Qx=xx
1 ways to do 1=3 with Q=xxx
1 ways to do 0=4
3 ways to do 1=3 with x=Qxx
3 ways to do 2=2 with xx=Qx
1 ways to do 3=1 with xxx=Q

Moreover, each of these ways are not equally likely. You get a hint of that from the pictures above where each of the 3-1 cases are 6.22% while each of the 4-0 cases are 4.78% and each of the 2-2 cases are about 6.78% (40.70% for all 6, 6.78% for each 1). In general, the more equal a split, the more likely it is.

That is, each specific 2-2 split (Q6 opposite 84) is more likely than each given 3-1 split (Q64 opposite 8). The suit is more likely to split 3-1 in either direction if you combine all the cases. But it is more likely to split 2=2 than 3=1. And also more likely to split 2=2 than 1=3.

[Michael000]

Mbodell, on 2015-January-24, 03:54, said:

ight, there are:

1 ways to do 4=0
3 ways to do 3=1 with Qxx=x
3 ways to do 2=2 with Qx=xx
1 ways to do 1=3 with Q=xxx
1 ways to do 0=4
3 ways to do 1=3 with x=Qxx
3 ways to do 2=2 with xx=Qx
1 ways to do 3=1 with xxx=Q

Moreover, each of these ways are not equally likely. You get a hint of that from the pictures above where each of the 3-1 cases are 6.22% while each of the 4-0 cases are 4.78% and each of the 2-2 cases are about 6.78% (40.70% for all 6, 6.78% for each 1). In general, the more equal a split, the more likely it is.

That is, each specific 2-2 split (Q6 opposite 84) is more likely than each given 3-1 split (Q64 opposite 8). The suit is more likely to split 3-1 in either direction if you combine all the cases. But it is more likely to split 2=2 than 3=1. And also more likely to split 2=2 than 1=3.

I have now found this feature and spent some hours playing with it and it is very nice. . . .my problem is (and I recognise that it may be just my problem) I can't see why any one possible split would be more or less likely than any other. For me that breaks all the rules of probability. If someone could explain that to me so that I can move on I'd be grateful?

[PhilKing]

Michael000, on 2015-January-25, 07:50, said:

I have now found this feature and spent some hours playing with it and it is very nice. . . .my problem is (and I recognise that it may be just my problem) I can't see why any one possible split would be more or less likely than any other. For me that breaks all the rules of probability. If someone could explain that to me so that I can move on I'd be grateful?

It's because each player is dealt thirteen cards. You can try this imaginary experiment as "proof". Construct the North South hands so that you are in a grand slam missing two cards in spades - the king and the two. Now deal West the ♠2.

Shuffle the remaing 25 cards and give 12 to West and 13 to East. Now you can see that the odds of East having the missing king are 13 to 12 (52% to 48%), so this is why each equal division is slightly more likely than each unequal division. So in mathematical terms, you now know why it is better (barring any table feel or other inference) to play for the drop in this situation, despite there being one case whee it succeeds and one where it fails.

[Michael000]

PhilKing, on 2015-January-25, 08:43, said:

It's because each player is dealt thirteen cards. You can try this imaginary experiment as "proof". Construct the North South hands so that you are in a grand slam missing two cards in spades - the king and the two. Now deal West the ♠2.

Shuffle the remaing 25 cards and give 12 to West and 13 to East. Now you can see that the odds of East having the missing king are 13 to 12 (52% to 48%), so this is why each equal division is slightly more likely than each unequal division. So in mathematical terms, you now know why it is better (barring any table feel or other inference) to play for the drop in this situation, despite there being one case whee it succeeds and one where it fails.

Thanks for that . . I can see what you say, I will use it to try and get results matching the Hand Analysis feature for 3, 4 and 5 missing cards.

[barmar]

Where's getting outside the BBO-specific topic and slipping into general bridge principles.

There are a number of bridge books that deal with probabilities of different distributions. A somewhat advanced one (whose title suggests otherwise) is "Expert Bridge Simplified". Another is "Bridge, Probability, & Information".