A maths problem

[SimonFa]

I thought I had a reasonable mathematical brain - maybe its age catching up.

 

I've been trying to understand distribution probabilities and have hit a brick wall and was hoping someone could help as it's doing my head in!

 

On page 67 of the Rodwell Files we get a great tip for working out the number of "cases" there for a missing number of cards - straight forward 2^n. This is nice for anyone who has done even a little bit of computing. So 6 missing cards gives of 64 cases.

 

In his book, Expert Bridge Simplified(pp 24), Jeff Rubens explains how combinations work c=(n,r) = n! / (r! * (n-r)!) and even gives a nice shorthand way of calculating them at the table. I get that and find it very useful. But when I put the two tips together it doesn't appear to work.

 

When I take six cards missing and look for a 4,2 split I get 64 cases and c = 15 ( checked this in excel and here, I always double check). These means that the probability of a 4-2 split is 15/64 = 23% for a specific hand to have 4 cards so 46% for either hand to have 4 cards ie 4-2 split, doesn't it?

 

When I check this with the table Jeff provides on page 35 he gives the answer as 48%, but with some rounding errors so I'm not losing any sleep at this point. But then I notice that for 5,1 split he says 15% and I make it 18%, now I'm not so happy. But the worst case is 3,3 split. By my calculations there are 20 combinations which is 31.25% but Jeff's table says 36%. That's a big difference and one I wouldn't want to rely on.

 

At my standard its probably academic as I only tend to look at the bigger picture but the more I play the more I start looking at different lines of play and thinking about probabilities, even if I don't go down to the decimal points.

 

So what am I missing?

 

As always, thanks in advance.

 

Simon

[Cascade]

I thought I had a reasonable mathematical brain - maybe its age catching up.

 

I've been trying to understand distribution probabilities and have hit a brick wall and was hoping someone could help as it's doing my head in!

 

On page 67 of the Rodwell Files we get a great tip for working out the number of "cases" there for a missing number of cards - straight forward 2^n. This is nice for anyone who has done even a little bit of computing. So 6 missing cards gives of 64 cases.

 

In his book, Expert Bridge Simplified(pp 24), Jeff Rubens explains how combinations work c=(n,r) = n! / (r! * (n-r)!) and even gives a nice shorthand way of calculating them at the table. I get that and find it very useful. But when I put the two tips together it doesn't appear to work.

 

When I take six cards missing and look for a 4,2 split I get 64 cases and c = 15 ( checked this in excel and here, I always double check). These means that the probability of a 4-2 split is 15/64 = 23% for a specific hand to have 4 cards so 46% for either hand to have 4 cards ie 4-2 split, doesn't it?

 

When I check this with the table Jeff provides on page 35 he gives the answer as 48%, but with some rounding errors so I'm not losing any sleep at this point. But then I notice that for 5,1 split he says 15% and I make it 18%, now I'm not so happy. But the worst case is 3,3 split. By my calculations there are 20 combinations which is 31.25% but Jeff's table says 36%. That's a big difference and one I wouldn't want to rely on.

 

At my standard its probably academic as I only tend to look at the bigger picture but the more I play the more I start looking at different lines of play and thinking about probabilities, even if I don't go down to the decimal points.

 

So what am I missing?

 

As always, thanks in advance.

 

Simon

 

The cases are not all equally likely.

 

A specific 3=3 split is more likely than a specific 4=2 split which in turn is more likely than a 5=1 split and again that is more likely than a 6=0 split.

 

Without any additional information the probabilities can be calculated as follows:

 

P(3=3 split) = (6C3) * (20C10) / (26C13) = 0.355 (3sf)

 

That is the combinations are calculated from choosing three of the six outstanding cards in the suit of interest and 10 of the 20 cards in the other three suits to make up the hand of 13 cards. This is divided by the total number of ways of choosing any 13 cards from the 26 cards that the opponents hold.

 

Similarly

 

P(4=2) split = (6C4) * (20C9) / (26C13) = 0.242 (double this to 0.484 to get the probability of a 4=2 split or a 2=4 split)

 

P(5=1) split = (6C5) * (20C8) / (26C13) = 0.073 (0.145)

 

P(6=0) split = (6C6) * (20C7) / (26C13) = 0.007 (0.015)

[S2000magic]

Cascade covered it quite well.

 

To simplify it a bit: when you're doing the calculation with combinations for the given suit, you're ignoring the remaining cards in the opponents' hands - treating all possible divisions as if they are equally likely. Jeff's calculations consider the remaining cards in the opponents' hands, so they explicitly include the fact that a (given) 4-2 division of six cards is more likely than a (given) 3-3 division of those same six cards.

[kenberg]

Possibly one further thought is useful Think of the simplest case where you have 11 cards between yourself and dummy. Say you are missing the 2 and the 3.There are four possibilities. Both on your left, both on your right,deuce alone on the left, deuce alone on the right. But books rate the 1-1 split as better than 50-50. That's because if they split 2-0 then the bridge god has to place 11 other cards on one side, the rest on the other side. There is a slightly larger number of ways to choose 12 cards to place on one side, the rest on the other.

 

In fact, the odds for 1-1 split are usually given as 52%. The fact that 52 is (13/12) times 48 is not a coincidence here.

 

The general idea is that if you calculate the odds simply be figuring the number of ways to place the missing cards in the suit of interest, you will be underestimating by a bit, by a couple of percentage points or more, the chances of an even (as possible) split. Thus, if you are missing 5 cards, there are 10 ways to place exactly two cards on the left, ten ways to place exactly two cards on the right, so at first glance the odds seem to be 20/32 =5/8 =0.625 for a 3-2 split. http://www.bridgehan...istribution.htm puts it at 67.8% I have not checked it, but I have no reason to doubt it.

[Statto]

I'll just add a buzzword: vacant spaces.

 

And a link to a useful web-based tool with some good background info: http://www.automaton.../en/OddsTbl.htm

 

Others have covered it well.

[paua]

I thought I had a reasonable mathematical brain - maybe its age catching up.

 

I've been trying to understand distribution probabilities and have hit a brick wall and was hoping someone could help as it's doing my head in!

 

On page 67 of the Rodwell Files we get a great tip for working out the number of "cases" there for a missing number of cards - straight forward 2^n. This is nice for anyone who has done even a little bit of computing. So 6 missing cards gives of 64 cases.

 

In his book, Expert Bridge Simplified(pp 24), Jeff Rubens explains how combinations work c=(n,r) = n! / (r! * (n-r)!) and even gives a nice shorthand way of calculating them at the table. I get that and find it very useful. But when I put the two tips together it doesn't appear to work.

 

When I take six cards missing and look for a 4,2 split I get 64 cases and c = 15 ( checked this in excel and here, I always double check). These means that the probability of a 4-2 split is 15/64 = 23% for a specific hand to have 4 cards so 46% for either hand to have 4 cards ie 4-2 split, doesn't it?

 

When I check this with the table Jeff provides on page 35 he gives the answer as 48%, but with some rounding errors so I'm not losing any sleep at this point. But then I notice that for 5,1 split he says 15% and I make it 18%, now I'm not so happy. But the worst case is 3,3 split. By my calculations there are 20 combinations which is 31.25% but Jeff's table says 36%. That's a big difference and one I wouldn't want to rely on.

 

At my standard its probably academic as I only tend to look at the bigger picture but the more I play the more I start looking at different lines of play and thinking about probabilities, even if I don't go down to the decimal points.

 

So what am I missing?

 

As always, thanks in advance.

 

Simon

 

Another interesting website :

http://www.durangobill.com/BrSplitHowTo.html

[paua]

Possibly one further thought is useful Think of the simplest case where you have 11 cards between yourself and dummy. Say you are missing the 2 and the 3.There are four possibilities. Both on your left, both on your right,deuce alone on the left, deuce alone on the right. But books rate the 1-1 split as better than 50-50. That's because if they split 2-0 then the bridge god has to place 11 other cards on one side, the rest on the other side. There is a slightly larger number of ways to choose 12 cards to place on one side, the rest on the other.

 

In fact, the odds for 1-1 split are usually given as 52%. The fact that 52 is (13/12) times 48 is not a coincidence here.

 

The general idea is that if you calculate the odds simply be figuring the number of ways to place the missing cards in the suit of interest, you will be underestimating by a bit, by a couple of percentage points or more, the chances of an even (as possible) split. Thus, if you are missing 5 cards, there are 10 ways to place exactly two cards on the left, ten ways to place exactly two cards on the right, so at first glance the odds seem to be 20/32 =5/8 =0.625 for a 3-2 split. http://www.bridgehan...istribution.htm puts it at 67.8% I have not checked it, but I have no reason to doubt it.

 

 

Or look at how four cards split. There are 16 possible combinations, so each should be 6.25%, but in fact the two 4-0 splits are only 4.783%. Each 3-1 split is 6.217%, and each 2-2 split is 6.783%.

The simple at-the-table method over-estimates the extreme splits, and under-estimates the even splits.

[Statto]

Or consider the case where you're playing in 4♠ after LHO opened 3♦ then led a ♥. You have 7 trumps in hand and a void in dummy. There are various possible splits for trumps, but one of the 6-0 splits is virtually impossible. There's obviously a bias towards RHO having more trumps, but I hope this illustrates that the more extreme splits have less 'space' to be plausible.

[SimonFa]

Thanks all. I was hoping I'd found a quick and simple method or approximation, but obviously not. Looks like a lot more reading and, more importantly, practice.

 

Simon

[phil_20686]

Thanks all. I was hoping I'd found a quick and simple method or approximation, but obviously not. Looks like a lot more reading and, more importantly, practice.

 

Simon

 

I think you misunderstood the rodwell tip. 2^{n} refers to the number of cases. If there are 6 cards in the suit of interest, and at least 6 cards in each hand, then clearly there are 2^{6} cases, as each card is dealt to either one hand or the other. So 6C0+6C1+6C2+6C3....6C6 = 2^{6}.

 

However, each case is not equally likely, but normally they are close enough that just counting and dividing by 2^{n} is reasonably close. Particularly if you can ignore the 6-0 and 5-1 breaks if you cant pick them up anyway.

 

 

so say for a 3-2 break, rodwells there are 20 3-2 breaks out of 32 = 62.5. If all cases are equally likely, but 68% if you include the likelyhood that 3-2 breaks are better. This means that case counting is a pretty effective way to evaluate the evaluation of certain lines, and is really pretty easy to do at the table.

[jogs]

Thanks all. I was hoping I'd found a quick and simple method or approximation, but obviously not. Looks like a lot more reading and, more importantly, practice.

 

Simon

 

You confused. 31.25 is probability two equally matched teams will each win 3 of the first 6 games in a 7 games series.

Don't bother calculating the probabilities. There are tables published. Just memorize them.