Holding no cards

[Cyberyeti]

Have just played a session that was pretty striking for us not holding any cards:

 

26 boards, I held 243 points, partner held 232, very low but not ridiculous, the striking thing was we only held more than 22 points between us 3 times (and 25 or more only once), a 29, a 24 and a 23.

 

How do you go about doing the calculation as to how rare this is ?

[HighLow21]

Have just played a session that was pretty striking for us not holding any cards:

 

26 boards, I held 243 points, partner held 232, very low but not ridiculous, the striking thing was we only held more than 22 points between us 3 times (and 25 or more only once), a 29, a 24 and a 23.

 

How do you go about doing the calculation as to how rare this is ?

 

It's a binomial problem. 16 cards contain points. What is the odds that your side will get each card? You have 26 cards out of 52. After the first point-containing card, place the second one. Assuming you get the first card, you have 25 slots left for 51 cards remaining, so 25/51.

 

Let's start with Aces. The probability that your side has all 4 Aces, then, is 26/52 * 25/51 * 24/50 * 23/49. 3 Aces would be as follows: there are 4 different ways to get 3 of the 4 aces, so 4 * 26/52 * 25/51 * 24/50 * 26/59. (Why is the last numerator 26 rather than 23? It is because we are assuming the opponents get the 4th ace, not your side). Similarly, there are 6 different ways you can get 2 of the 4 Aces, and it's calculated in the same way.

 

It gets more complicated when you start adding Kings, Queens and Jacks. But overall, there are a certain number of ways the cards can be distributed among the 4 hands, and a certain probability for each distribution of cards. You'd want to calculate the probability of each and then add up all the different scenarios that result in your partnership having each point count total:

 

0 points: only one way to do this. The opponents get all 16 point-containing cards.

1 points: only 4 ways to do this. You or your partner have 1 jack and that's it.

2 points: already there are 10 different ways to do this. 4 ways for your partnership to have exactly one Queen, and there are 6 different sets of 2 jacks your partnership could have.

3 points: 4 ways to have 3 jacks; 16 ways to have one jack and one queen; and 4 ways to have one king = 24.

 

As you can see, this math gets extremely complicated very quickly. I'm sure it has been done, and I just spent a minute browsing Richard Pavlicek's site to find out if he's published this (I'd be surprised if he hasn't), but there's the math behind it. This is how you would calculate the probability of any number of HCP, 0-40, for the partnership on a given deal.

 

(Note: here is the probability chart for a given PLAYER on a single deal, but the probabilities are much different from those of a partnership: http://www.bridgehands.com/P/Probability_HCP.htm)

 

Let's just say, hypothetically, that this table tells you you have a 30% chance of getting 23 HCP or more (this shouldn't be far off). You can calculate the probability of getting fewer than a certain number of 23-HCP hands as follows:

 

Take the probability of getting the good (23-HCP) hands upfront (30% apiece), and then the "bad" (22 or lower) hands after that (70% apiece). Multiply the probabilities together.

==> For example, getting 3 good hands out of 26, we would calculate 30% * 30% * 30% * 70% * 70% * 70% * 70% * .... = 0.000738%.

 

Call this statistic "Initial Probability."

 

This is the odds of getting exactly the following sequence: GOOD then GOOD then GOOD then 23 BAD hands. But we don't care what ORDER we get the good vs. bad hands in; we just want to compute how likely it is to get exactly 3 GOOD hands, and 23 BAD hands, in any possible sequence.

 

Since each sequence is equally likely, all we have to do is take this Initial Probability, and multiply it by the NUMBER OF DIFFERENT ORDERS of possible combinations of 3 Good and 23 Bad hands.

 

I'll spare you the details on this, but there is a factorial formula called "combinatorics" that will tell you the answer: the number of possible ways to order x objects, where x can be of the value "a" or "b" only, is: (x!) / (a!*(x-a)!) where:

 

a = the number of "good" hands

b = the number of total hands

x! = the product of all integers from x to 1.

 

Thus the right combinatoric coefficient for 3 good hands and 23 bad hands is: (26!)/(3! * (26-3)!) = 26!/(3! * 23!) = [(26 * 25 * 24)/(3 * 2 * 1)] * (23!)/(23!) = 26 * 25 * 4 = 2,600.

 

And the probability of getting exactly 3 good hands and 23 bad hands out of 26, assuming good hands occur 30% of the time, is 2,600 * 0.000738% = 1.92%.

 

Now, we need to compute this for other outcomes BELOW 3 good hands. In other words, we want to know, "How frequently will we get 3 OR FEWER good hands?" Thus we need to add the probability of getting 0 good hands, 1 good hand, and 2 good hands out of 26. This will tell you how rare your outcome was.

 

In 26 boards, you would expect to get, on average, 30% * 26 = 8.4 of these deals (i.e., 23+ HCP hands).

 

The odds of getting 0 hands that are 23 HCP or more is 1*(1 - 30%)^26 = 0.00094%

The odds of getting 1 hand that is 23 HCP or more is 26*(1 - 30%)^25*(30%)^1 = 0.10% (26 is the number of ways you can be dealt 1 good hand and 25 bad ones)

The odds of getting 2 hands that are 23 HCP or more is 325*(1 - 30%)^24*(30%)^2 = 0.56% (325 is the number of ways you can be dealt 2 good hands and 24 bad ones)

The odds of getting 3 hands that are 23 HCP or more is 2600*(1 - 30%)^23*(30%)^3 = 1.92% (2600 is the number of ways you can be dealt 3 good hands and 23 bad ones)

 

The sum of all these is 2.60%. In other words, IF we assume that 23-HCP hands occur 30% of the time, then on a board of 26 hands, we will get 3 or fewer of these good hands 2.6% of the time (or about 1 time in 40). That's fairly rare, but nothing egregious.

 

However, that 30% was an assumption on my part. The correct answer for how frequently you will get dealt 23+ HCP hands is likely to be different, and the right answer will depend on the EXACT probability of getting dealt 23 HCP or more.

 

I am virtually certain that the correct answer will be between 30% and 40% of the time, so I have provided a table for you, below, to tell you how rare your event was:

 

23-HCP Prob Your Event 1 chance in

30% 2.60% 39

31% 2.03% 49

32% 1.57% 64

33% 1.21% 82

34% 0.93% 108

35% 0.70% 142

36% 0.53% 188

37% 0.40% 252

38% 0.30% 339

39% 0.22% 460

40% 0.16% 628

 

As you can see, as the probability of getting dealt a 23-HCP hand varies from 30% up to 40%, the odds that you will get only 3 23+ HCP hands in your partnership (or fewer) starts to plummet. Assuming the probability is 40%, then you will be dealt 3 or fewer 23+ HCP hands (out of 26) only once in 628 boards -- i.e., once in a blue moon.

 

Any feedback from anyone out there would be most appreciated.

[Cyberyeti]

Thanks for that, basically it confirmed what I thought (I studied a fair amount of statistics at degree level) that it's not simple.

 

Was wondering if you could approach it slightly differently. Calculate the probability of the number of honours in the combined hands.

 

23+ points can be made up:

 

0-6 honours: no chance

12+ honours: 100% chance

 

7-11 honours you then have to calculate the combinations of honours that add to 23

[Siegmund]

30% was a very good guess.

 

150143814509712 times out of 495918532948104 =30.276%.

 

The latter number is 52C26, the number of ways your partnership can be dealt 26 cards, without caring which hand they are in.

Yes, I cheated and used Mathematica. I would hate to add up the probability of the 238 possible cases by hand:

 

ways[a_, b_, c_, d_] := Binomial[4, a] Binomial[4, b] Binomial[4, c] Binomial[4, d] Binomial[ 36, 26 - a - b - c - d];

points[a_, b_, c_, d_] := 4 a + 3 b + 2 c + d;

total = 0; good = 0;

Do[{If[points[a, b, c, d] >= 23, good = good + ways[a, b, c, d]];

total = total + ways[a, b, c, d]}, {a, 0, 4}, {b, 0, 4}, {c, 0,

4}, {d, 0, 4}];

 

It comes out to a 2.43% chance (~1 in 41) of seeing 3 or fewer hands in a 26-board session.

 

Great reason to go back to 32-board sessions, right?

[Cyberyeti]

If that 30% figure is accurate, then 2.4% is about what I'd expect (mean = 26x0.3 = 7.8, variance = 26x0.3x0.7=5.46, standard deviation = sqrt(variance) = 2.34) so a little over 2 standard deviations below the mean.

[Fluffy]

I averaged 8.27 with my partner having 8.83 yesterday. Was kind boring, I had a total yarborough and 4 hands where I made no trick or bid.

[Vampyr]

"Whingeing about poor cards" threads have been appearing a lot lately. Get over it everyone -- it's not rubber bridge!

[GreenMan]

I've told this story before: A friend described a night where he averaged about 3 HCP on his first several hands, and decided to keep track of his average holding for the whole session. He averaged about 5 HCP for the night, even with a 2♣ opener and a 2NT opener in the mix. He also won the session easily, and credited it to paying attention to the bad hands instead of losing interest.

[Cyberyeti]

"Whingeing about poor cards" threads have been appearing a lot lately. Get over it everyone -- it's not rubber bridge!

 

Yeah but it's really dull when you have no decisions to make and you're completely at the mercy of what opps do as to how you score. Particularly in a random club field where the normal number of tricks for declarer is worth 80%+ most of the time.

 

Example from Friday: Axxx, Jx, x, AKJ10xx opposite void, KQxx, AKQ10xx, Q9x, diamonds are 3-3, 6♣ is cold, 3N has 13 tricks on a non heart lead, a spade was led at every table, only 2 declarers made 13 tricks in 3N, one 5♣+1, one 2N+5 so we got very little for normal actions, even finding A♥ would be below average.

[blackshoe]

He also won the session easily, and credited it to paying attention to the bad hands instead of losing interest.

And that right there is the key, isn't it? :D

[Cyberyeti]

And that right there is the key, isn't it? :D

 

Defending really well also helps, recalls time at university where 2 junior world champions scored 57% with their revolutionary "pass throughout" system.

[Cthulhu D]

The hand records we get after the game at my club include HCP, 7+ card suits, singletons and voids for each of N/S/E/W so you can usually confirm whether it was bias or the fact that you where averaging below 8.5 HCP a hand with no voids that made that session boring. I like sessions where I have low HCP but very shapely hands, I tend to do better.

[PhilKing]

I remember a hand from the YC marathon where my partner held a yarborough with ♦832. We were defending a slam and his first discard was the ♦8 (reverse) to "clarify the position" and I was now squeezed for the contract. This cost us first place in a structure where first gets £1500 and second £300. :(

 

Of course, this was not the only critical hand, but I do like to mention this one occasionally. B-)

[billw55]

I'm no statistics whiz, but just subjectively the event described in the OP seems ordinary to the point that it isn't even worth mentioning. You held 243 points in 26 boards, that is 9.34 points per board. Your partner's 232 comes to 8.92 per board, and the combined 475 is 18.27 per board. These numbers are not nearly far enough from the expectation (10 points for one hand, 20 for both) to get me interested. At the table I probably wouldn't even notice. What were you expecting?

 

OK, maybe you were a little unlucky on game deals your way. Still, it's not like you got a mountain of yarbs. And if you have almost your share of points, but few games, then probably not many games went your opponents' way either. So it was a partscore session, it happens. 26 boards is very small sample. I would bet that in every such session, you can find a statistical oddity if you look for it. I didn't get my expected share of nines tonight!! Call the forum stat police!!

 

Cyberyeti, I know you didn't really mean it that way. I just got carried away writing ..;)

[Cyberyeti]

The hand records we get after the game at my club include HCP, 7+ card suits, singletons and voids for each of N/S/E/W so you can usually confirm whether it was bias or the fact that you where averaging below 8.5 HCP a hand with no voids that made that session boring. I like sessions where I have low HCP but very shapely hands, I tend to do better.

 

Only in Mitchell movements, we were moving around NS/EW

 

And I've very rarely seen averages outside 9-11 for the HCP by seat and 19-21 by partnership, sure enough this set by seat/partnership fell within those averages.

[Cyberyeti]

I'm no statistics whiz, but just subjectively the event described in the OP seems ordinary to the point that it isn't even worth mentioning. You held 243 points in 26 boards, that is 9.34 points per board. Your partner's 232 comes to 8.92 per board, and the combined 475 is 18.27 per board. These numbers are not nearly far enough from the expectation (10 points for one hand, 20 for both) to get me interested. At the table I probably wouldn't even notice. What were you expecting?

 

OK, maybe you were a little unlucky on game deals your way. Still, it's not like you got a mountain of yarbs. And if you have almost your share of points, but few games, then probably not many games went your opponents' way either. So it was a partscore session, it happens. 26 boards is very small sample. I would bet that in every such session, you can find a statistical oddity if you look for it. I didn't get my expected share of nines tonight!! Call the forum stat police!! ;)

It's statistically significant (the 5% level is often used by statisticians and 3 hands of 23 points or more occurs only 2.4% of the time).

 

I suspect only 1 over 25 points is also pretty rare.

 

Opps had 6 25+ and 4 23-24 by comparison

[kenrexford]

Play R.U.N.T. if nothing else, it keeps you entertained on the otherwise boring hands/sessions. LOL

[Vampyr]

The other night I held rubbish cards the whole session and then picked up a balanced 28-count on the arrow switch. I felt sorry for all the other Norths.

[jogs]

You must have spent your life as a lucky card holder. I average fewer than 9 points per hand all the time.

 

Average points per hand is 10 +/- 4.13

Ave points per 26 boards is 260 +/- 21.041

 

That means in a 26 board session you should hold 239 or fewer points about 16% of the time.

 

Your pard and you held 45 points fewer than expected. That should occur about 6.6% of the time.

Not that unusual.

[Cyberyeti]

You must have spent your life as a lucky card holder. I average fewer than 9 points per hand all the time.

 

Average points per hand is 10 +/- 4.13

Ave points per 26 boards is 260 +/- 21.041

 

That means in a 26 board session you should hold 239 or fewer points about 16% of the time.

 

Your pard and you held 45 points fewer than expected. That should occur about 6.6% of the time.

Not that unusual.

 

What are you quoting as the +/- ? Standard deviation, variance, 95% confidence intervals ? And where do you get the 6.6% figure from ?

 

Also as a subjective feeling with no real logical or statistical basis, I tend to find that the point distribution is more extreme when you move about between the directions, almost as if something in the computer dealing programs keeps the points for each seat within bounds.

[jogs]

+/- is standard deviation. Just one std dev. You should be within that range about 68% of the time.

95% is two std dev.

6.6% was for 45 points lower than expected for 52 boards. I probably should have made the calculation for partnership on 26 boards. Should be close to the same number.

[jogs]

Forgot that I had the distribution for points for a partnership.

The probability of the partnership having 45 points less than expected

was much less than I thought.

20 +/- 4.77 per board.

+/- 24.329 for 26 boards.

1.85 std dev for 45 points less than expected.

1.85 std dev is ~ 3%

 

The probability of both partners holding 'bad' cards was less than I expected.