Best-hand dealing?

[TimG]

I'm confused and I will probably regret asking, but...

 

It is my understanding that in Best Hand games, south has at least as many HCP as any other hand.

 

There are probably many ways to deal such hands, but two that come to mind quickly are:

 

1) Deal randomly, check to see if south has at least as many HCP as any other hand, discard if this condition is not met.

 

2) Deal randomly, check to see if south has at least as many HCP as any other hand, if not, then rotate* the deal and check again. Repeat as needed.

 

I don't see why these two methods wouldn't produce the same HCP average for South (and the same HCP average for NS).

 

Were you using a 3rd way to produce deals?

 

* Instead of rotating, you could find a hand that did meet the "at least as many HCP as any other hand" and switch that hand with south rather than rotating the whole deal. Should be the same result.

[allfail]

Yes it is indeed very confusing (at least to me) and I did think it is the same for a long time. But in the end

(1) will have S average 14.85 points and

(2) will have S average 15.1 points.

 

The reason is that in algorithm (1) you are less likely to discard a hand with 10-12 points than discarding a hand with a higher point count. Or to put in other words, the two distribution would be identical if (1) picks up any board distribution as often as (2) does. However, in the boards with the highest HCPs tied (1) is more likely to pick them up.

[Cascade]

This is indeed correct.

 

I can agree with allfail.

 

Here are the two probability distributions (calculated not simulated):

 

HCP  Rotating    Discarding
10	0.001054593	0.003827362
11	0.032779272	0.05152155
12	0.099157473	0.117524213
13	0.154414808	0.161255334
14	0.172892127	0.168641459
15	0.156873806	0.147463208
16	0.126468288	0.116571669
17	0.093131625	0.084991426
18	0.063996213	0.058159037
19	0.041428835	0.03759938
20	0.025741115	0.023355396
21	0.015114687	0.013713674
22	0.008401711	0.007622938
23	0.004476148	0.004061244
24	0.002236134	0.002028862
25	0.001057111	0.000959125
26	0.000466732	0.000423469
27	0.000196266	0.000178074
28	7.43E-05    6.74E-05
29	2.67E-05    2.42E-05
30	8.79E-06    7.98E-06
31	2.45E-06    2.22E-06
32	6.88E-07    6.24E-07
33	1.41E-07    1.28E-07
34	2.82E-08    2.56E-08
35	3.93E-09    3.57E-09
36	3.78E-10    3.43E-10
37	2.52E-11    2.29E-11

[allfail]

Thanks, Cascade.

 

It is quite interesting to note that if we look at the renormalization of the probabilities, say the 37 count, it shows a ~10% difference, which means the tied highest HCP deals occupied 10% of all the dealt hands! Of course not all of them would be bad for you (roughly 2/3 of them is bad), and this amounts to 1-2 deals every tournament, surprisingly confirming my estimate.

 

So here's something I'd like to ask: is it possible to change the dealing method of those besthand tournaments to the deal-and-rotate method? I just think that "whenever you get a 15 count or below you would be facing an anomalous high chance of somebody who has the same HCP as yourself" is not a very welcome thought.

 

Besides there's strategic inferences. I always think that there might be a way to setup GiB and double their contract to gain points. Knowing the distribution this way might slightly favor this approach when one have some 12-15 count perhaps...

[hanp]

Apparently allfail was at least correct that it wasn't obvious to all.

[TimG]

Or to put in other words, the two distribution would be identical if (1) picks up any board distribution as often as (2) does. However, in the boards with the highest HCPs tied (1) is more likely to pick them up.

I'm still not getting it.

 

When the highest HCP is tied, (2) always picks it up, because (2) never discards a deal.

 

Is the problem that (1) discards the ties where south in not involved in the tie? So that in (2) ties are more likely to happen. And, ties are more likely to involve higher HCP totals?

[hanp]

Say the chance of having two equally strong strongest hands is P. If you draw such a hand then the chance that it is kept is 1/2, while the chance that any other hand is kept is 1/4. So after discarding the hands where south does not have (one of the) strongest hands, the chance that there are two equally strong strongest hands is:

 

(P/2)/ (P/2 + (1-P)/4) = 2P / (1 + P/4).

 

For small P this is almost 2P.

 

By rotating every deal so that south has (one of the) strongest hands you just get P.

[Cascade]

Here are the probabilities of ties with the two methods:

 

Tied   Rotate      Discard
0    0.905157035	0.821256074
1    0.08857877  0.160736425
2    0.005209602	0.014180139
3    0.001054593	0.003827362

 

Where the tied column tells us how many other hands we are tied with.