khaggblo

Given that you have the 10 million deals used to generate the initial hcp distribution, I think there is a mucher easier way to obtain the statistics without using any "advanced" mathematics. 1. Arrange all deals into 6 deal sets (e.g. by putting every 6th deal into a new set). 2. Discard the sets that violate the desired constraints for the 6 last deals. 3. Use all remaining deals to calculate the new hcp distribution for deal nr 21 (all because any deal in any set could be the next one). 4. If you really want to delve into this, you can select other 6 deal sets (there is a HUGE number of possibilities) from the original data and repeat steps 1-3. By combining these results, you can obtain more reliable estimates and even some confidence bounds on the expected hcp distribution.

I agree completely. On the other hand, how come the min and max are not integers? Because it's the average of the 24 boards tourney.You are right. http://forums.bridgebase.com/style_images/1/icon8.gif In theory you are right,if I would simulate a infinite number of deals the result should be symmetrical.Now take a coin and throw it. Both sides are equally likely, but only one side can be up. So to get symmetry you have to throw it a 2nd time, and there is a 50% chance you will get the same side up. If the same side comes up again you will have to throw another 2 times to have a chance for symmetry again. So whenever you set a limit lower than infinite for the the number of tries/simulations, you will find deviations from the symmetry. Not necessarily. Just add up the NS statistics and the EW statistics and you have a completely symmetrical result. Why don't you think that would be more accurate? On average I think you are sitting EW as often as NS...

I agree completely. On the other hand, how come the min and max are not integers? Furthermore, they should be symmetrical with respect to 20. If the NS max is 20+n, the EW min must be 20-n (why consider only one side?). The same principle applies to the simulation of hcp distribution (see HotShot in a previous post). If there are N deals where NS have, say, 10 hcp in a given number of deals, there must be exactly N deals where EW have 30 hcp. In theory, 20-n hcp occurs exactly as often as 20+n hcp. One could thus improve the accuracy of simulations somewhat by using the mean value of the number of deals having 20-n and 20+n hcp for both 20-n and 20+n hcp deals.

Actually not, I am trying to say more than that. I think it should be clear from my previous post that I am considering all remaining deals after a certain deal, in this case deal nr 20. It is clear that the statistics of the remaining deals are different from random deals, and this difference is not negligible. In particular, the variance of the point distribution will be reduced. This, for example, means that various kinds of estimations and predictions concerning any of the remaining deals, even deal 21 when we start bidding it, can be made with greater accuracy than with random deals. On the other hand, it seems that you are mainly concerned with problems of the last deal and that one of your solutions is not to play the "last deal". Are you implying that a limit of 22.7 on 27 deals is equivalent to any limit between 21.86 and 23.36 on 26 deals? A don't buy that. If the limit 21.86 is enforced on 26 deals, I think most of the discarded deal sets would be within the limit 22.7 on 27 deals. And a par with an average of 17 hcp allowed by the 23.36 limit would have every reason to be disappointed and angry since you have promised to deliver quality checked deal sets...

I think this analysis is flawed. With the given constraints, each side knows they will have in total 520 +-70 hcp in a 26 deal set. If the NS average is 22.5 after 20 deals, they have had 450 hcp. Thus, they know with certainty that they will have 70 +-70 hcp, i.e. between 0 and 140 hcp, in the remaining 6 deals. They know, for example, that they cannot have the points for game on all remaining deals. That is knowing a lot compared to random deals. And of course, the situation could become more extreme closer to the last deal. Similarly, EW know they will have between 100 and 240 hcp in the remaining 6 deals. This means, for example, that balancing and competing for the part score will be pretty risk-free. EW know they will have some values and will not be unlucky in terms of hcp.