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True, but it's not a major city like NY, Philly, DC, Chicago, SF, Boston, etc. It's more on a par with Kansas City. If WWII is your thing, Honolulu is good. If WWI is your thing, KC is good. If neither world war is your thing, let's hope you like the restaurants.

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To really enjoy Hawaii involves really spending time, and that isn't compatible with playing bridge. On the other hand, in a city you can visit a museum before the afternoon session, have a nice dinner at break, and go to the theatre if you get knocked out and want to take a day off.

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Yes, the arithmetic used that model. So the sim's 43.1% would have converged to 43.4% in a longer sim. But at 50K the 95% confidence intervals from a sim are still almost a full point wide so 43.1% just means somewhere between 42.6% and 43.6%.

Since most people can't do that math, it might be best to give confidence intervals when reporting sim results.

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It's wasn't exactly my idea. I was suggesting a modification of Paul's idea to make it cost a lot less and still help people out. I was trying to do so without being rude to Paul. Perhaps that's what you could have learned from my post if you hadn't just wanted to ridicule it.

In the circles I hang out in, signing an affidavit is a big deal. So rather than saying you sound more obnoxious than usual, perhaps I'd say you are showing your true colors.

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Peg, I morphed that on purpose, since you morphed what I said from saying “a show of hardship” to “submitting a new sheet”. It's useless to point out red herrings so I simply return the favor.

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Peg, I think you are probably older than I am and don't care whether bridge is still around in 20 years. I do.

You don't ask people to give net sheets. You ask them to sign affidavits if they want a hardship discount (no one else needs to submit anything), and to agree to share some basic info if asked (which if they are actually poor, won't be much info). This isn't rocket science.

If the ACBL wants to run bridge your way, and basically let people pay what they want to pay, it won't be nearly adequate to eliminate free food. They'll have to stage nationals in trailer parks during tornado season.

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Perhaps the reduced fees would need to be based on a show of financial hardship? We have a lot of well-heeled octogenarians who need to keep paying for bridge to go on.

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A heart is different from a minor for empty spaces because RHO has “five or more” hearts. So you have to use a separate empty spaces argument for each number of hearts he can hold, and then take an average weighting by the probability RHO actually holds each specific number of hearts.

Taking your argument on your terms and showing that it leads to absurd conclusions:

What you are trying to say is that you can take the excess hearts, and lump them in with the other cards in a vacant spaces argument. If you weren't taught vacant spaces properly, so you don't see immediately that this is wrong, just follow it to its logical conclusion. In case C (4=2=x=y) you say you are distributing 9 spades, 6 hearts, and 19 minor suit cards among 34 vacant spaces. Since 8 of those spaces belong to RHO, you are saying he has on average 8*(6/34)=1.41 extra hearts. Now, I ask you, is it really your experience that when you hold 4=2=x=y and RHO opens 1♥, his average number of hearts is 6.41 ???? Of course it isn't. It's much less. (In fact it's 5.69.) But that puts more hearts in partner's hand than the naive application of vacant spaces, and therefore fewer spades.

You can't fix PART of a suit and then distribute the rest of it IN THE SAME HAND (RHO's) according to vacant spaces (not without naming the individual cards and thus making them separate suits anyway–but these aren't individual named cards–just “five or more of them”).

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To make it really simple, let's not worry about RHO opening something other than hearts if he has five hearts…let's say his bidding system is hearts first with 5 or more: If I am 4=8 in the major, then there are 25 minor suit cards and 9 spades to distribute, and partner gets 13 of them. So his average number of spades is 13*(9/34)=3.441. I HOPE WE CAN ALL AGREE ON THAT!

If I am 4=7 in the majors, there are two cases: RHO has 5 hearts or 6 hearts. If he has five hearts then the other hearts goes to partner half the time, so partner gets 12.5 of the 24 minor suit card and 9 spades to distribute, so his average number of spades is 12.5*(9/33)=3.409. If RHO has six hearts, partner gets 13 of the 24 minor suit and 9 spades to distribute, so on average he gets 13*(9/33)=3.545 spades. But we need to compute the odds of RHO having 5 or 6 hearts. There are 6 hearts and 33 non-hearts missing from your hand. There are (33C7)*(6C6)=4,272,048 ways to give RHO six hearts and (33C8)*(6C5)=83,304,936 ways to give RHO five hearts. Thus, he only has a sixth heart 4.878% of the time, and has only five hearts the other 95.122% of the time. Thus, we see that when I am 4=7 in the majors, the average number of spades partner holds is 0.95122*3.409 + 0.04878*3.545 = 3.416, which is less than 3.441.

The same math works for a smaller number of hearts in my hand, we just have more cases to handle (3 cases if I have 6 hearts, 4 cases if I have 5 hearts, etc.).

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Georgiana, you grew up at 71st and Jeffrey and no mention of the Oriental Institute on 58th street? I mean that's right up there with Chicago's other great museums. (A few years ago it was clearly better than the Science and Industry Museum, when S&I was in some disrepair, but the Science & Industry exhibits that were in bad shape seem to have been fixed now.)

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This is truly swatting flies with a Buick, John. You have to do some hard math to get the probability partner has at least four spades, but you can get the average number of spades with simple arithmetic using reasoning I provided or reasoning Gerben Dirksen provided. Until you go back and try to understand those explanations, no amount of sims are going to help you comprehend what is going on here.

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Your sim gives the general flavor, but your sampling variability is introducing errors. I gave the precise numbers above. Why do people want to simulate numbers that can be computed precisely?

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You have to do that much math to solve this problem, John. (Note that I changed the brackets to parentheses, so BW's markup language is no longer hiding the math.)

My 3 suit example is the same thing as the OP's example. Reds are playing the role of hearts, blacks the role of spades, and blues the role of the minor suits. You have a FIXED number of blacks (or spades); you know RHO has AT LEAST A CERTAIN NUMBER of reds (or hearts); and the number of reds (hearts) in your hand is the RELEVANT DIFFERENCE between hands A, B, and C in the post and between the 2=4=0 and 2=3=1 hands in my example. Which part of this analogy is escaping you?

Frankly the main difference between the OP's example and mine is that by having 5+ hearts concentrated in just 1 of *3* rather than 1 of 2 unseen hands, the effect is actually much more pronounced in the OP example. Mine doesn't exaggerate–it just makes the math easier. (The simple example also makes the math easier by not having anything analogous to hands with 5+ hearts that don't open 1♥. This simplifies things, and if John's logic were right, it wouldn't change the answer anyway. The main point is that you can have a problem like the OP's where you can do the math easily by hand and see that you don't get the same answers for the different cases.)

The reason for this phenomenon is that RHO will pretty strongly tend to have the minimum number of hearts that he can. He might hold more, but (considering case C) there are a lot more ways to give him five hearts and the other two hands six than the are to give him seven and the other two hands just four. And even when RHO has SIX hearts in case C, partner averages more spades than when RHO has FIVE hearts in case A. Surely you can see that.

Another way to put it is that your application of vacant spaces is invalid because you need to condition it on how many hearts partner holds. The HEARTS in his hand cannot be treated as vacant spaces the way you are trying to do.

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No, the hearts can be given to RHO. In particular, he can have more than 5 hearts if I have fewer than 8.

Let's do an example where you can follow all the math in a post rather than a spreadsheet.

Imagine there are three suits of six cards each, black, red, and blue, and three hands (you, RHO, pard). Imagine RHO has at least two reds. You have two blacks. What is the chance pard has at least two blacks when you are 2=4=0 vs when you are 2=3=1?

When you are 2=4=0, RHO has two reds and the other four blacks and six blues are split between RHO and pard. Pard can only fail to have two blacks if RHO has 3 or 4 blacks in 4 vacant spaces. The chance of this is ((4C3)*(6C1)+(4C4)(6C0))/(10C4) = 25/210, so partner has at least two blacks 185/210 of the time

When you are 2=3=1, RHO may have 2 or 3 reds. If he has 2 reds, pard has 1 and there are 4 blacks and 5 blues to fit into 9 spaces, of which 5 are partner's. Thus, partner fails to have 2 blacks when RHO has at least three blacks, ((4C3)*(5C1)+(4C4)*(5C0))/(9C4)=21/126, so partner has at least two blacks 105/126 of the time.

If you are 2=3=1 and RHO has 3 reds, then the remaining 4 blacks and 5 blues go into 9 spaces of which only 3 belong to RHO. Thus, partner fails to have two blacks only ((4C3)*(5C0))/(9C3) = 4/84 of the time, so partner in this case has two blacks with you 20/21 of the time.

Now, when you are 2=3=1, and RHO has two or more reds, there are (3C2)*(9C4)=378 ways for him to have two reds and (3C3)*(9C3)=112 ways for him to have three reds.

Doing all the resulting weighted averages, the chance partner has two or more blacks with you when you are 2=4=0 is 88.10%, and when you are 2=3=1 is (83.33%*378+95.24%*112)/(378+112)=86.05%.

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Another thought: Define some events that take place in the US but not at the time of nationals. (It could even be any regionals not within one week of nationals.) If you haven't played in a sufficient number of those events, triple entry fees for you at the NABC. After all, the international players will be the least price-sensitive, because card fees are small compared to their travel costs. And many of them are playing professionally, anyway.

Christopher Monsour

Christopher Monsour

Christopher Monsour

Christopher Monsour

Since most people can't do that math, it might be best to give confidence intervals when reporting sim results.

Christopher Monsour

Christopher Monsour

In the circles I hang out in, signing an affidavit is a big deal. So rather than saying you sound more obnoxious than usual, perhaps I'd say you are showing your true colors.

Christopher Monsour

Christopher Monsour

You don't ask people to give net sheets. You ask them to sign affidavits if they want a hardship discount (no one else needs to submit anything), and to agree to share some basic info if asked (which if they are actually poor, won't be much info). This isn't rocket science.

If the ACBL wants to run bridge your way, and basically let people pay what they want to pay, it won't be nearly adequate to eliminate free food. They'll have to stage nationals in trailer parks during tornado season.

Christopher Monsour

Christopher Monsour

Taking your argument on your terms and showing that it leads to absurd conclusions:

What you are trying to say is that you can take the excess hearts, and lump them in with the other cards in a vacant spaces argument. If you weren't taught vacant spaces properly, so you don't see immediately that this is wrong, just follow it to its logical conclusion. In case C (4=2=x=y) you say you are distributing 9 spades, 6 hearts, and 19 minor suit cards among 34 vacant spaces. Since 8 of those spaces belong to RHO, you are saying he has on average 8*(6/34)=1.41 extra hearts. Now, I ask you, is it really your experience that when you hold 4=2=x=y and RHO opens 1♥, his average number of hearts is 6.41 ???? Of course it isn't. It's much less. (In fact it's 5.69.) But that puts more hearts in partner's hand than the naive application of vacant spaces, and therefore fewer spades.

You can't fix PART of a suit and then distribute the rest of it IN THE SAME HAND (RHO's) according to vacant spaces (not without naming the individual cards and thus making them separate suits anyway–but these aren't individual named cards–just “five or more of them”).

Christopher Monsour

If I am 4=7 in the majors, there are two cases: RHO has 5 hearts or 6 hearts. If he has five hearts then the other hearts goes to partner half the time, so partner gets 12.5 of the 24 minor suit card and 9 spades to distribute, so his average number of spades is 12.5*(9/33)=3.409. If RHO has six hearts, partner gets 13 of the 24 minor suit and 9 spades to distribute, so on average he gets 13*(9/33)=3.545 spades. But we need to compute the odds of RHO having 5 or 6 hearts. There are 6 hearts and 33 non-hearts missing from your hand. There are (33C7)*(6C6)=4,272,048 ways to give RHO six hearts and (33C8)*(6C5)=83,304,936 ways to give RHO five hearts. Thus, he only has a sixth heart 4.878% of the time, and has only five hearts the other 95.122% of the time. Thus, we see that when I am 4=7 in the majors, the average number of spades partner holds is 0.95122*3.409 + 0.04878*3.545 = 3.416, which is less than 3.441.

The same math works for a smaller number of hearts in my hand, we just have more cases to handle (3 cases if I have 6 hearts, 4 cases if I have 5 hearts, etc.).

Christopher Monsour

Christopher Monsour

Christopher Monsour

Christopher Monsour

Christopher Monsour

My 3 suit example is the same thing as the OP's example. Reds are playing the role of hearts, blacks the role of spades, and blues the role of the minor suits. You have a FIXED number of blacks (or spades); you know RHO has AT LEAST A CERTAIN NUMBER of reds (or hearts); and the number of reds (hearts) in your hand is the RELEVANT DIFFERENCE between hands A, B, and C in the post and between the 2=4=0 and 2=3=1 hands in my example. Which part of this analogy is escaping you?

Frankly the main difference between the OP's example and mine is that by having 5+ hearts concentrated in just 1 of *3* rather than 1 of 2 unseen hands, the effect is actually much more pronounced in the OP example. Mine doesn't exaggerate–it just makes the math easier. (The simple example also makes the math easier by not having anything analogous to hands with 5+ hearts that don't open 1♥. This simplifies things, and if John's logic were right, it wouldn't change the answer anyway. The main point is that you can have a problem like the OP's where you can do the math easily by hand and see that you don't get the same answers for the different cases.)

The reason for this phenomenon is that RHO will pretty strongly tend to have the minimum number of hearts that he can. He might hold more, but (considering case C) there are a lot more ways to give him five hearts and the other two hands six than the are to give him seven and the other two hands just four. And even when RHO has SIX hearts in case C, partner averages more spades than when RHO has FIVE hearts in case A. Surely you can see that.

Another way to put it is that your application of vacant spaces is invalid because you need to condition it on how many hearts partner holds. The HEARTS in his hand cannot be treated as vacant spaces the way you are trying to do.

Christopher Monsour

Christopher Monsour

Let's do an example where you can follow all the math in a post rather than a spreadsheet.

Imagine there are three suits of six cards each, black, red, and blue, and three hands (you, RHO, pard). Imagine RHO has at least two reds. You have two blacks. What is the chance pard has at least two blacks when you are 2=4=0 vs when you are 2=3=1?

When you are 2=4=0, RHO has two reds and the other four blacks and six blues are split between RHO and pard. Pard can only fail to have two blacks if RHO has 3 or 4 blacks in 4 vacant spaces. The chance of this is ((4C3)*(6C1)+(4C4)(6C0))/(10C4) = 25/210, so partner has at least two blacks 185/210 of the time

When you are 2=3=1, RHO may have 2 or 3 reds. If he has 2 reds, pard has 1 and there are 4 blacks and 5 blues to fit into 9 spaces, of which 5 are partner's. Thus, partner fails to have 2 blacks when RHO has at least three blacks, ((4C3)*(5C1)+(4C4)*(5C0))/(9C4)=21/126, so partner has at least two blacks 105/126 of the time.

If you are 2=3=1 and RHO has 3 reds, then the remaining 4 blacks and 5 blues go into 9 spaces of which only 3 belong to RHO. Thus, partner fails to have two blacks only ((4C3)*(5C0))/(9C3) = 4/84 of the time, so partner in this case has two blacks with you 20/21 of the time.

Now, when you are 2=3=1, and RHO has two or more reds, there are (3C2)*(9C4)=378 ways for him to have two reds and (3C3)*(9C3)=112 ways for him to have three reds.

Doing all the resulting weighted averages, the chance partner has two or more blacks with you when you are 2=4=0 is 88.10%, and when you are 2=3=1 is (83.33%*378+95.24%*112)/(378+112)=86.05%.

Christopher Monsour

Christopher Monsour