[Vampyr]

Back to the article:

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Since you already knew that at least one of the other two doors held a booby prize, you have learned nothing. You still have the same one chance in three that you started with.

This is very obviously wrong, since there are only two doors left. Even if the author doesn't understand that switching changes your odds to 2/3, he ought to be able to see that you can't have a one in three chance if there are only two doors.

[Antrax]

I may have misstated my question. I'm actually not that interested in the original Monty Hall, in the birthday paradox and in the 99% accurate test.
Solution 1's reasoning seems off to me, for the exact reason kenberg gives. It's irrelevant that west has more *spades* than his partner, just that he has more *known cards not containing the ♣Q* than his partner.

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Suppose your opponents at the other table somehow reach three notrump from the North hand. East leads a red suit and (surprise!) he has more cards in that suit than West. Is your opponent supposed to finesse against West for the club queen while you finesse against East?

Yes, if E shows ♦ are 3-5, then the odds favor finessing against W. If E shows diamonds are 3-5 and W shows spades are 5-3 on the same deal, then it's a toss.

Quote

Given that spades is West's longest suit, the expected spade break is roughly four and a half-three and a half. West rates to have one more spade than East. So the actual five-three break is only one card away from expectation. It is equivalent to a random suit's breaking four-three.

That means you can counter the bias by pretending that East has only one extra unknown card instead of two. You cash the club king and lead toward the ace. East follows. Now he has zero extra unknown cards. So it's a toss-up.

This just seems wrong to me. The first sentence is mumbo-jumbo, and everything else hinges on that. After cashing the ♣K and leading a club from N, you've eliminated the possibility E has Q doubleton or any singleton. How that swings the pendulum back towards the drop is beyond me.

[GreenMan]

kenberg, on 2013-August-19, 19:09, said:

But in the current situation, if I understand it correctly, W willl lead from a five card suit anytime he has one and the trick 2 play will always disclose it, and he will lead from his four card suit anytime he does not have a five card suit, and the trick 2 play will disclose that it was from only four. If so, W is basically a robot. He agrees to tell declarer that he has a five card suit whenever he has it, and he agrees to tell declarer that he does not have it when he does not have it. I see no issue of Restricted Choice.

The article addresses those concerns. In the initial condition:

"West leads a low spade. You duck in both hands; East continues spades. It appears from the carding that spades are five-three."

Similarly for the other two cases.

Also at the end:

"These arguments assume that West can be relied upon to have led his longest suit. If the auction makes certain leads unattractive, or if West has led from a sequence, or if West is simply known to be perverse, then he might have a longer suit and none of this applies."

Obviously the opps may be able to fool you, but sometimes you do have a good indication, even against humans, and then the article's arguments will apply.

[awm]

I think the basic line of thought is right, but would disagree with the actual odds here.

The basic theory is that on certain hand patterns, the person on lead might've had a choice of equally appealing leads. These should be discounted in the probability distribution because "if he had that hand, he might've lead something else." If we assume that LHO would always lead his longest suit or pick randomly from suits of equal length, the the odds worked out at the end of "solution one" are likely correct.

However, in reality I think LHO is virtually always leading a spade from 5-5 in the pointed suits, because the auction given calls for a major suit lead. And if he was 5-5 in the majors he might've bid (and in any case, 5-5 majors is much less likely because we can see a lot of hearts between declarer and dummy). This swings the odds to the point where finessing looks better than the drop.

The inference is stronger if LHO leads from a four-card spade suit. Presuming that he'd prefer a five-card suit, the only way he can have singleton club is if he's precisely 4441. Further, he would have a choice of equally appealing leads from that hand. Playing LHO for a singleton is really quite poor in this case! If he has doubleton club, he must be 4432 or 4342, whereas with three clubs he can be 4333 or 4243 or 4423. Note that 4432 and 4423 have equally appealing heart leads, whereas 4333 is presumably always leading a spade. It follows that we should finesse LHO for the queen of clubs (4333 or 4243 or half 4423 more likely than half 4432 or 4342).

[GreenMan]

Vampyr, on 2013-August-19, 20:08, said:

This is very obviously wrong, since there are only two doors left. Even if the author doesn't understand that switching changes your odds to 2/3, he ought to be able to see that you can't have a one in three chance if there are only two doors.

You started with a 1/3 chance. You knew that at least one of the other doors had no prize behind it. Nothing has changed. The chance your door was the right one is still 1/3.

[ArtK78]

GreenMan, on 2013-August-19, 22:16, said:

You started with a 1/3 chance. You knew that at least one of the other doors had no prize behind it. Nothing has changed. The chance your door was the right one is still 1/3.

That is true.

And the chances that your door is the wrong one is 2/3.

And you also know which of the other two doors is the wrong one.

So, if given the opportunity to switch, you should switch, since the chances that the switch is right is 2/3.

[Antrax]

Greenman, in the original MH you can choose a strategy that wins if your door has the prize (not switch) or one that wins if your door doesn't have it (switch). Which do you think has higher odds?

awm, can you explain where the vacant spaces logic goes wrong?

[GreenMan]

Antrax, on 2013-August-19, 22:22, said:

Greenman, in the original MH you can choose a strategy that wins if your door has the prize (not switch) or one that wins if your door doesn't have it (switch). Which do you think has higher odds?

Depends on the specific conditions, as I've already said. I'm not sure what you're asking.

[Antrax]

Trying to end the discussion diverting the thread about how this applies to Bridge, mostly.

[sfi]

Antrax, on 2013-August-19, 21:30, said:

I may have misstated my question. I'm actually not that interested in the original Monty Hall, in the birthday paradox and in the 99% accurate test.
Solution 1's reasoning seems off to me, for the exact reason kenberg gives. It's irrelevant that west has more *spades* than his partner, just that he has more *known cards not containing the ♣Q* than his partner.

The problem with your analysis is that you are assuming that the choice of suit is random. but it's not. I haven't worked through the author's arithmetic, although he's normally spot on. However, start with the assumption that opening leader will normally lead their longest suit unless there is a compelling reason not to do so. Therefore, when West leads a spade, we assume that they have at least 4 cards in the suit (the average will be somewhere around 4.5). Once we find out that West holds 5, it really doesn't change the probability of holding any other specific card all that much. And if we find out that West has led from a 4 card suit, it increases the expected length in the other suits and makes it more likely that West holds the CQ than East.

In short:
- We know West has length somewhere, and we will find out where on the opening lead. Chance of holding the club queen is approx. 50%.
- We happen to find out it's in spades. Chance of holding the CQ still approx. 50%.
- If we find out that West holds longer than expected spades, chance of holding CQ < 50%. If West holds shorter than expected spades, chance of holding CQ > 50%.

Distributions in other suits are more likely to affect the odds, but again it's not straightforward. If, for instance, we find out hearts break 3-4, that's pretty much expected. Since West chose a spade rather than a heart, we can estimate their heart holding at around 3 if seven cards are missing. If we find out they are 2-5, then that would change the odds.

To summarise, vacant spaces is a good theory. But there are caveats to how it is applied, and those caveats can be tricky.

[Antrax]

Quote

when West leads a spade, we assume that they have at least 4 cards in the suit (the average will be somewhere around 4.5)

This bit I have a problem with. Given that West's longest suit is spades, the probability of it being four cards is significantly lower, no? It seems counter-intuitive that 5332 would be as likely as 4333 and 4432 now, even ignoring the possibility of a six-card suit.

The argument that makes sense to me is that if we narrow west down to eight cards, finessing E for CQ plays him to have a 4-3-1 pattern, where 4-2-2 or 3-3-2 are still more likely? But that doesn't change if W randomly leads a suit and it happens to be five cards, so it's probably not the right way to look at this.

[gwnn]

gnasher, on 2013-August-19, 16:45, said:

There are two hands. Which one are you worried about? This is one of my favourite bridge articles, so I'd prefer not to find out that it's wrong.

Having said that, there is one bit of analysis in the second deal that is incorrect:

If East had not opened, you would still know that clubs were three-five (He would not have played the nine at trick one from king-queen-nine.) and it would be clear to finesse West for the spade queen.

If a strong East played 9-Q-K, you should be wondering why he so helpfully told you the club distribution. If you also know that he knows you're good enough to wonder, you should be wondering if he wants you to wonder.

Remarkably, I didn't see his mathematical reasoning, let alone that there were two hands in the website (young people nowadays and their attention span?). Interesting stuff, of course it has little to do with Monty Hall but still quite interesting.

[sfi]

Antrax, on 2013-August-20, 01:13, said:

This bit I have a problem with. Given that West's longest suit is spades, the probability of it being four cards is significantly lower, no? It seems counter-intuitive that 5332 would be as likely as 4333 and 4432 now, even ignoring the possibility of a six-card suit.

I don't know what the actual value is, but the longer the suit is the more likely it is that West will bid. Surely it's below 5, and you're right that it may well be above 4.5. I would be quite surprised if it's more than 4.7, but I'm happy enough to accept a value someone else works out. 4.5 makes any table estimates easier though.

[gnasher]

kenberg, on 2013-August-19, 19:09, said:

All I am saying so far is that I don't see that this involves Restricted Choice.

The only place where the article mentions Restricted Choice is in the discussion of when LHO leads a four-card suit. "In theory, he can't have five of either suit and, by restricted choice, he is less likely than normal to have four." What's wrong with that?

[Zelandakh]

ArtK78, on 2013-August-19, 15:19, said:

I don't remember the exact probability for 30 random people, I just know that if you have 23 random people in a room, the chances are greater than 50% that two of them have the same birthday.

This was one of the first problems posed in my basic probability and statistics course in my freshman year in college.

This is a university level problem in America?! In England, I was given it to solve at a maths seminar somewhere around 13 or 14. I could see it as a first task in a "Basic Computer Science for Mathematicians" course perhaps but as a probability problem it is way too simple for university undergraduates. On the same seminar course was the problem of finding the expected win/loss for the game of craps - that was far more difficult (and interesting).

[kenberg]

Edit: OK, I now see how restricted choice plays a role in a spade lead from four cards. Holding 4=4=4=1 he might have led a heart instead of a spade, holding 4=3=4=2 he would lead a spade, except if the diamonds were something like KQJx.. So restricted choice says that the lead of a spade from a four card holding is evidence for a doubleton club. It's not just a matter of counting possible hands, we must also consider that a spade was led instead of a heart. The reason may well be that he has only three hearts.
This is in the "he is known to have four spades" situation, I started off looking only at the "known to have five spades" case.I still believe RC, or Monty Hall, or whatever you wish to call it, does not apply in the fivee card case.

Quote

The only place where the article mentions Restricted Choice is in the discussion of when LHO leads a four-card suit. "In theory, he can't have five of either suit and, by restricted choice, he is less likely than normal to have four." What's wrong with that?

Well,his title refers to Monty Hall, which is the classical example of restricted choice. My own history with this is: I was at a restaurant when a former student came up and gave me the Monty Hall problem This was back shortly after the Ask Marilyn article appeared in Parade. He explained it carefully, I thought for a bit, and said "The answer is 2/3, this is just like a problem in bridge where it is referred to as restricted choice"


A couple more points.

I had not gotten as far as Adam. And here is where restricted choice type arguments really have a role. If he led a four card suit, and if we can safely (for the purpose of the problem) infer from this that he has no five card suit, then indeed he is unlikely to have a stiff club.One can reason that he then might have led one of his other four card suits, or more simply (and preferably imo) simply reason that two clubs and 4-4-3 in the other suits can happen in more ways than one club and 4-4-4 in the other suits. This is because two clubs opens up the possibility of 4=4=3=2, or 4=3=4=2. And really we should just count leading from four spades as leading from a four card major.


It seems to me that one can (again for purposes of analysis) think of the situation as equivalent to the following: INstead of W leading, the game goes as follows. Before a card is led, dummy comes down. Declarer plays a club to the board and a club back, and sees three spot cards. Before he plays from hand, he gets to ask W "Do you hold at least one five card suit?" and W will answer truthfully. Basically, as I read the problem, that's what happens. At crunch time, declare knows the answer to this hypothetical question, and knows nothing else exacept that if declarer has a five card suit, one of those suits is spades.


Now again Adam observes that the choice between two five card suits is not random. Correct. Unlike the usual restricted choice situation, I think this situation is way too loaded with unlikely hypotheticals to be very useful.


One further thought occurs to me: With no information at all to go on, we play for the drop. But if knowledge that W has no five card suit makes it more likely that we should play for the drop, then knowledge that he has a five card suit makes it less likely we should play for the drop. The books have to balance here.The initial prbability of Qx in W's hand is the sum of the probability of Qx when holding five times the probability that he has at least one five card suit and the probability he has Qx when not holding five times the probability of no five card suit. If the probability of Qx when he has no five card suit is less than the initial probability of Qx, then the other has to be more.It takes more effort than i have expended to figure out how much more.
I realize I have phrased the above rather badly, but the point is that the books have to balance.

Bottom line: Play from the drop if the original lead was from four, finesse if it was from five. Which is pretty much what anyone would do, I think.

[spaderaise]

Zelandakh, on 2013-August-20, 05:10, said:

This is a university level problem in America?! In England, I was given it to solve at a maths seminar somewhere around 13 or 14. I could see it as a first task in a "Basic Computer Science for Mathematicians" course perhaps but as a probability problem it is way too simple for university undergraduates. On the same seminar course was the problem of finding the expected win/loss for the game of craps - that was far more difficult (and interesting).

(birthday problem)

Here you can find it on a Cambridge University maths exam paper from 2009: http://www.statslab....osQuestions.pdf

And here you can find it on the first problem sheet for the current first-year probability course for maths undergraduates in Oxford:
http://www.maths.ox..../19623/material

You'll find it in university courses and textbooks the world over. And why not?! It's an important and surprising example and plenty of undergraduates won't have seen it before. And even for who have already seen it, there's plenty of depth to the question if you want to generalise it. The fact that a bright 13-year old might understand it doesn't mean that it's not worth a bright 19-year old thinking about it too.

[billw55]

Zelandakh, on 2013-August-20, 05:10, said:

This is a university level problem in America?! In England, I was given it to solve at a maths seminar somewhere around 13 or 14. I could see it as a first task in a "Basic Computer Science for Mathematicians" course perhaps but as a probability problem it is way too simple for university undergraduates. On the same seminar course was the problem of finding the expected win/loss for the game of craps - that was far more difficult (and interesting).

I think this has not much to do with a difference between the USA and the UK, and much more to do with the difference between you and most other people. I suspect you are a more or less an outlier in terms of mathematical literacy, or at least you were at age 14.

[kenberg]

I know it is off topic, but I must comment. Some years back I was in the coffee lounge, about to go off and teach calculus. A European visitor saw the text I was carrying and asked if I had taken on a part time job teaching high school.

Why don't we just agree to the following. Europeans think we are all idiots over here. They believe that no opportunity to point this out should be ignored. Got it, message heard and understood. It does not have to be endlessly repeated.

[Free]

kenberg, on 2013-August-20, 07:52, said:

I know it is off topic, but I must comment. Some years back I was in the coffee lounge, about to go off and teach calculus. A European visitor saw the text I was carrying and asked if I had taken on a part time job teaching high school.

Why don't we just agree to the following. Europeans think we are all idiots over here. They believe that no opportunity to point this out should be ignored. Got it, message heard and understood. It does not have to be endlessly repeated.

...and then you give us the opportunity to quote your post and keep pointing it out...