bonanova

Are you saying that his choice of non-winning door affects your chances of winning?

Hi CR, You say If B is not the winner, they are more likely to open door A next (since C will be the winner 3/5 times when it isn't B). That isn't necessarily true. Monty Hall is always certain to open a non-winning door (since he knows the winning door). Are you saying that his choice of non-winning door affects your chances of winning?

This puzzle has been posted. Please read this spoiler. Since this thread has achieved a reasonable life, I'll keep it open.

Well put. Is this your card puzzle?

When the first boy comes across, the other girls don't have their b/f's with them.

Construct a truth table of the premise "One child is a girl" for the 4 cases BB/BG/GB/GG. What you find is that "One is a girl" is false for BB and true otherwise. "One is a girl" is logically equivalent to "they are not both boys". Included among the things that premise does not do is to assign the gender of a selected child. Both of your methods scrutinize individual children, which ends up trivializing the OP to: "The probability that a child is a girl is 1/2. What is the probability that child X is a girl?" Because "one is a girl" means only "not both are boys" we're told [only] about gender combinations. And we're asked about combinations. Specifically, the case "both are girls." [one + the other = both] The expectation for 100 two-child families is: 25 have 2 girls; 25 have 2 boys; 50 are mixed gender. Removing "both are boys" shrinks the sample space to 75 -- 25 with two girls, and 50 with mixed gender. These probabilities are 1/3 and 2/3, respectively.

You know the drill. Only one of the doors is a winner; but this time there are four doors. Behind three of them stand bill collectors. Behind the other is your high-school sweetheart. Monty Hall gives you the rules: You point to a door. Of the other three doors, at least two have bill collectors. I will open one of them. You decide whether to stay with your choice or pick another door. I will then open another bill collector door that is not your current pick. You then decide whether to stay with your choice or pick the only other unopened door. What are the best odds for kissing once again your HS sweetie, and how do you get them? Should you stick / stick, stick / switch, switch / stick, or switch / switch? You pick door 1. MH opens door 2: Bill Collector. You switch your pick to door 3. MH opens door 1: Bill Collector. You decide whether to stick with door 3 or switch to door 4. Or: [*]You pick door 4. [*]MH opens door 1: BC. [*]You stick with door 4. [*]MH opens door 2: BC. [*]You decide whether to stick with door 4 or switch to door 3.

That does it. There is a 17-move solution.

When 3 drops off C and returns, C is left with 1 and 2. This violates the agreement.

Nope. When B gets D and rows past the island, C is left with boys F and H but without her boyfriend D.

Four couples, not two. Keep trying.

Tearz, It was posted here a year ago. It has a long discussion thread and similar groups of intuitants who absolutely know it's a wash.

Four young couples go for a stroll and come upon a river. They find a boat, able to carry two persons at a time. The girls, of course can row. But there is a problem: the girls are extremely jealous. So there is an agreement: No girl may stay be in the company of another's boy friend unless her own b/f is present as well. How do they all get across? They can't. But one of the boys spots a small island in the river, which can be used for intermediate loading and unloading. Now is it possible?

You say: ----------------- "There is a 2 child family who have at least one girl. What is the probability that they are both girls?" Then this is an open-and-shut case. 1/3. Now look at our problem again: "They have two kids, one of them is a girl, what is the probability that the other kid is also a girl?" Not the same is it? Same intention, different wording, different answer. ----------------- No it's not a different answer. Consider: "They have two kids ... " This means that with equal probability the kids are same or mixed gender. "... one of them is a girl ..." This means that the kids are either both girls or mixed gender, the second option having twice the probability of the first. "... what is the probability that the other kid is also a girl?" This means they are both [one + the other = both] girls, and that probability [against the twice as likely mixed case] is 1/3.

Are the numbers necessarily different? Are you sure they're both two-digit numbers?

I don’t quite follow Bona’s "__X" notation. X denotes initial cells. I missed one of your 3x3 configurations [lower left]; your other [lower right] doesn't paint. Including the 3x3 ones I missed, and using "o" instead of "_" for unpainted squares: 2x2: 2 Sets Xo [2] oX 3x3: 14 sets Xoo [2] Xoo [4] [color="#8B0000"]Xoo [4] oXo [4][/color] oXo ooX [color="#8B0000"]ooo ooo[/color] ooX oXo [color="#8B0000"]XoX XoX[/color] 4x4: 54 sets - probably missed some Xooo [2] oXoo [2] Xooo [4] Xooo [4] Xooo [4] Xooo [2] Xooo [4] oXoo Xooo oooX oXoo oooX ooXo ooXo ooXo oooX ooXo oooX oXoo oXoo oooX oooX ooXo oXoo ooXo ooXo oooX oXoo [color="#8B0000"]Xooo [4] Xooo [4] Xooo [4] Xooo [4] Xooo [4] Xooo [4] Xooo [4] Xooo [4][/color] [color="#8B0000"]oXoo oXoX oXoX oooX ooXo oXoo oXoX oooX[/color] [color="#8B0000"]oooo oooo oooo oooo oooo oooX oooo oXoo[/color] [color="#8B0000"]oXoX oXoo oooX oXoX oXoX oXoo ooXo oooX[/color] [/codebox]

This is the crux of your reasoning, which I believe that I follow. Taken in isolation, it's true. It's a premise of the OP. And it's true, before the conditions of the OP have been stated. But consider: the same is true of whichever child is referred to as the "one child". But the OP changes that probability from 1/2 into certainty when it says "one child" is a girl. At the same time, and with equal force, it changes the probability of the gender of whichever child is referred to as "the other kid". You say the two are not related, that they can be taken separately. But logic does not support it - the conditions in the OP affect them both. You're inspecting the trees, so to speak, as if there were no forest. But if one is a girl, then if the other is a girl, they are both girls. You somehow want to not see that the OP is asking a question that is logically equivalent to both being girls. But there simply is not room to have that difference: one of them + the other of them = both of them. So, yes, it's not the same. And if it were consistent with the OP it would lead to a different answer.

One child and the other child do not refer to specific children, who then can be discussed separately, as would be the case if it said the older child is a girl, what is the probability that the younger child is a girl. One does not give the gender of a specific child. One gives the count of the minimum number of children who are not boys. One child is a girl has no logical consequence other than to eliminate the case of two boys. What is the probability the other is a girl is logically identical to what is the probability of two girls. I'll grant that my statements are a semantical interpretation of the puzzle. But they are more than just a description of what personally makes sense. They are supportable by truth tables. They have logical stance as well.

One is a girl is equivalent to at least one is a girl. If the more restrictive One and only one is a girl meaning is assumed, then the probability of another girl is 0, not 1/2.

Punch 1 - no counter-punch thrown. Punch 2 - no counter-punch thrown. Punch 3 - no counter-punch thrown. 3. What does "one of them is a girl" mean? It means that it is not the case that the children are both boys. And nothing else. Therefore, If it's a same-gender family, the children are girls. If you disagree, say why, and call it Counter-punch 3. One is a girl changes nothing except to eliminate two boys. Nothing in the counter=punch says otherwise, so the punch stands. Punch 4 - no counter-punch thrown. Punch 5 - no counter-punch thrown. Punch 6 - no counter-punch thrown. Score: Blue 6 Red 0.

I'll start it off

Yeah. I thought you'd already shown statement 1, so I just restated it. Here's the proof. If the cell to be painted shares two edges with painted cells, painting it removes two units of perimeter and adds two units - the perimeter remains the same. If the cell to be painted shares three edges with painted cells, painting it removes three units of perimeter and adds one unit - the perimeter decreases. If the cell to be painted is surrounded on four sides with painted cells, painting it removes four units of perimeter and adds none - the perimeter decreases. These are the only cases. Therefore, when a cell becomes painted by other cells, the perimeter of the painted region does not increase.

In the Blue corner we have the 1/3 guys. And in the Red corner we have the 1/2 guys. This is a 6-round fight, and the 3-knockdown rule has been suspended because no one could agree on the probability that a third knockdown could even occur, given that two knockdowns had preciously occurred. Now let's hear the rules: In each round, the Blue camp will lead with a punch and the Red camp will be given the opportunity to counter-punch. In the event the Red team fails to throw even one legal counter-punch, the Blue team shall be declared the winner. If the Red team shall lose, but insist on a rematch, the fight will be held at a new venue. Now, go to your corners, and let's have a clean fight. The bell sounds ...! 1. What does the OP say? A family has two children. Boys and girls have equal birth probability. One of them is a girl. Then it asks: What is the probability that the other is a girl? Pertaining to the question, the OP says this, and only this. If you disagree, say why, and call it Counter-punch 1. 2. What does "boys and girls have equal birth probability" mean? It means that, before any other conditions were stated, the gender of each child is boy or girl, with equal likelihood. As a consequence, [do the math, it's easy] same-gender families and mixed-gender families have equal likelihood. half of the same-gender families will be boys; the other half, girls. If you disagree, say why, and call it Counter-punch 2. 3. What does "one of them is a girl" mean? It means that it is not the case that the children are both boys. And nothing else. Therefore, If it's a same-gender family, the children are girls. If you disagree, say why, and call it Counter-punch 3. 4. What gender distributions now exist, other than 2 girls or mixed-gender? No other gender distribution exists. If you disagree, say why, and call it Counter-punch 4. 5. What is the likelihood of a 2-girl family relative to that of a mixed gender family? [see points 2 and 3 above] The likelihood of a 2-girl family is 1/2 the likelihood of a mixed-gender family. If you disagree, say why, and call it Counter-punch 5. 6. If there are two and only two possible outcomes, say A and B, and B has twice the likelihood of A, what is the probability of A? It is 1/3. If you disagree, say why, and call it Counter-punch 6. If you cannot counter any of the six punches but still think the 2-girl case has probability of 1/2, then create your rematch in the Others venue, and name it Alternative approaches to logic.

Hi AdrianSecci Welcome to the Den. You're absolutely correct. Each hat can be red or black without regard for the colors of the other hats.

Prime has it.