Error in Marilyn's Explanation

(Corrections invited.)

This week's "Ask Marilyn" column has a problem for which Marilyn gives the correct answer for the wrong reason. The question is, "Is being dealt two aces or an ace and a deuce more probable." There are a couple of unstated assumptions: using ordinary deck and dealing exactly two cards are the main ones. Marilyn's explanation is to use the tableau:

AA
22

then to point out that being dealt two aces requires being dealt the top row and being dealt two deuces is being dealt one of the columns. This explanation is wrong on two points. First, it implies that the odds for the A2 combination are twice that of the AA combination (actually, the odds are 8:3 in favor A2 over AA) and it uses an arrangement of the ex post dealt cards rather than ex ante undealt deck.

The correct explanation: for two aces, the probability of an ace on the first card is 4/52 and the probability of an ace on the second card is 3/51; as these both must occur, the probability of both occurences is 12/2652. For an ace then a deuce, one has a probability of 4/52 for the ace and 4/51 for the deuce giving 16/2652 plus the (identical) probabilities of a deuce then an ace giving 32/2652 for the probability of the A2 pair.

Of course, naming the cards makes the probabilities equal. That is, the ace of spades and deuce of hearts has the same probability as the ace of hearts and the ace of diamonds.

Marilyn's problem in the above calculation is that after the first card is dealt, she isn't playing with a full deck.

Marilyn's problem in the above calculation is that after the first card is dealt, she isn't playing with a full deck.

Is this possibly the first public critique of this error? If so, it could or should be blogged. We can do this simply by copying the first post to the Best of DC thread. The blog article will be created by SenatorScript.

You'd also have to state whether the second card was drawn with or without replacement of the first card.

"Being dealt" pretty strongly implies "without replacement".

There's another comment (I just found with GOOGLE) but it suggests that the deck consists of four cards, in which case that guy's analysis is correct. It's on stconsultant.blogspot.com. I've tossed the magazine so I don't remember. Marilyn's probabilities would still be off.

I just dug up the Parade. No deck is described, so the assumption is that a normal deck is used.

(Corrections invited.)

This week's "Ask Marilyn" column has a problem for which Marilyn gives the correct answer for the wrong reason. The question is, "Is being dealt two aces or an ace and a deuce more probable." There are a couple of unstated assumptions: using ordinary deck and dealing exactly two cards are the main ones. Marilyn's explanation is to use the tableau:

AA
22

then to point out that being dealt two aces requires being dealt the top row and being dealt two deuces is being dealt one of the columns. This explanation is wrong on two points. First, it implies that the odds for the A2 combination are twice that of the AA combination (actually, the odds are 8:3 in favor A2 over AA) and it uses an arrangement of the ex post dealt cards rather than ex ante undealt deck.

The correct explanation: for two aces, the probability of an ace on the first card is 4/52 and the probability of an ace on the second card is 3/51; as these both must occur, the probability of both occurences is 12/2652. For an ace then a deuce, one has a probability of 4/52 for the ace and 4/51 for the deuce giving 16/2652 plus the (identical) probabilities of a deuce then an ace giving 32/2652 for the probability of the A2 pair.

Of course, naming the cards makes the probabilities equal. That is, the ace of spades and deuce of hearts has the same probability as the ace of hearts and the ace of diamonds.

Actually, if the deuce is wild, then, 2-2 is two aces, so you can add the probabilities for 2-2, 2-A and A-2 together as opposed to A-A. It would be an easy bet.

Marilyn's explanation is to use the tableau:

AA
22

then to point out that being dealt two aces requires being dealt the top row and being dealt two deuces is being dealt one of the columns.

Oops.

The rest of your analysis is good though.

Also, while her analysis is oversimplified by reducing it to a "deck" of 4 cards in order to give readers a "look-see" proof, she didn't even do her simplified analysis right, as the Ace-deuce combo can be achieved by being dealt "one of the columns", as she correctly states, but she *overlooks* that one can also achieve it by being dealt *one of the diagonals*.

Thus in her "4 card-deck" example (consisting of 2 aces and 2 deuces), getting an A-2 combo from a 2-card deal is actually FOUR times as likely as an A-A deal, not twice as likely.

To follow your own mathematical treatment, in the 4-card deck example, the probability of getting dealt A-A is 2/4*1/3 = 2/12, whereas the probability of getting an A-2 (or 2-A) deal is 2*(2/4 * 2/3) = 8/12.

A classic, not-exactly-but-kinda related probability puzzle/paradox: A woman mentions that one of her two children is a boy. What is the probability that her other child is also a boy?

(Yes, it's a straight question, it's not a "gotcha" base on the ambiguity of the word "one". Just assume that you know she has two children, and when mentioning one of them she incidentally revealed that the child under discussion was male. Now, is her other, unmentioned child likely to be a boy or a girl?)

I can't find an earlier posting on the internet. I guess after correcting the typo (about two deuces, thanks Ichy), it's worth blogging. I cannot edit the post anymore as it's too old.

There's another comment (I just found with GOOGLE) but it suggests that the deck consists of four cards, in which case that guy's analysis is correct. It's on stconsultant.blogspot.com. I've tossed the magazine so I don't remember. Marilyn's probabilities would still be off.

I just dug up the Parade. No deck is described, so the assumption is that a normal deck is used.

I googled it, too. Here's a little discussion, "next card after the Ace", but yours makes more sense, Doc.

http://forumserver.twoplustwo.com/25/pr ... ce-569206/

That's not the same problem. But it's a pretty interesting question on its own.

A classic, not-exactly-but-kinda related probability puzzle/paradox: A woman mentions that one of her two children is a boy. What is the probability that her other child is also a boy?

(Yes, it's a straight question, it's not a "gotcha" base on the ambiguity of the word "one". Just assume that you know she has two children, and when mentioning one of them she incidentally revealed that the child under discussion was male. Now, is her other, unmentioned child likely to be a boy or a girl?)

The companion to this is when the woman mentions that he eldest child is a boy.

I'd say the probable sex of any random child is male, just because more boys are born than girls. IIRC, the ratio is ~1.05 to 1 in favor of boys...

A classic, not-exactly-but-kinda related probability puzzle/paradox: A woman mentions that one of her two children is a boy. What is the probability that her other child is also a boy?

(Yes, it's a straight question, it's not a "gotcha" base on the ambiguity of the word "one". Just assume that you know she has two children, and when mentioning one of them she incidentally revealed that the child under discussion was male. Now, is her other, unmentioned child likely to be a boy or a girl?)

A monster, Ichneumon, a little [or not so little, if grown up] monster.

I'd say the probable sex of any random child is male, just because more boys are born than girls. IIRC, the ratio is ~1.05 to 1 in favor of boys...

For the sake of my puzzler, assume an exact 50/50 chance for boy/girl offspring.

Also, I worded the problem poorly, in a way that trivializes the paradox. Instead, consider the following questions as equivalent, then state your answer:

1. "Mr. Smith says: ‘I have two children and at least one of them is a boy.' Given this information, what is the probability that the other child is a boy?"
2. "Mr. Smith says: ‘I have two children and it is not the case that they are both girls.' Given this information, what is the probability that both children are boys?"

I'd say the probable sex of any random child is male, just because more boys are born than girls. IIRC, the ratio is ~1.05 to 1 in favor of boys...

For the sake of my puzzler, assume an exact 50/50 chance for boy/girl offspring.

Also, I worded the problem poorly, in a way that trivializes the paradox. Instead, consider the following questions as equivalent, then state your answer:

1. "Mr. Smith says: ‘I have two children and at least one of them is a boy.' Given this information, what is the probability that the other child is a boy?"
2. "Mr. Smith says: ‘I have two children and it is not the case that they are both girls.' Given this information, what is the probability that both children are boys?"

The chance is 1/3.

Given that there is at least one boy the possibilities are BB, BG, GB. All are equally likely so the chance of two boys given that there is at least one boy is 1/3.

The chance is 1/3.

Given that there is at least one boy the possibilities are BB, BG, GB. All are equally likely so the chance of two boys given that there is at least one boy is 1/3.

Ok, I'm no statistician, but that statement doesn't make sense to me.

Given that the one boy already exists, the only choices are BB or BG. GB requires a sex change on the part of the existing child.

The chance is 1/3.

Given that there is at least one boy the possibilities are BB, BG, GB. All are equally likely so the chance of two boys given that there is at least one boy is 1/3.

Ok, I'm no statistician, but that statement doesn't make sense to me.

Given that the one boy already exists, the only choices are BB or BG. GB requires a sex change on the part of the existing child.

Both BG and GB have "one boy"; but they are not equivalent. There are two ways of generating "one boy." One must be very careful in reading these problems. "One boy" is different from "eldest boy."

There are physical implications. Distinguishable particles obey different statistics (Maxwell-Boltzmann) from indistinguisyable particles (Bose-Einstein). The lambda point of liquid helium is one effect. The Monte Hall Problem is another.

A parlor trick illustrates the effect too. The performer shows three cards; one is black on both sides, one is red on both sides, the other is red on one side and black on another. (They're easy to make.) The trick is to have the Mark shuffle the cards under a table and show the upper side (keeping the lower side hidden.) Which ever side is up, the performer predicts that the other side will be of the same color. Fair bet?

The chance is 1/3.

Given that there is at least one boy the possibilities are BB, BG, GB. All are equally likely so the chance of two boys given that there is at least one boy is 1/3.

Ok, I'm no statistician, but that statement doesn't make sense to me.

Given that the one boy already exists, the only choices are BB or BG. GB requires a sex change on the part of the existing child.

Yes, but the catch is that if you consider the universe of all households (or parents) with exactly two children, there will be *twice* as many "BG" households as "BB" households....

There are equal numbers of the following four categories:

1. Eldest child girl, youngest child boy. (GB)

2. Eldest child girl, youngest child girl. (GG)

3. Eldest child boy, youngest child boy. (BB)

4. Eldest child boy, youngest child girl. (BG)

Thus, there are equal numbers of household in which the children are the same sex (BB or GG) versus different sex (GB or BG). Nothing surprising about that.

But this means that if you eliminate the "GG" households from being a possibility for Mr. Smith's family (by hearing him make one of the above announcements), then his kids are either case #1, case #3, or case #4. Each case occurs in the population of 2-child households with equal probability, and in two of those three he has a daughter, thus the probability that his "other" child is a girl, given that he has "at least" one son, is 2/3, not 1/2.

Ok, I admit that I was thinking a family where the first (eldest) child was known to be a boy, and the second was not known. My bad...

[GSlob]

...
If the question is, what is the P[two boys / given one is a boy], then P[BB] = 1/4

If the question is, what is the P[two boys/ given the FIRST one is a boy], then P[BB] = 1/4...

1. [BB vs BB + BG+GB]=1/3 .
2. [BB vs BB+BG] =1/2
And in China or India the answers are seriously different.

Last edited by GSlob on Sun Jan 10, 2010 5:43 pm, edited 1 time in total.

Note, however, that it's easy for the conditions which produce this paradox to break down. If the sex of a *specific* child is known, the chance of the other child being a boy reverts to 1/2, even though knowing the sex of the specified child (as being a boy, for example) sounds like it gives you the same information as in the paradox ("at least one child is a boy").

The key to producing the paradox is to have examined a (presumably randomly encountered) two-child household and, by examining *both* children, having determined that there is at least one boy child (i.e. the children are not both daughters). In *this* case the probability of a daughter in the mix is 2/3, not 1/2.

You *don't* get the same probabilities if you examine *ONE* child of a two-child household and find it's a boy. In *that* case, the probability of the unexamined child being a girl is still 1/2.

Most cases in real life match the latter case, not the former. For example, if a woman says, "one of my two kids is valedictorian of his class", you know that one of her children is a boy, and that she has two children, but since the gender information is for a *specific* child (the one who is valedictorian), this matches the latter case, and the odds of her other child being a girl are still 50/50.

Likewise for meeting a guy who says, "I only have one sibling". His sibling still has 50/50 odds of being one gender or another.

Note, however, that it's easy for the conditions which produce this paradox to break down. ...Likewise for meeting a guy who says, "I only have one sibling". His sibling still has 50/50 odds of being one gender or another.

You had to specify the absence of monozygotic twins in all the calculations, Ichneumon.

Note, however, that it's easy for the conditions which produce this paradox to break down. ...Likewise for meeting a guy who says, "I only have one sibling". His sibling still has 50/50 odds of being one gender or another.

You had to specify the absence of monozygotic twins in all the calculations, Ichneumon.

Wait here while I go look for my baseball bat...

Actually, that's a good point. Even after postulating an exact 50/50 ratio for boy/girl births, the phenomenon of identical twinning would (slightly) increase the probability of someone's sibling being the same sex as themselves. Roughly 3 out of every 1000 births worldwide involve identical twins.

Interesting to note that as of 2007, UC Berkeley is still using the same textbooks that I used many years ago at UC Santa Barbara. There are even homework problems at the link, in case anyone wants to have some fun. The solution set brings back some fond memories.

STAT 200A: Introduction to Probability and Statistics at an advanced level (Fall 2007)

http://www.stat.berkeley.edu/~sourav/st ... all07.html

As if the probability theory has changed much since then. 2x2 is still 4, unless one is talking about obamic budgeteers.