mmiguel Posted September 3, 2012 Report Share Posted September 3, 2012 Remove all cards form a standard deck except for the Aces and Kings. Randomly select 2 cards from this and deal them to a friend. She looks at the two cards and says: "After counting the number of red Aces in my hand, I can tell you that this number is greater than zero." What is the probability that both of her cards are Aces of any color? Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted September 3, 2012 Report Share Posted September 3, 2012 5/13 The Kings and Aces form T7 = 28 hands. 13 hands have at least 1 red ace. 5 of them have two aces. Quote Link to comment Share on other sites More sharing options...

0 mmiguel Posted September 7, 2012 Author Report Share Posted September 7, 2012 5/13 The Kings and Aces form T7 = 28 hands. 13 hands have at least 1 red ace. 5 of them have two aces. To make this short-lived thread more interesting, part B is below: You decide to repeat this game, and follow the same procedure as above in order to deliver her two cards to her. This time, she looks at the two cards and says: "I'm thinking of one of the cards in my hand right now, and it is a red Ace!" What is the probability that both of her cards are Aces of any color? Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted September 7, 2012 Report Share Posted September 7, 2012 To make this short-lived thread more interesting, part B is below: You decide to repeat this game, and follow the same procedure as above in order to deliver her two cards to her. This time, she looks at the two cards and says: "I'm thinking of one of the cards in my hand right now, and it is a red Ace!" What is the probability that both of her cards are Aces of any color? This depends on how she selects the card to talk about, and that otherwise she would not speak. That is, she "thinks about" one of the cards, and the statement is true of that card. Possibilities include: 1. the higher-ranking card, 2. the lower-ranking card, 3. the first card dealt to her, 4. the second card dealt to her, 5. a random selection, e.g. one based on the result of a coin toss, 6. her way of saying I just looked at both cards and it turns out that at least one of them is a red ace, 7. her way of responding to my question: is at least one of your cards a red ace? When a reporter describes some aspect of an outcome, the conditional probabilities about that outcome are affected by the algorithm the reporter follows when she makes her report. For example, see 1. 3/11 2. 1 (certainty) 3. 4. and 5. 3/7 6. and 7. 5/13 (same as first puzzle) Quote Link to comment Share on other sites More sharing options...

0 mmiguel Posted September 7, 2012 Author Report Share Posted September 7, 2012 (edited) This depends on how she selects the card to talk about, and that otherwise she would not speak. That is, she "thinks about" one of the cards, and the statement is true of that card. Possibilities include: 1. the higher-ranking card, 2. the lower-ranking card, 3. the first card dealt to her, 4. the second card dealt to her, 5. a random selection, e.g. one based on the result of a coin toss, 6. her way of saying I just looked at both cards and it turns out that at least one of them is a red ace, 7. her way of responding to my question: is at least one of your cards a red ace? When a reporter describes some aspect of an outcome, the conditional probabilities about that outcome are affected by the algorithm the reporter follows when she makes her report. For example, see 1. 3/11 2. 1 (certainty) 3. 4. and 5. 3/7 6. and 7. 5/13 (same as first puzzle) each of those adds some assumptions about her method of selection which were not given in the problem. what is the fundamental premise uniting 3, 4, and 5 to make them have the same answer? (3/7 is the correct answer to part B btw). is it possible to use that same premise without adding additional detail other than what was stated explicitly in the problem? Edited September 7, 2012 by mmiguel Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted September 7, 2012 Report Share Posted September 7, 2012 each of those adds some assumptions about her method of selection which were not given in the problem. what is the fundamental premise uniting 3, 4, and 5 to make them have the same answer? (3/7 is the correct answer to part B btw). is it possible to use that same premise without adding additional detail other than what was stated explicitly in the problem? OP says "I am thinking about one of my cards." That is an act of selection. It followed an algorithm of some sort. Flipping a coin, shuffling them face down then turning one over, taking the one closest to her as I tossed them across the table, saying eenie meenie money mo ... not thinking at all, praying, in some way, deterministic or not, she made a selection. And, inescapably, how that selection was made affects the answer. So the answer as the OP stands is, "depends." Without knowing her selection algorithm, how does one arrive at 3/7 as 'right'? If she thought of the lower-ranking card, the probability of both cards being aces is unity: spade ace plus one of the red aces are the only possibilities. OP doesn't say the thought-of card was the lower one in rank, true; it says nothing about the thought-of card. 3/7 is 'right' if the thought-of card was the result of a coin toss. But OP does not say she tossed a coin. I understand your basic question; I have wondered it myself. Absent any description of the reporter's algorithm, is there a least-restrictive default assumption that becomes preferable? I could, along with you, make a case for random choice (coin toss) to be that preferred, assumed algorithm. But unless that premise is understood and generally accepted among problem writers and solvers, and not just a few of us, it does leave answers uncertain. In my boy-girl puzzle I had some fun making this point, by imposing a random selection of reporter algorithms. Added in edit: 3., 4., and 5. I believe, all apply to picking either card with equal probability, along with the fact that the answers are the same for the two cards on that basis of choice: of the 56 permutations, of 14 hands the statement could have been made; and of those 14, six have two aces of any color. Quote Link to comment Share on other sites More sharing options...

0 mmiguel Posted September 7, 2012 Author Report Share Posted September 7, 2012 (edited) OP says "I am thinking about one of my cards." That is an act of selection. It followed an algorithm of some sort. Flipping a coin, shuffling them face down then turning one over, taking the one closest to her as I tossed them across the table, saying eenie meenie money mo ... not thinking at all, praying, in some way, deterministic or not, she made a selection. And, inescapably, how that selection was made affects the answer. So the answer as the OP stands is, "depends." Without knowing her selection algorithm, how does one arrive at 3/7 as 'right'? If she thought of the lower-ranking card, the probability of both cards being aces is unity: spade ace plus one of the red aces are the only possibilities. OP doesn't say the thought-of card was the lower one in rank, true; it says nothing about the thought-of card. 3/7 is 'right' if the thought-of card was the result of a coin toss. But OP does not say she tossed a coin. I understand your basic question; I have wondered it myself. Absent any description of the reporter's algorithm, is there a least-restrictive default assumption that becomes preferable? I could, along with you, make a case for random choice (coin toss) to be that preferred, assumed algorithm. But unless that premise is understood and generally accepted among problem writers and solvers, and not just a few of us, it does leave answers uncertain. In my boy-girl puzzle I had some fun making this point, by imposing a random selection of reporter algorithms. I like this topic, and those boy-girl puzzles made me think about this a lot a couple years ago. Here is what I think of it, and you may agree with me or not, either way, I find this thought provoking. I think: The fundamental difference between part A and part B above is that in that in one case, a statement was made about a specific object (part B), while in the other case, no statement was made about any specific object. I see this manifest in your solutions 3, 4, and 5 as well. In 3, we identify one of the cards i.e. the card identified as the first one drawn, and make a statement about that object. In 4, we also identify one of the cards, i.e. the card identified as the last card drawn, and make a statement about that object. In 5, we identify one of the cards i.e. the card identified as the one corresponding to heads (for example) In 6 and 7, the statement is not applied to any specific card. What I was trying to get at, is that by her "thinking" about a specific card, she has made a selection, and has thus made a statement about a specific card, i.e. the card identified as the one she thought about at this specific point in time. This concept, that making a statement about a specific object changes the probability problem is very neat to me. The real idea behind it, is being able to distinguish between objects, and the impact of that on the probability model. Here is how I think about it in a more general context than this problem: We are talking about two objects, which must differ in some way (for if they did not, how would you even know they are two objects and not one?). We can represent them as an ordered 2-tuple (x,y). When we say something about a specific object, we distinguish them in some way, and we can interpret the meaning of the index of the tuple as an indicator of any differing characteristic. Let's think of some differing characteristics, and apply them here: x is short and fat y is tall and lean you can say the index here (index 0 in the tuple is position x, index 1 in the tuple is position y), can be interpreted as an indicator of fatness. if the index is 0, the object is fat, if 1 the object is lean. you can also interpret it as an indicator of shortness. if the index is 0, the object is short, else the object is tall. the reason i'm bringing this up, is that you can apply this concept to any distinguishing characteristic. one such characteristic might be, (friend of problem solver from above problem is thinking about the object). in general terms, you can take any differing characteristic and map it into an ordering scheme for this 2-tuple. if in a probability problem, we make a statement about a specific object, then we can define a sample space of outcomes, which can be represented as tuples containing the potential states of the two objects, ordered by whatever distinguishing characteristic may be inferred from the statement about the specific object. let's say that specific object was x (i.e. the object in index 0). We may modify the sample space to remove all possible outcomes in which the element in index 0 of the tuple does not satisfy the the statement. now assume no distinguishing statement is made about either object -- this is like Part A above. for lack of a better ordering interpretation, let's use whatever ordering system we used from the case in which a statement about a specific object was made. What outcomes are we allowed to eliminate now from the sample space, given the generic statement about no specific object? We eliminate all outcomes where neither x nor y (i.e. object in index 0 nor object in index 1) satisfy the statement. This is a completely different change to the sample space than the other case, and the probability conclusions will be different in general. Thus, a small, subtle detail, changes everything. And it all comes down to being able to distinguish between objects, which in most cases is satisfied by making a statement about one concrete object, and not making the same statement about the other. Maybe that wasn't the cleanest explanation, but I think I got the meat of it in there. Edited September 11, 2012 by bonanova Quote Link to comment Share on other sites More sharing options...

0 ~andy~ Posted October 23, 2012 Report Share Posted October 23, 2012 You look at the cards left in your hand and just know what cards she is holding Quote Link to comment Share on other sites More sharing options...

## Question

## mmiguel

Remove all cards form a standard deck except for the Aces and Kings.

Randomly select 2 cards from this and deal them to a friend.

She looks at the two cards and says: "After counting the number of red Aces in my hand, I can tell you that this number is greater than zero."

What is the probability that both of her cards are Aces of any color?

## Link to comment

## Share on other sites

## 7 answers to this question

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.