Some probability puzzles

[transformer_99]

For the last one, there is zero probability you will pick the cup with the coin under it. Go to the carnival or state fair and you may pay a $ 1 to find that out. The hand is quicker than the eye. The gamer may lure you in with an easy win, but when your $ 1 is on the line, you'll lose every time. It's a shell game and slight of hand.

 

[B_dumbcow]

This may be futile

Seems so. we will all have to agree to disagree, since everything has been explained clearly in many different formats many different times :smile:

If I go on that gameshow, I'm swapping :smile:

• The original pick is more likely to be bad.
• Remaining bad choice is removed
• Swapping will always result in a win if original pick is bad (2/3)
• Sticking will always result in a loss if original pick is bad (2/3)

I don't see the flaw in that explanation. There is not a 50/50 chance of originally picking a good one.

 

[Jovial]

The best part of this article, Monty Hall problem - Wikipedia, the free encyclopedia, is

"These and other problems involving unequal distributions of probability are notoriously difficult for people to solve correctly, and have led to numerous psychological studies. Even when given a completely unambiguous statement of the Monty Hall problem, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief."

QED

 

[ManlyBanisters]

Seems so. we will all have to agree to disagree, since everything has been explained clearly in many different formats many different times :smile:

If I go on that gameshow, I'm swapping :smile:

• The original pick is more likely to be bad.
• Remaining bad choice is removed
• Swapping will always result in a win if original pick is bad (2/3)
• Sticking will always result in a loss if original pick is bad (2/3)

I don't see the flaw in that explanation. There is not a 50/50 chance of originally picking a good one.

That is the best and most succinct explanation I have seen so far.

The people arguing against the 66.6(rec)% v's 33.3(rec)% odds are wrong. They are over-complicating matter, sometimes in stupendous ways - but that does not change the essential wrongness of their positions.

The best part of this article, Monty Hall problem - Wikipedia, the free encyclopedia, is

"These and other problems involving unequal distributions of probability are notoriously difficult for people to solve correctly, and have led to numerous psychological studies. Even when given a completely unambiguous statement of the Monty Hall problem, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief."
QED

Ain't that the truth!

Guys, really - I have no dog in this fight. The 'paradox' holds - greater minds than any on this thread have agreed upon that. If you can't get your head around the numbers then stop trying.

I'll try to explain it one last time with a different 'prize'.

Man1, Man2 and Man3 are standing in front of you with their hands behind their backs. One of these men is holding a million dollars - 2 of these men have their fist balled up ready to hit you.

You are asked to stand in front of one man. You pick Man2. At this stage Man2 is 66.6% likely to hit you (and 33.3% likely to hand you a million bux).

Man1 says "I was gonna hit you" and walks away.

You are still standing in front of Man2.

Man2 is STILL 66.6% likely to hit you - THAT DOES NOT CHANGE WHEN Man1 WALKS AWAY.

You now have the choice of staying with the 66.6% chance of getting hit OR swapping to the 33.3% chance of getting hit.

Is that helping anyone?

 

[B_dumbcow]

My attempt to try and (hopefully) show how the Monty Hall game works in practice:

http://www.lpsg.org/100619-lets-play-monty-hall-game.html

 

[Calboner]

Seems so. we will all have to agree to disagree, since everything has been explained clearly in many different formats many different times :smile:

It makes no sense to "agree to disagree" on a point of mathematics on which the correct solution has been found and proved. There are people who get it; there are people who don't get it, but who study the solution and try to get it; and then there are people like Stapledshut and BoyCordoba who not only don't get it but who ceaselessly generate bogus objections, repeat fallacies, and accuse you of "changing the rules" or "changing your assumptions" when you try to explain the solution to them. It makes no sense for someone who understands the solution of the problem to "agree to disagree" with people of that sort. You agree to disagree when you can credit the reasons that your opponent has for disagreeing with you. That is not the case here.

I don't know if either one of them is coming back here, but just for good measure I am going to repeat what I said to them in my last post:

This may be futile, but I will make one more attempt to present the solution to the problem, but this time without using the idea of successive trials.

Imagine that 300 people play the game, each getting just one round, and all playing without knowing anything about what the other players are doing, so each operates independently of the others. (Each in a separate room, say, with a different host, a different prize, and so on.) Suppose that half of them (150) choose to stay with their first guess and half of them (150) choose to switch.

Here is what I claim: Of the people who stay with their first guess, approximately 1/3, or 50, will win; of the people who switch, approximately 2/3, or 100, will win.

First question: Do you grant this claim? I certainly hope that you do, as I don't see how you can grant the equivalent claims made in terms of successive trials and reject this one in terms of simultaneous trials.

Second question: If you grant this claim, do you still maintain that, when you play just once, the odds of winning are 1/2 if you stay and 1/2 if you switch? If so, how do you reconcile that with the fact that 1/3 of the people who stay win while 2/3 of the people who switch win?​

 

[BoyCordoba]

I am coming back here. Just been out of town for a day.

Like I said on my last post, I give up on trying to make my point. I already said what I think about having a choice at the second pick. Probably is the fact that English is not my first language or that I'm just incapable of coming accross to show you my point of view in the case of repeated instances of the game, but I will try no further.

I already said that I agree that always sticking with the original choice has 1/3 chance of winning and switching 2/3. It's not the maths I'm arguing about here, just the mechanics of the game. If I played the game let's say 10 times, I would sometimes stick with the first option and some other times switch, so in the end it would be closer to 50/50 and not 33/66.

 

[BoyCordoba]

And Calboner, please, do not insult me. I have not done so with you.

I do not generate "bogus objections" nor "repeat fallacies" and, specially, I'm not "people of that sort". Alright?

Take care man. I'm not opponent to anyone, just another guy here at the forum trying to have a good time reading some shit, looking at some pics, reading some stories and learning a lot of things that are helping me to feel better about myself in many aspects. Peace.

 

[Calboner]

My apologies for the "people of that sort" business. I was provoked by your accusing me of changing the assumptions or rules of the game to suit my purposes, after Stapled had made the same (unjust) charge.

I still don't understand how you can admit that switching doors has a 2/3 probability of winning yet claim to prefer a strategy of switching half the time and staying half the time. But I can understand your being simply tired of the whole discussion.

 

[D_Tintagel_Demondong]

As for the first question: it's ambiguous and may have two answers:

1. A priori.

Assumptions aside, the likelihood of having a female child is always 50%

Mem, BoyCordoba, Rubi, JustAsking, Crescendo69, uniqueusername, Gl3nn, StapledShut are all correct.

2. A posteriori.

I'll forgo a truth table for an equation:

P (B) = (1/4|bb) + (1/4|bg) + (1/4|gb) + (0|gg)
P (B) = 1/3

Calboner and Jovial are correct.

 

[Calboner]

As for the first question: it's ambiguous and may have two answers:

1. A priori.

Assumptions aside, the likelihood of having a female child is always 50%

Mem, BoyCordoba, Rubi, JustAsking, Crescendo69, uniqueusername, Gl3nn, StapledShut are all correct.

2. A posteriori.

I'll forgo a truth table for an equation:

P (B) = (1/4|bb) + (1/4|bg) + (1/4|gb) + (0|gg)
P (B) = 1/3

Calboner and Jovial are correct.

I take it that these comments both refer to my first puzzle. But I don't understand what you are doing with the terms "a priori" and "a posteriori" here, or how answer no. 1 can be said to be correct, "a priori" or otherwise.

If your point is that I was making an unstated and empirically inaccurate assumption about the incidence of male and female children, then I agree with you: I was assuming and expecting others to assume that in general a child selected at random has a probability of 1/2 of being male and 1/2 of being female. I ought to have included that assumption in posing the problem.

But the only sense I can make of applying the terms "a priori" and "a posteriori" to the case would be to say that I was assuming a priori that in general the probability of a child's being male and the probability of a child's being female are both 1/2, while the fact of the matter a posteriori (i.e., as known from observation) is a slightly different figure. But none of this seems to me relevant to the divergent answers that were offered.

I do not see any respect at all in which the answer "1/2" is a correct answer to the question that was posed. If by "assumptions aside," you mean "aside from the assumption that a family contains two children, one of which is a girl," then you are not talking about the problem that I stated.

 

[D_Tintagel_Demondong]

I take it that these comments both refer to my first puzzle. But I don't understand what you are doing with the terms "a priori" and "a posteriori" here, or how answer no. 1 can be said to be correct, "a priori" or otherwise.

If your point is that I was making an unstated and empirically inaccurate assumption about the incidence of male and female children, then I agree with you: I was assuming and expecting others to assume that in general a child selected at random has a probability of 1/2 of being male and 1/2 of being female. I ought to have included that assumption in posing the problem.

But the only sense I can make of applying the terms "a priori" and "a posteriori" to the case would be to say that I was assuming a priori that in general the probability of a child's being male and the probability of a child's being female are both 1/2, while the fact of the matter a posteriori (i.e., as known from observation) is a slightly different figure. But none of this seems to me relevant to the divergent answers that were offered.

I do not see any respect at all in which the answer "1/2" is a correct answer to the question that was posed. If by "assumptions aside," you mean "aside from the assumption that a family contains two children, one of which is a girl," then you are not talking about the problem that I stated.

Question: A certain family contains two children. One of them is a girl. What is the probability that the other is a girl?

Without adequate assumptions, expectation values of boy/girl distributions must be drawn as a pre-defined constant. In this case, the expectation of a boy's birth may be assumed to be 0.5, 0.33, 0.67, or any other value. Since expectations are set before the experiment, this can be regarded as an a priory possibility.

If you added, "assuming that the boy/girl distribution is 1:1," then you'd disallow this a priory variable. Otherwise, you must allow for an epistemological argument. :tongue:

I know that I'm nitpicking, but I'm just in that mood.

 

[cockoloco]

Thank you all for making me feel like an idiot.

 

[Calboner]

Question: A certain family contains two children. One of them is a girl. What is the probability that the other is a girl?

Without adequate assumptions, expectation values of boy/girl distributions must be drawn as a pre-defined constant. In this case, the expectation of a boy's birth may be assumed to be 0.5, 0.33, 0.67, or any other value. Since expectations are set before the experiment, this can be regarded as an a priory possibility.

If you added, "assuming that the boy/girl distribution is 1:1," then you'd disallow this a priory variable. Otherwise, you must allow for an epistemological argument. :tongue:

I still don't think I understand you. In your previous post, you said, "Assumptions aside, the likelihood of having a female child is always 50%." You gave this as a reason for saying that the answer given by Mem, BoyCordoba, Rubi, JustAsking, Crescendo69, uniqueusername, Gl3nn, and StapledShut -- namely, that the chance that the other child is a girl is 1/2 -- was correct. But now you say that "without adequate assumptions . . . the expectation of a boy's [or, I assume, a girl's] birth may be assumed to be 0.5, 0.33, 0.67, or any other value."

I don't see how to reconcile these two statements. If my statement of the problem leaves us free to adopt any expectation that we like with regard to the sex of a child, then the problem has no correct answer; for different assumptions will yield different answers. If we adopt an expectation of 1/2 for a girl, then we get the answer 1/3 for the other child, while if we adopt the expectation of 2/3, we get the answer 1/2,* and so on for other possible values.

(*Again, working with the four possible patterns bb, bg, gb, and gg: Pr(bb) = (1/3)(1/3) = 1/9; Pr(bg) = (1/3)(2/3) = 2/9; Pr(gb) = (2/3)(1/3) = 2/9; Pr(gg) = (2/3)(2/3) = 4/9. So for the families that have a girl in them, we get the ratio 2:2:4 for bg:gb:gg. The last, gg, is the family in which both children are girls, so its probability is 4/8 = 1/2.)

 

[D_Tintagel_Demondong]

I still don't think I understand you. In your previous post, you said, "Assumptions aside, the likelihood of having a female child is always 50%." You gave this as a reason for saying that the answer given by Mem, BoyCordoba, Rubi, JustAsking, Crescendo69, uniqueusername, Gl3nn, and StapledShut -- namely, that the chance that the other child is a girl is 1/2 -- was correct.

Well, I first said, "assumptions aside," just to parse the algorithm. I wanted to be clear that if there are no assumptions, then you may use a constant based on expected probability distributions. This would include all non-defined assumptions in the postulation. However, if you include all possible expectation variables, then you can safely move on to an empirical a posteriori argument.

But now you say that "without adequate assumptions . . . the expectation of a boy's [or, I assume, a girl's] birth may be assumed to be 0.5, 0.33, 0.67, or any other value."

Without adequate assumptions, you may arbitrarily assign your pre-defined constants. The uncertainty, or ambiguity, of any postulation encourages a priori reasoning. I was just giving an example of how ambiguity prevents you from logically claiming that, "There can only be one answer."

(*Again, working with the four possible patterns bb, bg, gb, and gg: Pr(bb) = (1/3)(1/3) = 1/9; Pr(bg) = (1/3)(2/3) = 2/9; Pr(gb) = (2/3)(1/3) = 2/9; Pr(gg) = (2/3)(2/3) = 4/9. So for the families that have a girl in them, we get the ratio 2:2:4 for bg:gb:gg. The last, gg, is the family in which both children are girls, so its probability is 4/8 = 1/2.)

So you are supporting my a priori claim that the probability of the family having two girls is 1/2 and not 1/3?

I still assert that the question is ambiguous, and may have two answers.

 

[Calboner]

Rec3000, I think I just don't understand your language. Maybe there are other people reading this who understand it better than I do. But if so, I don't think that there can be many of them.

So you are supporting my a priori claim that the probability of the family having two girls is 0.5?

Well, for the "rec," as it were tongue, I was only saying that you can get the result of 1/2 (as the probability that the other child in the family is a girl) if you make the assumption that in general the probability of a child being born female is 2/3. My main point was that if we are free to assume any probability we like for the birth of a female child, then there is no such thing as the correct answer to the problem. But my choice of that particular example (i.e., Pr(g) = 2/3) shows that, in order to get the answer "1/2" to the problem, we have to assume that twice as many girls as boys are born; and it is a safe bet that no one who gave the answer of "1/2" to the problem was assuming any such thing.

*********************************

Let's move on to A NEW PROBLEM.

This one is much simpler and easier than the Monty Hall problem, but still very easy to get wrong. I'll call it problem no. 4:

4. There is a lottery. Exactly 1,000 tickets are sold. Two tickets will be drawn at random: the holder of the first ticket drawn will win first prize; the holder of the second ticket drawn will win second prize. If you hold one ticket, what is your chance of winning second prize?

Just to be clear: the arrangement is that once a ticket is drawn, it is out of the lot; so no ticket can get drawn twice.

Rec3000, you will probably figure this one out pretty easily, so please let the less sophisticated have a go at it before you post an answer!

 

[D_Tintagel_Demondong]

Rec3000, you will probably figure this one out pretty easily, so please let the less sophisticated have a go at it before you post an answer!

No problem. As for the "(rec)" joke, you really stooped. :smile:

 

[Irish]

I'm so late to the party...

I admittedly went back to read a lot of stuff before actually submitting my reply, but... there seem to be some really common things happening here that I believe to be problematic with statistics.

1. 50% - The sex of one child does not affect the sex of the other. There was an example like this with a coin flip in both my statistics book and my psychology book. Pulled from the ever popular Wikipedia... (Assuming no genetic anomalies and an even birth rate among both sexes. There ARE more women than men in the world, but... for the sake of the problem [or converting people to coins] then it's still 50%.)

Statistic dearth is a common misuse of probability. According to this theory, the longer a particular outcome doesn't happen, the more likely that outcome will become. It is impossible to back up with logical evidence, and does not interfere with chance. An example of this theory is; flipping a coin and having it land on tails will heighten the chance of the next flip being heads. This is just one example of Statistical misuse.

2. 5% - Too much inverting.

3. Swap. In my data structures and algorithms course not only did we work the problem out, but were forced to right a simulation of the problem. I feel very comfortable with the answer by this point.

4. Dependent events... not winning first, then winning second. 999/1000 * 1/999 = 1/1000. Strange.

 

[Calboner]

1. 50% - The sex of one child does not affect the sex of the other. There was an example like this with a coin flip in both my statistics book and my psychology book. Pulled from the ever popular Wikipedia... (Assuming no genetic anomalies and an even birth rate among both sexes. There ARE more women than men in the world, but... for the sake of the problem [or converting people to coins] then it's still 50%.)

A lot of people have asserted that same maxim in support of the same answer: "The sex of one child does not affect the sex of the other," therefore the probability that the other child is a girl is 0.5. But the maxim is ambiguous. If it means "There is no causal connection between the sex of one child and the sex of the other," then it is true but irrelevant. The "influence" of the specification that one of two children in a family is a girl on the probability that the other is a girl is a matter of sampling, not of causation. If the maxim means that there is no such influence, then it is relevant but false.

Here is why. We start out with the specification: "A family has two children." We are assuming that the probability of a randomly selected child being female is 1/2, and the same for its being male. So, if we looked at, say, 800 families of two children each, the makeup of which exactly conformed to a probability of 1/2 for each sex, then we would find the following distributions of family compositions:

Type 1: families of two boys (BB): 200
Type 2: families of one boy and one girl (BG or GB): 400
Type 3: families of two girls (GG): 200
(Total number of families: 800)

Now we add the further specification: "One child in the family is a girl." This tells us that we can strike out families of type 1. Our sample is then as follows:

Type 2: families of one boy and one girl (BG or GB): 400
Type 3: families of two girls (GG): 200
(Total number of families: 600)

Finally, we pose the question: "What is the probability that the other child is a girl?" To say that the other child is a girl is to say that the family is of type 3. The incidence of families of type 3 in the sample in question -- i.e., families of type 2 and type 3 -- is 200/600, or 1/3. Therefore, the probability that the other child in the family is a girl is 1/3.

4. Dependent events... not winning first, then winning second. 999/1000 * 1/999 = 1/1000. Strange.

That is different from how I thought about the problem, but quite correct. In fact, it's much neater and more cogent than my way of thinking about the problem. I was hoping that we would get some wrong answers first! Well, we may get some yet.

 

[Irish]

The girl in question has one sibling either older or younger than her.

She could have an older sibling that's a girl, or a younger sibling that's a girl.
She could have an older sibling that's a boy, or a younger sibling that's a boy.

This also handles this possibility of twins since they are not born simultaneously (one twin is either older or younger).

Of the four possible variations for this girl's sibling, two of them find the sibling to be male, the other two find the sibling to be female.

I understand the logic of your explanation, but I feel there's a problem somewhere.

I'm going to read all the arguments on this first problem as it seems to be the most debated.

Edit: The controversy over the first problem seems to stem from ambiguity. Reading back I see that even you reversed your answer on it at least twice, Calboner. For the sake of a definitive answer, could you restate the problem without ambiguities?

From the set of all families with two children, a child is selected at random and is found to be a girl. What is the probability that the other child of the family is a girl?

or

From the set of all families with two children, a family is selected at random and is found to have a girl. What is the probability that the other child of the family is a girl?

Double Edit: Thanks for the props on number 4.