# Some probability puzzles

Discussion in 'Et Cetera, Et Cetera' started by Calboner, Aug 24, 2008.

1. ### Calboner Gold Member

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I've been learning about probability lately and finding out some interesting things. I would reckon it not probable that many people will find these puzzles interesting enough to post their attempts at solutions in this thread, but I will give it a try. The second one is just complicated (hence the use of a multiple-choice format); the third is comparatively simple but potentially extremely difficult.

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

2. Multiple-choice question: There is a certain disease which one person in a thousand has. There is a test to detect the presence of this disease, but it is not infallible. In 99% of cases of persons tested who have the disease, it gives a positive result (a true positive); in the other 1% of cases, it gives a negative result (a false negative). Also, in 2% of cases of persons who do NOT have the disease, it gives a positive result (a false positive). You have taken the test and gotten a positive result. How probable is it, given this information, that you have the disease? Approximately

(a) 5%
(b) 75%
(c) 98%
(d) 99%

3. You and I are going to play a game. This is how it works: I have placed a valuable gold coin underneath one of three cups. I know which cup it is under, but you don't. If you can guess which cup hides the coin, you get the coin; but the game proceeds in two steps. First, you make an initial guess as to which cup hides the coin. I will then lift one of the other two cups, to show you that the coin is not under that cup. You will then have the opportunity to switch your guess to the other remaining cup, or to stick with your original guess. For instance, suppose you guess that the coin is under cup 2, and I show you that it is not under cup 1, and give you the choice of sticking with cup 2 or switching to cup 3. Question: Do you improve your chance of winning by switching (and if so, how)?

(People who are already familiar with the third problem, which is commonly presented in different terms, are asked not to identify it by name.)

2. ### Mem Gold Member

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My guess.
1. 50%
2. D
3. You always switch your choice.

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3. ### BoyCordoba VerifiedGold Member

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Well, let's see....

1) 50%

2) I would say 98%

3) You would think that switching may change the odds from 33% to 50% but I don't think you would improve the chances of winning.

Of course, I may be wrong in all of them. Shame on me for not taking Probability and Statistics class.

4. ### D_Gunther Snotpole Account Disabled

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I'm gonna say:

1) 50 percent
2) roughly 98 percent, so C.
3) switch to cup 3
(I point out that I was twice a statistic in Statistics.)

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I agree with Mem and Rubi.

I always forget the explanation about the switching thing, but each time it is explained to me, I realize it is true. So now I just rely on the memory that I have understood it and agreed with it a number of times.

6. ### tiggerpoo Gold Member

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Dear Calboner

I'm a mathematician specializing in probability theory. So, I won't spoil the fun. Great stuff to see you post questions like this. Great stuff to see members attempt an answer.

Kind regards

Tiggerpoo

7. ### crescendo69 Gold Member

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A)About 50%; each child born has the same chances as the last of being a boy or girl. There is a small difference in those odds - I forget what.

B)I believe only the last statistic, i.e., 2% false/positive, applies here, so the answer is 98%.

C)Since I choose the correct cup 1/3 of the time, the other two times you have left the ball in the other cup, leaving a 2 out of 3 chance it will be there. So switch!

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50%

5%.

Explanation: Since 1/1000 people have the disease, you can assign these probabilities to a population. Say you have a million people. 1000 will have the disease, 999000 will not. Of those who do, 990 will test positive and 10 will test negative. Of those who do not, 19980 will test positive and 979020 will test negative.

Taken in reverse, 20970 of the million will test positive. 990 of those will actually have the disease, or about 4-5%. This is why, if you test positive for a rare disease, you have to take the test again.

I've heard this problem before, but I don't remember the answer. It's difficult either way

EDIT: I looked it up on Wikipedia, but I won't spoil it here.

9. ### BoyCordoba VerifiedGold Member

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I have a question here: Is the exercise in item 3 going to be repeated or just played once?

I mean, if we're going to play for that gold coin based on a 2 out of 3, 3 out of 5 times or something like that, then you'd be better of changing your pick, but if you're just going to do it once to win the gold, I will stick to my first answer. Anyway you're getting a 50% chance since you're picking from 2 cups (provided it happens just once).

10. ### Gl3nn Gold Member

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1) 50%
2) C
3) You switch because then you increase your chances with 17%

Am I right?

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11. ### Calboner Gold Member

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I am delighted to see that there is interest in these questions. And thank you, Tiggerpoo, for not spilling the proverbial beans. (Of course, Tiggers don't like beans anyway, do they?)

Regarding no. 2:

Well done, "unique." This is the correct answer &#8212; both the correct result and a correct (as well as lucid and concise) way of arriving at it. One can arrive at the answer by using Bayes's theorem, but it makes sense only in terms of the probability calculus and not intuitively &#8212; unless one is so adept in the calculus that such moves themselves seem "intuitive" (as they certainly do not to me!).

The crucial point is that what I believe is called the "base rate" &#8212; the rate of incidence of the disease that we are trying to detect in the whole population &#8212; has to be taken account of. Given that the rate of false positives is non-zero, the more the people there are who do not have the disease (i.e., the rarer the disease), the more false positives there are [added in editing: I mean as a proportion of the positives); and the more false positives there are [added in editing: again, I mean proportionately], the less probable it is that a positive test result is a true one. If the rates for true positives and false negatives were the same but the base rate of the disease were 1 in 10,000 rather than 1 in 1,000, the probability that your positive result is a true positive would be not 990/20,970 but 99/20,079, which equals a minuscule 0.5%.

This kind of thing is, of course, of vital relevance to medical decision making. What is more, I have read that surveys have shown that medical doctors are just as likely to draw false conclusions in such matters as the medically untrained are.

Thanks!

Now about question no. 1: I may be making a mistake here, but this is how I reasoned: If a family has two children, then there are four possible ways in which boys or girls can be born to it (assuming, of course, that every child is exclusively either male or female and that there is an equal probability of either sex):

1) BB
2) BG
3) GB
4) GG

with 'B' for boy and 'G' for girl, of course. If all that we know about a family is that it has two children, then each of these four options has a probability of 1/4. However, since we are given the information that one of the children is a girl, we can infer that (1) is not an option. We therefore have the three options &#8212;

2) BG
3) GB
4) GG

Each of these is equally probable, and since they divide up the entire space of possibilities, they must have a probability of 1/3 each. Since we already know that one child is a girl, we can represent the three options merely with respect to the one that is known to be a girl, thus:

2) The other child is a boy
3) The other child is a boy
4) The other child is a girl

By this reckoning, the probability that the other child is a girl is 1/3. (Just corrected this in editing: I first said the probability that the other child is a boy is 2/3. That is perfectly correct, but it does not correspond to the question that I posed.)

Now I did not invent this problem: I saw it in a book. But I did not see the solution! So I could be making a mistake here. But I am sure someone will show me what my mistake is if I have made one.

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12. ### Calboner Gold Member

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This is an interesting response. In one respect, the fact that the procedure is repeatable is essential to the problem: that is, it is essential that you know going into the game that when I lift up one of the cups to show you that there is nothing underneath it, I am not doing this in reaction to your choice. Of course, my choice of that particular cup is in reaction to your choice: I have selected a cup that you did not pick and that I knew to have nothing underneath it. But if you see me lift up a cup and offer you a chance to switch your bet, and I did not tell you ahead of the game that I would do this, you would be bound to suspect that your guess was correct and that I am trying to lure you away from the cup with the coin underneath it. But that is not the setup. The setup is that the whole procedure is established and known to both participants in advance.

Beyond that, I don't see why it should matter if you play the game just once or a series of times.

13. ### Gl3nn Gold Member

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Concerning question 1...

I think the solution is 50%. As you all know, it depends on the father if it is a girl or a boy since he gives an X or a Y chromosone, the mother always gives an X chromosone. The selection of the X or Y chromosone always happens randomly. Of course there are exceptions of guys whose X or Y chromosone is stronger or so, but neglecting that small part of the guys... it happens randomly and therefor it's 50%

The BB, BG, GG, GB you're talking about are the fenotypes (the way a gen shows, eg: hair colour, eye colour, ...) and then we have to look at recessive and dominant etc etc.

But I'm not sure. If there's an expert on this, plz tell me. I remember this vaguely from my biology classes but I'm not sure at all.

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14. ### Ickday Member

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Well, to the third one, I'd say choose one, and then change your answer once one is eliminated. It was in a movie that recently came out =D

15. ### Calboner Gold Member

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Interesting angle, but surely the problem is not affected by the distinction between genotype and phenotype. The underlying causes of sex difference don't affect the probabilities, which were assumed in the setup of the problem to be 50/50 for each birth.

I think that the reason why people say that the probability that the other child is a girl is 50% is that they are thinking of permutations &#8212; sequences of items &#8212; rather than combinations &#8212; mere collections, regardless of order. If we were told that the first child born to the family was a girl, and the question were "What is the probability that the other (i.e., the second) child of the family is a girl?", the answer would be 50%, because there are two equally probable outcomes, GG and GB. But that is not the setup of the problem! The setup is: one child, who could be either the first or the second, is a girl. This determines not two but three equally probable options, as indicated above, in only one of which is the other child a girl.

16. ### D_Thoraxis_Biggulp New Member

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#1 is easy. 50%. One child's gender has no bearing on the other.

#2 is deceptive. It's actually only 5%, or 4.72% to be more accurate. 0.1% of the population have it, and with 99% showing a true positive, it means that 0.099% of the population would show such. Of the 99.9% who don't have it, 2% show a false positive, which means 1.998% of the population shows such. Add it up, you have 2.097% of the population showing a positive. On average, of 2097 positives, 99 will be true and 1998 will be false. Probability of truth in the positive is 4.72%.

#3 is slightly less deceptive. The probability of guessing right is 33.3% before disproving one option. Afterwards, it's 50%, but it applies to your original choice and the other cup. The only reason to change is superstition.

17. ### D_Thoraxis_Biggulp New Member

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The problem with saying there's a 1/3 chance of the other child being a girl is that you're bringing unnecessary factors into biological probabilities. Even without knowing if the mentioned daughter is first or second, it has no bearing whatsoever on the other child. Thinking otherwise is just the first step on a very slippery slope. And I'll prove it:

Actual population counts show that 51% of children born are boys, for reasons yet to be determined. Also, of those whose first born was a boy, 70% tried for another child. Of those whose first born was a girl, 68% tried for another. So, of 138 two-child homes, 70 started with a boy, 68 with a girl. 50.7% to 49.3%. Combine that with those first stats for determining the second child's gender, and you have 25.857% BB, 24.843% BG, 25.143% GB and 24.157% GG. Divide GG by GG+GB+BG, and you get 32.6%.
However, this is basing the possibility of a second girl on census statistics and not biological possibility. Furthermore, the probability of the first child's gender is being based only on the statistics of parents who tried for a second child, disregarding the original 51/49 spread. I'm not a statistician, so I have no idea how to incorporate the two to get a more accurate measure of the probability of a two-child home starting with one or the other. Point being, bringing in more population statistics calls for still more to clarify the ones we have. But when it comes down to it, aside from rare cases of men with XXY or XYY instead of XY, determining gender is a biological coin toss, regardless of when another child was born and of what gender.

18. ### Jovial Gold Member

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The answer to #1 is 1/3 chance of the other child being a girl. Calboner's reasoning is correct. If we were told that the older child was a girl, then the answer would be 1/2. But that's a different problem then. In this problem we're just told that either the older or younger sibling is a girl. This is an exercise in conditional probabilities.

Here is another problem:
#4. From families with 3 children, a child is selected randomly and found to be a boy. What is the probability that he has an older brother?

19. ### D_Thoraxis_Biggulp New Member

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50%

1/3 chance you picked 2nd-born, times 1/2 that the 1st is a Boy.
Plus 1/3 chance you picked 3rd-born, times 1/2 that the 1st is a Boy.
Plus 1/3 chance you picked 3rd-born, times 1/2 that the 2nd is a Boy.
1/6 + 1/6 + 1/6 = 1/2
You could incorporate the odds of picking 3rd-born and both 1st and 2nd being Boys, and come up with 7/12, but you'd be repeating scenarios in a problem in which each selection has no bearing on the other. Now if it involved picking different colored objects blindly from a finite set of both colors and objects, you would need to consider the odds of a color repeat.

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That's certainly a roundabout way of figuring this out, but here is your mistake:

Each of these is equally probable, and since they divide up the entire space of possibilities, they must have a probability of 1/3 each.

Each of those is NOT equally probable. BG and GB each have a 25% chance of occuring, while GG has a 50% chance. This is because there are really two GGs. Instead of using simply GG, let's call the girl we already know about "G1" and the possible other girl "G2". Your new combinations look like this:

BG1
G1B
G1G2
G2G1

Notice that you have four combinations, not three. The probabiliy for each of those is .25, but since the first two and the last two are essentially the same, you have .50 for BG1 and .50 for G1G2.

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