You know that a couple has two children. You go to the couple’s house and one of their children, a young boy, opens the door. What is the probability that the couple’s other child is a girl?
Oops, I changed it to a more unintuitive one right after you replied! In my original comment, I said “you flip two coins, and you only know that at least one of them landed on heads. What is the probability that both landed on heads?”
And… No! Conditional probability strikes again! When you flipped those coins, the four possible outcomes were TT, TH, HT, HH
When you found out that at least one coin landed on heads, all you did was rule out TT. Now the possibilities are HT, TH, and HH. There’s actually only a 1/3 chance that both are heads! If I had specified that one particular coin landed on heads, then it would be 50%
No. It’s still 50-50. Observing doesn’t change probabilities (except maybe in quantum lol). This isn’t like the Monty Hall where you make a choice.
The problem is that you stopped your probably tree too early. There is the chance that the first kid is a boy, the chance the second kid is a boy, AND the chance that the first kid answered the door. Here is the full tree, the gender of the first kid, the gender of the second and which child opened the door, last we see if your observation (boy at the door) excludes that scenario.
1 2 D E
B B 1 N
B G 1 N
G B 1 Y
G G 1 Y
B B 2 N
B G 2 Y
G B 2 N
G G 2 Y
You can see that of the scenarios that are not excluded there are two where the other child is a boy and two there the other child is a girl. 50-50. Observing doesn’t affect probabilities of events because your have to include the odds that you observe what you observed.
Yes! They responded to my comment before the edit, where I gave the coin example: “I flipped two coins, at least one of them was heads. What is the probability that both of them are heads?”
Before I read their reply, I edited it to the more confusing and infuriating two kids example. It’s annoying because it seems like it should be the same as saying “I have two children, and at least one is a boy. What is the probability that I have two boys?” In both the coin case and this one, the answer is 1/3, but when one child answers the door, it’s like sliding one quarter out from behind my hand. Now you know a particular (child, coin)'s (sex, face), and the answer is 50% again.
The issue is that you weren’t told a particular one was heads, only that at least one was heads. If I flipped a nickel and a dime, then the four possibilities are NtDt, NtDh, NhDt, and NhDh. If I say that at least one of them is heads but don’t tell you which one, then there are three possibilities: I flipped NhDt, NtDh, or NhDh. It’s only when I tell you that the nickel landed on heads that it collapses to NhDt and NhDh.
Sorry if the acronyms are hard to read, they’re much faster than typing something like “Heads (nickel) tails (dime)”
Again, you are assuming that every occurrence has the same chance. When in fact, they have not. There are 3 random events happening here:
Flip of one coin (50% chance each)
Flip of the other coin (50% chance each)
The coin that you told me (let’s say it’s 50% nickel 50% dime for simplicity’s sake)
Also, I am assuming that these 3 events are completely unrelated. That is, the result of a coin flip won’t determine whether you tell me the nickel or dime. A complete list of events is as follows:
T T N
T T D
H T N
H T D
T H N
T H D
H H N
H H D
After telling me that one of them is heads, the list is as follows:
H T D
T H N
H H D
H H N
H H is 50% chance, and the sum of HT + TH is the other 50%
This is very easily simulated. I use Tasker, you can use Python or something. Assume 1 is heads and 2 is tails
Randomize variable %dime to 1 or 2
Randomize variable %nickel to 1 or 2
This flips two coins. At this point there’s a 25% chance of each TT, TH, HT, and HH occurring.
Goto 1 IF %dime != 1 AND %nickel != 1
This is the point where I tell you that at least one of them is heads, but not which. It flips the coins again only if they landed TT, which means there are three possible sets of coins now: TH, HT, or HH.
Add 1 to variable %HeadsTails IF %dime = 2 OR %nickel = 2
Add 1 to variable %HeadsHeads IF %dime = 1 AND %nickel = 1
This keeps track of what sets made it past line 3. If either is tails, it adds one to %HeadsTails. If neither is tails, then it adds one to %HeadsHeads
Add 1 to variable %Counter
Goto 1 IF %Counter <1000
You can set your counter as high as you like. Python would probably handle 10,000 flips faster than Tasker can handle 1,000, but I’m on my phone and also python illiterate.
Press play, and I get 357 sets with two heads, and 643 sets that have a tails. The longer it runs, the closer to a ratio of 1:2 it’ll get.
If you simulate it like that, it leads to a contradiction.
According to the problem, 2 coins are flipped, and we all agree that it leads to an event pool of {TT,TH,HT,HH}, where all 4 events are equally as likely.
In the simulation, however, you just ignore the “TT” situation, which leads to a total event pool of {HT,TH,HH}. Where all events are equally likely.
The way the problem was phrased was “2 coin flips happen, and I have a machine that tells me either if both are Tails or not, this time it turns out that there is at least 1 heads”. But the way you simulated it is “I will make coinflips until I have at least 1 heads”.
But you do know which one it is, because you said “other child”. As soon as you ask the question, you assign a specific outcome to a specific child eliminating HH and HT (or in the new example, BB and BG). “What are the odds they have a female child” and “what are the odds the other child is female” are not the same question.
it depends on how you know this information. did you look at one of the coins? then it’s 50/50. did you ask a friend to look and yell you whether at least one of them is heads? then it’s 2/3.
This is a ridiculous argument when taken to the extreme. Say you have three bags. Bag A contains 100 blue marbles. Bag B contains 99 blue marbles and 1 red marble. Bag C contains 100 red marbles. You reach into a random bag and draw a red marble. You’ve only eliminated bag A. Would you say it is a 50-50 whether you are left with a bag now containing 99 blue marbles or 99 red marbles? No, the fact that you drew a red marble tells you something about the composition of the bag you drew from. The odds that you drew out of bag B is 1/101, the total number of red marbles in bag B divided by the total number of red marbles across all bags. The odds that you are dealing with bag C is 100x that.
Now let’s say you have 4 bags. BB, BR, BR, and RR. You draw an R. There is a 50% chance you are dealing with bag 2 or 3 because together they contain 2 out of 4 R. There is also a 50% chance you are dealing with bag 4. So it is equally likely that you draw either color of marble if you take the remaining marble out of the bag you randomly selected despite there being twice as many BR bags as RR bags.
It is 50-50, though. The remaining possible states are BG and BB. Both are equally likely. Any further inference is narrative… not statistics.
The classic example of this is flipping 100 coins. If you get heads 99 times in a row… the last coin is still 50-50. Yes, it is obscenely unlikely to get heads 100 times in a row. But it’s already obscenely unlikely to get heads 99 times in a row. And it is obscenely unlikely to alternate perfectly between heads and tails. And it is obscenely unlikely to get a binary pattern spelling out the alphabet. And it is obscenely unlikely to get… literally any pattern.
Every pattern is equally unlikely, with a fair coin. We see 99 heads in a row versus 1 tails at the end, and think it narrowly averted the least-probable outcome. But only because we lump together all sequences with exactly one tails. That’s one hundred different patterns. 1-99 is not the same as 99-1. We just treat them the same because we fixate on uniformity.
Compare a non-binary choice: a ten-sided die. Thirty 1s in a row is about as unlikely as 100 heads in a row. But 1 1 1… 2 is the same as 1 1 1… 3. Getting the first 29 is pretty damn unlikely. One chance in a hundred million trillion. But the final die can land on any number 1-10. Nine of them upset the pattern our ape brains want. Wanting it doesn’t make it any more likely. Or any less likely.
It would be identically unlikely for a 10-sided die to count from 1 to 10, three times in a row. All the faces appear equally. But swap any two events and suddenly it doesn’t count. No pun intended.
If this couple had eight children, for some god-forsaken reason, and you saw seven boys, the eighth kid being another boy is not less likely for it. The possibility space has already been reduced to two possibilities out of… well nine, I suppose, if order doesn’t matter. They could have 0-8 boys. They have at least 7. The only field that says the last kid’s not a coin toss is genetics, and they say this guy’s chromosome game is strong.
You’re right, but it’s not a subversion of the Gambler’s Fallacy, it’s a subversion of conditional probability. A classic example is that I have two kids, and at least one of them is a boy. What is the probability that I have two boys?
The intuitive answer is 50%, because one kid’s sex doesn’t affect the other. But when I told you that I have two kids, there were four possibilities: GG, GB, BG, or BB. When I told you that at least one of them is a boy, all I did was take away the GG option. That means there’s only a 1 in 3 chance that I have two boys.
But by having one child answer the door, I change it yet again–now we know the sex of a particular child. We know that the child who opened the door is a boy. This is now akin to saying “I have two children, and the eldest is a boy. What is the possibility that I have two boys?” It’s a sneaky nerd snipe, because it targets specifically people who know enough about statistics to know what conditional probability is. It’s also a dangerous nerd snipe, because it’s entirely possible that my reasoning is wrong!
Two more for funsies! I flipped two coins. At least one of them landed on heads. What is the probability that both landed on heads? (Note: this is what my comment originally said before I edited it)
I have two children. At least one of them is a boy born on a Tuesday. What is the probability that I have two boys?
watches the people with basic math skills fight to the death over the answer
If you really wanna see a bloodbath, watch this:
You know that a couple has two children. You go to the couple’s house and one of their children, a young boy, opens the door. What is the probability that the couple’s other child is a girl?
50%, since the coins are independent, right?
Oops, I changed it to a more unintuitive one right after you replied! In my original comment, I said “you flip two coins, and you only know that at least one of them landed on heads. What is the probability that both landed on heads?”
And… No! Conditional probability strikes again! When you flipped those coins, the four possible outcomes were TT, TH, HT, HH
When you found out that at least one coin landed on heads, all you did was rule out TT. Now the possibilities are HT, TH, and HH. There’s actually only a 1/3 chance that both are heads! If I had specified that one particular coin landed on heads, then it would be 50%
No. It’s still 50-50. Observing doesn’t change probabilities (except maybe in quantum lol). This isn’t like the Monty Hall where you make a choice.
The problem is that you stopped your probably tree too early. There is the chance that the first kid is a boy, the chance the second kid is a boy, AND the chance that the first kid answered the door. Here is the full tree, the gender of the first kid, the gender of the second and which child opened the door, last we see if your observation (boy at the door) excludes that scenario.
1 2 D E
B B 1 N
B G 1 N
G B 1 Y
G G 1 Y
B B 2 N
B G 2 Y
G B 2 N
G G 2 Y
You can see that of the scenarios that are not excluded there are two where the other child is a boy and two there the other child is a girl. 50-50. Observing doesn’t affect probabilities of events because your have to include the odds that you observe what you observed.
Yes! They responded to my comment before the edit, where I gave the coin example: “I flipped two coins, at least one of them was heads. What is the probability that both of them are heads?”
Before I read their reply, I edited it to the more confusing and infuriating two kids example. It’s annoying because it seems like it should be the same as saying “I have two children, and at least one is a boy. What is the probability that I have two boys?” In both the coin case and this one, the answer is 1/3, but when one child answers the door, it’s like sliding one quarter out from behind my hand. Now you know a particular (child, coin)'s (sex, face), and the answer is 50% again.
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You assume that the probability of TH = HT = HH
When In fact, the probabilities are as follows:
P(HT)+P(TH) = 50% P(HH) = 50%
For all the probabilities being equal, you’d have to consider 4 cases:
HT, TH, HH (1) and HH (2).
The difference between HH (1) and HH (2) is which one you were told that was heads.
Then P(HH) = P(HH (1)) + P(HH (2)) = 2/4 = 50%
The issue is that you weren’t told a particular one was heads, only that at least one was heads. If I flipped a nickel and a dime, then the four possibilities are NtDt, NtDh, NhDt, and NhDh. If I say that at least one of them is heads but don’t tell you which one, then there are three possibilities: I flipped NhDt, NtDh, or NhDh. It’s only when I tell you that the nickel landed on heads that it collapses to NhDt and NhDh.
Sorry if the acronyms are hard to read, they’re much faster than typing something like “Heads (nickel) tails (dime)”
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Again, you are assuming that every occurrence has the same chance. When in fact, they have not. There are 3 random events happening here:
Also, I am assuming that these 3 events are completely unrelated. That is, the result of a coin flip won’t determine whether you tell me the nickel or dime. A complete list of events is as follows:
T T N
T T D
H T N
H T D
T H N
T H D
H H N
H H D
After telling me that one of them is heads, the list is as follows:
H T D
T H N
H H D
H H N
H H is 50% chance, and the sum of HT + TH is the other 50%
This is very easily simulated. I use Tasker, you can use Python or something. Assume 1 is heads and 2 is tails
Randomize variable %dime to 1 or 2
Randomize variable %nickel to 1 or 2
This flips two coins. At this point there’s a 25% chance of each TT, TH, HT, and HH occurring.
This is the point where I tell you that at least one of them is heads, but not which. It flips the coins again only if they landed TT, which means there are three possible sets of coins now: TH, HT, or HH.
Add 1 to variable %HeadsTails IF %dime = 2 OR %nickel = 2
Add 1 to variable %HeadsHeads IF %dime = 1 AND %nickel = 1
This keeps track of what sets made it past line 3. If either is tails, it adds one to %HeadsTails. If neither is tails, then it adds one to %HeadsHeads
Add 1 to variable %Counter
Goto 1 IF %Counter <1000
You can set your counter as high as you like. Python would probably handle 10,000 flips faster than Tasker can handle 1,000, but I’m on my phone and also python illiterate.
Press play, and I get 357 sets with two heads, and 643 sets that have a tails. The longer it runs, the closer to a ratio of 1:2 it’ll get.
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If you simulate it like that, it leads to a contradiction.
According to the problem, 2 coins are flipped, and we all agree that it leads to an event pool of {TT,TH,HT,HH}, where all 4 events are equally as likely.
In the simulation, however, you just ignore the “TT” situation, which leads to a total event pool of {HT,TH,HH}. Where all events are equally likely.
The way the problem was phrased was “2 coin flips happen, and I have a machine that tells me either if both are Tails or not, this time it turns out that there is at least 1 heads”. But the way you simulated it is “I will make coinflips until I have at least 1 heads”.
You’ve also ruled out TH by knowing one is heads. So the only possibilities are HT and HH. Is that not 50/50?
At least one is heads, but unless you know which one it is, you haven’t ruled out HT or TH
But you do know which one it is, because you said “other child”. As soon as you ask the question, you assign a specific outcome to a specific child eliminating HH and HT (or in the new example, BB and BG). “What are the odds they have a female child” and “what are the odds the other child is female” are not the same question.
it depends on how you know this information. did you look at one of the coins? then it’s 50/50. did you ask a friend to look and yell you whether at least one of them is heads? then it’s 2/3.
This is a ridiculous argument when taken to the extreme. Say you have three bags. Bag A contains 100 blue marbles. Bag B contains 99 blue marbles and 1 red marble. Bag C contains 100 red marbles. You reach into a random bag and draw a red marble. You’ve only eliminated bag A. Would you say it is a 50-50 whether you are left with a bag now containing 99 blue marbles or 99 red marbles? No, the fact that you drew a red marble tells you something about the composition of the bag you drew from. The odds that you drew out of bag B is 1/101, the total number of red marbles in bag B divided by the total number of red marbles across all bags. The odds that you are dealing with bag C is 100x that.
Now let’s say you have 4 bags. BB, BR, BR, and RR. You draw an R. There is a 50% chance you are dealing with bag 2 or 3 because together they contain 2 out of 4 R. There is also a 50% chance you are dealing with bag 4. So it is equally likely that you draw either color of marble if you take the remaining marble out of the bag you randomly selected despite there being twice as many BR bags as RR bags.
Cheeky bastard.
It is 50-50, though. The remaining possible states are BG and BB. Both are equally likely. Any further inference is narrative… not statistics.
The classic example of this is flipping 100 coins. If you get heads 99 times in a row… the last coin is still 50-50. Yes, it is obscenely unlikely to get heads 100 times in a row. But it’s already obscenely unlikely to get heads 99 times in a row. And it is obscenely unlikely to alternate perfectly between heads and tails. And it is obscenely unlikely to get a binary pattern spelling out the alphabet. And it is obscenely unlikely to get… literally any pattern.
Every pattern is equally unlikely, with a fair coin. We see 99 heads in a row versus 1 tails at the end, and think it narrowly averted the least-probable outcome. But only because we lump together all sequences with exactly one tails. That’s one hundred different patterns. 1-99 is not the same as 99-1. We just treat them the same because we fixate on uniformity.
Compare a non-binary choice: a ten-sided die. Thirty 1s in a row is about as unlikely as 100 heads in a row. But 1 1 1… 2 is the same as 1 1 1… 3. Getting the first 29 is pretty damn unlikely. One chance in a hundred million trillion. But the final die can land on any number 1-10. Nine of them upset the pattern our ape brains want. Wanting it doesn’t make it any more likely. Or any less likely.
It would be identically unlikely for a 10-sided die to count from 1 to 10, three times in a row. All the faces appear equally. But swap any two events and suddenly it doesn’t count. No pun intended.
If this couple had eight children, for some god-forsaken reason, and you saw seven boys, the eighth kid being another boy is not less likely for it. The possibility space has already been reduced to two possibilities out of… well nine, I suppose, if order doesn’t matter. They could have 0-8 boys. They have at least 7. The only field that says the last kid’s not a coin toss is genetics, and they say this guy’s chromosome game is strong.
You’re right, but it’s not a subversion of the Gambler’s Fallacy, it’s a subversion of conditional probability. A classic example is that I have two kids, and at least one of them is a boy. What is the probability that I have two boys?
The intuitive answer is 50%, because one kid’s sex doesn’t affect the other. But when I told you that I have two kids, there were four possibilities: GG, GB, BG, or BB. When I told you that at least one of them is a boy, all I did was take away the GG option. That means there’s only a 1 in 3 chance that I have two boys.
But by having one child answer the door, I change it yet again–now we know the sex of a particular child. We know that the child who opened the door is a boy. This is now akin to saying “I have two children, and the eldest is a boy. What is the possibility that I have two boys?” It’s a sneaky nerd snipe, because it targets specifically people who know enough about statistics to know what conditional probability is. It’s also a dangerous nerd snipe, because it’s entirely possible that my reasoning is wrong!
i hate it when ppl do nb erasure for their stupid math text problems. use anything else pls
What’s the probability of the other kid being non-binary?
And don’t forget that there’s always a slim chance that no matter the gender, the other child is GOAT.
Two more for funsies! I flipped two coins. At least one of them landed on heads. What is the probability that both landed on heads? (Note: this is what my comment originally said before I edited it)
I have two children. At least one of them is a boy born on a Tuesday. What is the probability that I have two boys?
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