Apologies are not good enough. You know better than this, and I’m officially tired of it.
This is your last warning. The very next time you level insults in General Questions, you will lose your posting privileges.
Scr4 -good job.
Knappy - your statement has the same problem as the OP. As I posted a little earlier(and have heard no one comment on) there’s a problem with saying “one announces she’s female”. This is just like “he hears from one, the East side”.
Do you mean “one and only one announces she’s female” or do you mean “both announce their sexes, and at least one of them is female”.
I think the first interpretation is more conventional, and leads to the 1/2 answer. The second meaning leads to the 1/3 answer. The 2 children probability problem statement is usually stated in the manner of the second meaning.
The key is, do we know one AND ONLY ONE sex/destination, or does someone know both, and just tells us that at least one is East/female?
Knappy - sorry, I re-read your post, and you made it clear that only one person spoke. That’s what makes it different from the (usual) 2 child problem.
It’s calld a sense of humour manhattan. I was not insulting anybody, it was simply a … shall we say “humorous” way of expressing my frustration to a question which I can so clearly see the answer too. That is why after each so calld “insult” I put some brackets, and quoted a particular Simpsons Episode in which Homer does the same thing to his Co-Workers, Lenny and Carl. That’s all it was, quoting Homer. Everyone here (especially Simpsons’ fans) would have realised that I meant it as a total joke and no harm intended.
Next time, I guess I will keep my jokes to myself, and I apologise for you lack of sense of humour.
I’m afraid, DeutschFox, that you’re missing the point, or perhaps are caught up in the semantics. The original problem is phrased thusly:
The bolded sentence is the bone of contention here. If you had instead said, “The passengers whisper to each other, and then one of them says, ‘At least one of us wants to go to the East side.’” then your answer of 1/3-2/3 is correct. I think that this is the scenario that you meant to convey in your OP, but, as written, that is not the meaning that comes across.
I interpret the bolded sentence as “The first passenger to answer says ‘The East side.’” in which case the 1/2-1/2 answer is correct.
Why? Because the passengers are now differentiated. They are not differentiated by left side of cab/right side of cab, but by speaking first/speaking second. Thus, of the four possibilities existing prior to that statement:
[ul][li]First speaking passenger goes east/second goes east.[/li][li]First speaking passenger goes east/second goes west.[/li][li]First speaking passenger goes west/second goes east.[/li][li]First speaking passenger goes west/second goes west.[/ul][/li]You can eliminate the last two, leaving a 50/50 chance that both passengers go east.
Arnold, I disagree. If you know that one child is a girl, and you can differentiate between the two children in any way, then the odds are 50/50.
I think you’ll agree that if the children example were rephrased: «There is a family with two children. You have been told that the oldest child is a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal?» then the answer is 1/2. (In fact, this variation is usually contrasted with the one above, within the quotation, to show how the extra information changes the odds from 2/3 to 1/2.)
Now, what if the problem was: «There is a family with two children. You have been told that the tallest child is a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal?» What is the answer? Still 1/2, because the children are differentiated; i.e., of the following four choices, equally possible before you’re told that the tallest child is a girl:
[ul][li]Tallest child = girl, shortest = girl.[/li][li]Tallest child = girl, shortest = boy.[/li][li]Tallest child = boy, shortest = girl.[/li][li]Tallest child = boy, shortest = boy.[/ul]two can be eliminated.[/li]
If you disagree with this answer, you’ll have to explain why differentiating children by height has a different outcome than differentiating children by age. If you agree with the answer, then you can see that any bit of information that allows you to place which child is the girl then changes the problem to one with a 50/50 answer. For example, stating that the child with the largest room is a girl, or the child with the darkest hair is a girl, or the child who has the best teacher is a girl, all give enough additional information to differentiate between the children, and thus leave a 50/50 chance that the family also has a son.
No … I don’t know what they did prior to the first passenger saying “East” … but they were 1) independent passengers, 2) the first passenger to answer was a random event, and 3) who answered “East” was indistinguishable to the cab driver …
Also, write this code in any language, plays=count=0:
(1) Flip two coins with C(1)=Rnd(0,1), C(2)=Rnd(0,1)
(2) If a East/Heads/1 is not “contained” in C(1) or C(2),
reflip the coins, e.g. goto Step (1)… otherwise
(3) play = play + 1
(4) If there is a Heads/1; if 1 is “contained” in C(1),C(2) and C(1)=C(2) … count=count+1 , goto Step (1)
(5) Otherwise don’t count if C(1)<>C(2), goto Step(1)
Only have to run a a couple of hundred loops. What is the count divide by plays …
The APL code for this is:
R “assign” TWO_COINS 10000
[1] X “assign” "neg"1 + ?10000 2 “rho” 2
[2] WW “assign” ^/0 = X
[3] X “assign” WW/[1]X
[4] EE “assign” +/^/ 1=X
[5] R “assign” 100 x EE “divide” 1 “take” “rho” x
R is the % of EE/Plays for 10,000 flips
Thanks … Andi
And here is the crux of the matter: you say that the passenger who answered “East” was indistinguishable to the cab driver. I say that the passenger was distinguished because he answered first. Since the first passenger answered “East”, then, to represent the scenario computationally, you must alter your coding like so:
(2) If a East/Heads/1 is not “contained” in C(1) only, reflip the coins, e.g. goto Step (1)… otherwise…
With this change, the 1/3-2/3 division is apparent.
In your program, C(1) represents the first passenger to speak, and C(2) represents the second passenger to speak. Since they are distinguishable, you can’t “cheat” by testing C(2) also.
If you disagree with this, tell me what C(1) and C(2) represent in reference to the OP. If you say C(1) represents the left-hand passenger (or whatever), you must alter your code to take into account which passenger speaks first.
So in other words, you’re checking both coins/passengers/whatever to see which results to eliminate.
Whereas in the original problem the cabbie only hears from one of the passengers. That’s why your coin-flipping experiment doesn’t model the problem as you originally stated it.
If you only know the destination of one of the passengers (as in the OP) then there is no way to eliminate just the tails-tails cases.
[QUOTE]
*Originally posted by Math Geek *
**
Hi Math Geek, not at all. If “ONE” of the passengers says “East” then the “West/West” is eliminated! … in code: if 1 is not “contained” in C(1), C(2), reflip …
You can write your code anyway you want to represent the effect of “one of the passengers saying East” … eliminating the WW/00 combination … Thanks
Zut answered this admirably.
For the people who feel 1/3rd are going the same way as the person who spoke up, consider this: Before anyone speaks, the probability they are going the same way is 1/2. The first person who speaks up will say either “east” or “west”. If the first person saying “east” means the probability they are going the same way is now 1/3rd, and if the first person saying “west” means the probability they are going the same way is now 1/3rd, how do you add these two possibilities together to get a probability of 1/2 that they are going the same way in the original case?
To put it another way, what if the cab driver thinks to himself “If I ask which way they want to go, one of them will answer me. Regardless of what he says, there is only a 1/3rd chance they are both going the same way. Therefore I don’t even have to ask the question to know there is only a 1/3rd chance they are going the same way.” Clearly, this reasoning is faulty.
OK, what the hell, I’ll try my hand at this. Here’s a rundown of all the possibilities of two people getting into a cab and one of them announcing his/her destination, along with their corresponding probabilities:
- Passengers are W and W; “W” is announced. 1/4
- Passengers are E and W; “E” is announced. 1/8
- Passengers are E and W; “W” is announced. 1/8
- Passengers are W and E; “E” is announced. 1/8
- Passengers are W and E; “W” is announced. 1/8
- Passengers are E and E; “E” is announced. 1/4
We’re given that “E” is announced, so we restrict our attention to cases 2, 4, and 6. In case 6 only, both are going east. So the probability both are going east is:
(1/4) / [(1/8) + (1/8) + (1/4)] = 1/2.
Note that this problem is distinctly different from the following: Two passengers enter a cab, at least one of them is going east; what is the probability both are going east?
Here we restrict our attention to cases 2, 3, 4, 5, and 6 since those are the ones where at least one of the passengers is going east (the “announcements” are irrelevant here). Again, only in case 6 are both going east, so the probability both are going east is:
(1/4) / [(1/8) + (1/8) + (1/8) + (1/8) + (1/4)] = 1/3.
As the original problem was worded, the answer is 1/2; the confusion, of course, results from the latter (incorrect) interpretation of the problem.
Here’s a script in perl to simulate the problem as described:
#/usr/local/pin/perl -w
#initalize some variables
@direction = ("EAST", "WEST");
@people = ("a", "b");
$trials = 100000;
$both_east = 0;
$east_then_west = 0;
sub coin_toss {return int(rand(2));} #this subroutine return 0 or 1 at random
for ($i=1;$i<=$trials;$i++){
#flip the coins
$person_a_wants = $direction[coin_toss];
$person_b_wants = $direction[coin_toss];
$first_speaker_is = $people[coin_toss];
if ($first_speaker_is eq "a") { #if person a goes first
next if ($person_a_wants eq "WEST"); #if he wanted to go west, ignore
if ($person_b_wants eq "EAST") {
$both_east++; #increment the both_east counter
}
else { #b wanted to go west
$east_then_west++; #increment the E-W counter
}
}
else {
next if ($person_b_wants eq "WEST"); #do the same stuff if b speaks first
if ($person_a_wants eq "EAST") {
$both_east++;
}
else {
$east_then_west++;
}
}
}
#print the results
$decided_trials = $east_then_west + $both_east;
print "$trials trials total.
";
print "$decided_trials trials decided (i.e West not picked first)
";
print "$both_east trials resulted in EAST-EAST. (", $both_east/$decided_trials, ")
";
print "$east_then_west trials resulted in EAST-WEST. (", $east_then_west/$decided_trials, ")
";
Note, this is by no means the most efficient way to solve this problem in perl. I was intentionally verbose so that non-perl programmers (and hopefully non-programmers) can follow.
Here’s the results of running the program for 100,000 trials, discarding all where the first person to speak selected “WEST”:
100000 trials total.
50025 trials decided (i.e West not picked first)
24781 trials resulted in EAST-EAST. (0.495372313843078)
25244 trials resulted in EAST-WEST. (0.504627686156922)
As you can see, the result are nearly 50%/50%. Sorry Deutsch and the other 1/3ers, you’re wrong. Greg, MadHatter, Math Geek et al. are right.
Pardon me for jumping into this discussion late. (And since I’ve skimmed at least a few of the many posts, I ask for special pardons of someone has already suggested this…)
Given the seemingly intractable dissension here, I’m surprised no one has gotten together with two friends and actually tried this experiment at home…no computer program where the problem might be slightly misunderstood by the coder, but an experiment using human beings.
Get two friends (A and B) together, and give then each a coin. Ask them to each flip their coins: “heads” means they want to go east, and “tails” means they want to go west. Now roll a die. If it comes up odd, friend A tells you where he wants to go. If it comes up even, friend B tells you where he wants to go.
If the die-chosen friend answers “west,” scrap the trial and start over.
If the die-chosen friend answers “east,” note the directional desires of both friends.
Seems like 50 trials should be enough to verify whether it’s 50/50 or 33/66.
Or am I missing something?
-Fezzik
Hold it right there pardner. Let’s look at your OP. You originally said,
Now, tell me, what does the other passenger do? Is the other passenger silent? Or does the other passenger say something also? This is the crux of the semantics problem here.
The problem statement implies that one passenger spoke, saying “East.” In this case, the correct answer is 50/50 because both West/West and West/East are eliminated.
If you meant the problem statement to be: “The passengers both speak at once, but the cabbie can’t quite hear, and only knows that at least one of them said ‘East’”, then the correct answer is 33/66 because, as you say, only the West/West combination is eliminated.
If you don’t understand the distinction I’m drawing here, please say so.
Well, yes and no. The bone of contention is exactly how to set up such an experiment. I would argue that the experiment you describe correctly reflects scenario the OP. However, [I think], others would argue that the experiment ought to be run as follows:
You will, for sure, get different results from these two tests. With a little reflection, you should see why these two experiments result in a different answer. With a little more reflection, you should, hopefully, see why Fezzik’s proposal is more in accordance with the OP.
Geez Feezik, where have you been through out this saga! Exactly right, first, the die toss is random and a “West” requires a replay, an “East” is play on … because in the OP “one of the passengers said East” … a great way to improve the experimental and code proof of this problem … danke … Andi
… except everytime one of your friends says “West” on the die toss, which was not said in the OP, it would eliminate the EW or WE outcomes 1/2 of the time they were tossed as well as the EE outcome 1/1. In the OP, “East” was the response, … thus the EW, WE, EE outcomes were never eliminated … only the WW outcome was … Andi
[QUOTE]
*Originally posted by zut *
**
Well I’m saying so.
If both passengers speak up, but the cabbie only hears one of them (and he says east) then the chances are still 50/50.
Because the cabbie ‘didn’t hear quite hear the 2nd guy’, why is the 2nd guy now more likely to have said west? There is no logic in your statement.
If I’m getting it right, I think he means that both passengers might have said ‘East’, so there is now no distinction between the two. In the other cases (and presumably the original, if we’re all reading it rightly), it was clear that one and only one person spoke, thus the distinction is made.
I think you’ve made it the most clear, MadHatter. One passenger has already spoken. The passengers are independent, so the other passenger remains at 50/50. A further inspired example :
Entirely unbeknownst to passengers, cabbie, or airport security, there is a tiny death-ray device in the cab that is activated whenever someone says ‘East’, and will kill them (and only them). So two passengers get in, unaware of the danger that awaits them. The cabbie doesn’t turn around, but asks where they want to go. Someone says ‘East’ and is promptly dispatched by the death ray. The cabbie is completely oblivious; in fact, he can’t distinguish between a dead body and a live one. Now, what are the odds that the other passenger wants to go East (assuming he or she is not too upset by the first passenger’s demise to be superstitious about going East)?