Killing Mind

I feel so bad for you. You are absolutely correct in your analysis of this very poorly worded. The wording of the question leads us to infer that one of the dice fell within your view and was seen to be a four, and the other fell out of your view and is unknown. In this case, your answer is in fact 1/6 as you have made clear.

Your professor obviously wants you to answer "what are the odds of rolling a seven given that AT LEAST ONE of the dice shows a four." I say "obviously" only because the 1/6 answer is too easy, and the 2/11 tests you on conditional probability, which is probably what you were studying. You were so right to give them the answer they wanted. What you did is called "playing the game" and represents many schools, but by no means all. I'm so happy to have gone to a school where the professors actually knew their stuff, and in the rare instance that they made a mistake, they came clean.

In your case, the professor should have used the phrase "Given that at least one of the dice is a four" in his question, because those magic words, "at least one," are the industry standard way to make sure that everyone understands the special (and unusual) case your professor has in mind.

My pet peeve is to always describe the game fully in these questions. This is why you are still wondering what game your professor has in mind. The game is an unrealistic one: "You roll a pair of fair dice into an area where you cannot see the results. An impartial observer will examine the result and tell you if at least one of the dice shows a four. You roll, and the observer tells you yes, you have rolled at least one four. What are the odds that you have rolled a seven?

As you can see, this game is much more contrived than the simple game you have in mind, where one of the dice fell within your view. This is because the magic number "four" has to be decided before you roll the dice, while the more reasonable game you have in mind simply reveals one of the dice and leaves the other to chance.

It also explains your paradox as to why the probabilities don't add up. They are not mutually exclusive. It is possible to have "at least one four" and "at least one five" on the same roll.

"Two dice are rolled simultaneously. Given that one die shows a "4", what is the probability that the total on the uppermost faces of the two dice is "7"?"

Actually, the teacher/question *did* say "given that *one* die shows a 4." He did not say "at least one die shows a 4." That is your (the blog writer's) error - if the question had instead been "given that at least one die shows a 4,"you would have been correct with 1/6th.

So, you were wrong. However, if the professor did not explain why you were wrong (like I just did), then that is definitely bad on his part - it is integral that math teachers can explain their reasoning well.

From the blog writer: thanks for taking the time to write. There's certainly food for thought there.

However, your interpretation is not what the lecturer meant either! Remember that the lecturer's answer was "2/11". If you suppose the lecturer meant that only one 4 appears, then the sample space is restricted to just 10 possibilities (the original 11 minus the double 4) and the calculated probability would be 2/10 = 1/5. Now we have another reasonable answer, different from both the lecturer's (2/11) and my own (1/6)!

Given that your interpretation is quite valid, and leads to a different answer, I believe an important point to make here is that there is simply far too much ambiguity in the original question - when it comes to probability you must be quite explicit.

I agree with all of the above. Basically the question is ambiguous.

The difference between real life and college is in real life you have to resolve the ambiguity too get the right answer, in college you have to know which answere is being sought.

I came accross a vaguely similar situation with a book question. You get 5 letters three are bills, one is not and you loose the last one. What is the probability the lost one was a bill. I answered 50:50 given you knew absolutely zero about the lost letter. The book then proceeded to use the bill / non bill probability distribution of the observed letters to estimate the likelihood it was a bill. You could argue for ever about whether the unseen bill is governed by the same statistics as the others. My parting shot was it was a crap qustion because it allowed the arguament to take place.

Best of luck
David Devoy, PhD,
Principal Design Engineer
Selex Sensors and Airborne Systems

Apologies about comments being broken. If this appears, all is fixed again! Comment away...

"In that case, the announcement that a "4" appears restricts our sample space not to 11 cases, but to 6. They are: 4-1, 4-2, 4-3, 4-4, 4-5, and 4-6, because a distinction has been made between the dice."

Yes, but these six cases are *NOT* equally likely. For example, there are two ways to roll a 4-1 -- you can roll the 4 and then the 1, or you can roll the 1 and then the 4.

The probability of rolling a 4 first is (1/6) and the probability of rolling a 1 next is (1/6). So the probability of rolling 4 followed by 1 is 1/36.

Along the same lines, rolling a 1 first is (1/6) and a 4 second (1/6), so 1 followed by 4 is also 1/36.

So the probability of rolling a 4 and a 1 (where order does not matter) is 2/36.

However, the probability of rolling 4/4 is only 1/36, because you have to roll a 4 (1/6) followed by another 4 (1/6). That is the only way to get two 4s so the probability is 1/36.

So the probability of rolling something with a 4 in it is:

4-1, 2/36
4-2, 2/36
4-3, 2/36
4-4, 1/36
4-5, 2/36
4-6, 2/36
---------
11/36

And the probability that you rolled a 4 and a 3, when you know you rolled at least one 4 is:

(2/36) / (11/36) --> 2/11.

All of your other arguments can be proven faulty as well.

The only thing that is ambiguous about the question is if you rolled 'at least one 4' or 'exactly one 4'. However, in the case of the latter, I think the correct answer is 2/10. So, that ambiguity does not help your case anyway.

"For example, there are two ways to roll a 4-1 -- you can roll the 4 and then the 1, or you can roll the 1 and then the 4."

You've misunderstood my description. The "4-1" statement explicity specifies that the 4 comes first. Re-read my post (and the comments so far) and you'll see that all your points have been addressed. Making the same argument using a different denominator wont convince me of much ;)

Two coins have been flipped. If I were to make a guess that they differ, I would have half a chance of being right. Say just before making my guess, I'm given reliable information that one of the coins is heads. According to the lecturer, now that I have more information about the situation I can be more confident that they differ. That is, my odds of being right about their differing has increased from 1 in 2 to 2 in 3.

Suppose for a moment that I was told one of the dice shows tails. Again, according to the lecturer, my chances that they differ have increased to 2 in 3. Just as if I was told that one of the dice shows heads. By the same logic, in both of these scenarios, armed with the information of what appears on one of the coins, I can bet my pretty pennies with extra confidence, because the chances that they differ have increased from 1 in 2 to 2 in 3. So knowing beforehand that a head appears, or a tail appears, appears to increase my chances of guessing they differ, compared to not knowing that a head or a tail appears. Next time I'm flipping two coins then, all I need someone to do is let me know "there's a head or a tail appearing!" and my chances of guessing they differ are better. Couldn't I just tell myself that? Looks like a paradox doesn't it? Well it's only a paradox if you believe that the chances ever increased to 2 in 3. I maintain they did not.

It may be easier to understand this "paradox" if we reduce the number of sides of the die from six to two, and recast the problem as a common-sense one.

Imagine a family with two children. What is the probability that they are of different sex? It's clearly one half, counting boy-girl or girl-boy out of the four possible combinations. If I told you that at least one of the children was a girl, then that only eliminates the combination boy-boy and means that the probability that they are of different sex increases from one half to two-thirds. The same happens if I said at least one were a boy.

I find conditional probability difficult to reason about. Sometimes it's easier to try. Simulating a million dice rolls repeatedly gives me a ratio of about 0.182 7's for each pair containing at least one 4. That's 2/11.

fizzle, I like the recast to a coin problem - the same argument now looks exceedingly obvious.

Paul, I actually addressed the family with two children problem here: http://heath.hrsoftworks.net/archives/000047.html You'll see that the conclusion still holds, but the reasons take a lot more shape.

Steve, your simulation is begging the question. Mikael Johanssons has done something similar, and you can read his results and my response on his blog.

PS. I just enabled HTML in the comments because it's hard to be precise without it.

Sorry, but your professor was absolutely correct. The only ambiguity is if the question means "at least one dice shows a 4" (in which case the answer is 2/11) or "exactly one dice shows a 4" (in which case the answer is 1/5).

The misunderstanding is your interpretation of how someone would answer the question "Does [at least] one dice show a 4?". The person must consider both dice in answering this, not just one like in your solution.

Next time I'm rolling two dice then, all I need someone to do is let me know "there's a number from 1 to 6 appearing!" and my chances of guessing the 7 total are better.

No, only if you state a specific number that will appear do the chances raise to 2/11. You could make this argument, for example, if you had two rigged dice that had sensors in them and whenever you rolled them at least one would always show a 4.

This sounds awfully similar to the infamous Monty Hall problem.

Incidentally, you say that "a distinction has been made between the dice" in saying that one of them is a 4. I believe this is not the case -- it is only specified that one of the dice is 4, not *which* one.

(addendum)

I think your error is unfamiliarity with the (admittedly bizzare) conventions of probability word problems. :p

Sorry, but your professor was absolutely correct. The only ambiguity is if the question means "at least one dice shows a 4" (in which case the answer is 2/11) or "exactly one dice shows a 4" (in which case the answer is 1/5).

The misunderstanding is your interpretation of how someone would answer the question "Does [at least] one dice show a 4?". The person must consider both dice in answering this, not just one like in your solution.

No, only if you state a specific number that will appear do the chances raise to 2/11.

You could make this argument, for example, if you had two rigged dice that had sensors in them and whenever you rolled them at least one would always show a 4.

The statement "one die shows a 4" can be interpreted in a few ways, all of which yield different answers:

* "I see at least one 4 among the two dice" - answer: 2/11

* "I see exactly one 4 among the two dice" - answer: 1/5

* "The first of the two dice I rolled shows a 4" - answer: 1/6

Personally, I would have chosen the first one. I think the third is a somewhat nonstandard interpretation of the phrase.

fizzle, your argument is incorrect. It boils down to:

P(A|B) = P(A|C)

B union C = everything

therefore P(A|B) = P(A)

where A = coins differ

B = at least one heads

C = at least one tails

The problem is that the last line does not follow from the first two unless B and C are mutually exclusive events.

I think your error is unfamiliarity with the (admittedly bizzare) conventions of probability word problems.

By the way, you're right about the Monty Hall problem. It in turn, is similar to the "women with two children" problem, which I've addressed and compared to this problem here: http://heath.hrsoftworks.net/archives/000047.html

Wow.. way to misunderstand a simple question and make it complicated and waste hours on it. Sorry dude, but it's not such a deep concept. The question implies "at least one."

I think your understanding of statistics is fine, but your understanding of the word 'given' is flawed. Here is how I think you are supposed to read the problem:

"Two dice are rolled simultaneously."

That means that there are 36 equally likely outcomes.

"Given that one die shows a '4'"

Looking in some dictionaries, we see that given means:

- Specified; fixed:
- assumed as actual
- Granted as a supposition

In other words, "Consider only the cases where at least one of the dice is a 4". This phrase does *not* tell us which die is 4, just that at least one is. So, we can only prune our list down to 11 equally likely possibilities.

"what is the probability that the total on the uppermost faces of the two dice is '7'"?

Of those 11 cases we are considering, 2 of them add up to 7, so the answer is 2/11.

Your problem is that you imagined the observer in the wrong way. In your example, the observer looks at the thrown dice, picks one of them by some method (left-most, random, etc), and tells you the number on it (which happens to be 4). In that case, you know that the observer can only be seeing one of 6 things, 4-1, 4-2, 4-3, 4-4, 4-5, 4-6. So the best guess you can make is 1/6.

However, the problem tells you 'given that one die shows 4', aka, 'it is specified that one die equals 4'.

In this case, the observer is *not* picking a die and telling you the number on it. Instead, he is supervising the rolls, and making sure you only get to make a guess when at least one of the die comes up 4.

It is somewhat ambiguous as to whether the question means 'at least one die shows 4' or 'exactly one die shows 4'. But the answer is either 2/11 or 2/10.

To show that the answer is 1/6, you would need to show that your interpretation of the word 'given' in that context is valid.

Your argument using the idea that some observer chose at random which number to tell you about would work, but that's not what's being stated here. I think you're right that this makes things problematic. We should conditionalize on all the information we have here, which includes the fact that the observer told us something, and not just what the observer told us. Unfortunately, if we don't know what strategy the observer is using, then we don't know how to do this. For instance, if she was using the strategy to always announce the lower of the two dice, then the probability of 7 is 0! And if she was always announcing the larger, then it's 2/7. So given that we lack this information, we have to appeal to our prior distribution over strategies, which is never explicitly stated, so there is no one right answer to this question.

However, one argument you make is definitely flawed. You point out that things should work out the same whether the number announced was 1, 2, 3, 4, 5, or 6. This is correct. However, when you thereby infer that this must be the same as the unconditional probability you are making a mistake. This inference is only valid if the possibilities you consider are mutually exclusive as well as exhaustive - but in this case, the sets of information conditionalized on in the different cases are not exclusive. Unless doubles are rolled, two of these events happen, not just one.

In the case where the events form an exclusive, exhaustive partition of the entire space, this principle is called "conglomerability" by people working in this field, and is close to one called "reflection" that was introduced by Bas van Fraassen in his paper "Belief and the Will" in 1984, I think. These principles play an important role in my dissertation that I'm writing, on the proper formal account of conditional probability.

Let us say that you flip two coins and are given the information that a head appears. Would you bet a penny they differ?

Let us repeat this procedure but let the information be that a tail appears. Would you bet a penny they differ?

Now consider this procedure being repeated many times, sometimes the information given being that a head appears and other times that a tail appears. If you bet each time according to the reasoning you used in the above two case would you expect to win money?

This is an elementary blunder.

In other words, in the case of 4-4, we are twice as likely to have a "4" announced than if a 3-4 was rolled. When we draw the sample space, given the announcement of a "4" then, we must make sure 4-4 appears twice as many times as each of the other combinations. That is: 1-4, 4-1, 2-4, 4-2, 3-4, 4-3, 4-4, 4-4, 5-4, 4-5, 6-4 and 4-6. There's 12 combinations there, and 2 that give us the favourable outcome of totalling 7. 2 in 12 is 1 in 6, as I originally proposed.

fizzle:
If I flip two coins once, don't see the result, but am told that one of them is a head, I would bet a penny that they are different. Same reasoning as with the Monty Hall problem.

If I repeat it, and this time, I'm told that one of them is a tail, I would bet a penny that they are different. Same reasoning as with the Monty Hall problem.

If we make an iterated game out of it, I wouldn't be able to make a reliable analysis without some sort of codified discussion of the observer's exact algorithm for deciding what to tell me.

That I would be willing to admit! Please direct me straight to the published conventions on describing probability problems with words ;-)

For that you have to go to class. :)

do you know any programming languages? if yes, write a simple program:

1. generate two random integers from 1-6.
2. if neither one is a 4, go back to 1 and repeat. else, continue.
3. record whether the sum of the die equals to 7.

run this a few thousand times and count the proportion of times you are returned that the sum equals 7. it will be very close to 2/11, rather than 1/6.

Further to alex's post, I get 18.18% - not a mathematical proof, but a very strong suggestion towards where the correct answer lies.

int roll()
{
return (rand() % 6) + 1;
} /* roll */

void main( )
{
long
Total=0,
Count=0;
int
Die1,
Die2;

while(1)
{
Die1 = roll();
Die2 = roll();

if( Die1==4 || Die2==4 )
{
if( Die1+Die2 == 7 )
{
Count++;
} /* if */
Total++;
} /* if */

printf( "%f% \r", 100.0*Count/Total );
} /* while */
} /* main */

As I think most of the commentors have agreed, the dispute is about translating the English question into a precise mathematical problem; not about solving the problem once you have it. The professor is following the standard conventions for how statisticians use language. This is a reasonable thing for him to test; although I have no idea whether it is his fault or yours that you haven't learned it in the course of the term. I've found this to be a very hard thing to pick up myself, but I can try to explain this point.

"Random event R takes place. Given condition C, find the probability of condition D"

means to compute

Probability of (D and C)/Probability of (C).

This is what is often denoted P(D|C), the conditional probability of D given C.

At times, it is difficult to imagine a physical scenario which would lead to this computation. In this case, what we have to imagine is that there is a maniac sitting in the room with the dice roller, who can be counted on to shriek at the top of his lungs if any four is rolled. (And he shrieks just as loudly if two fours are rolled.) Out of your sight, a man is rolling dice over and over. Suddenly you hear a shriek, and a bookie comes up and offers to bet that the dice came up 7. What odds should you accept?

While the definition of conditional probability is peculiar, it has the virtue that (a) it is well defined without specifying any information beyond the probability space and the two events and (b) it is always, always true that P(C and D)=P(C|D) P(D).

Indeed, conditioning on highly nonphysical things is, in my limited experience the key to proving all sorts of peculiar theorems. It is also, in my opinion, why probability papers are so darn hard to read :).

Regarding the coin flip paradox raised by fizzle, if the person deciding what to tell me is the coin flipper then I would not bet in this game. He is presumably deciding what information to reveal according to his own best interests. (This same ambiguity plagues the Monty Hall problem.) On the other hand, if the coin flipper promises ahead of time that he will tell me whether or not a head appears, and he then flips the coins and tells me that a head has occurred, I would certainly bet that the coins are different.

Hope this helps.

Maybe its me, but I am not fully comfortable with the preceding analyses.

"Given one of the dice shows a 4" is surely equivalent to "At least one dice shows a 4", at least that is what I would take it to mean straightaway.

Also, information from throwing the dice in succession may not be quite the same as information from throwing them simultaneously. Given the information "The first die shows a 4", then the probability that the second is 3 is of course 1/6. This is equivalent to labelling the dice and giving the information "Die 1 shows a 4" in the case of simultaneous throws. But that is not the same as "At least one is a 4".

Change the problem to an urn model. Two urns each have a set of chips numbered 1 thru 6. Someone makes a blind draw from urn A and urn B. The is a difference between "The chip from urn A shows a 4 (Heath Raftery's case)" and "At least one of the chips shows a 4 (the lecturer's case)" or equivalently "One of the chips shows a 4".

So you made the right choice of your answer, maybe for the wrong reasons.

Don't feel bad though, there is a relationaship between this apparent paradox and the famous one known as the Monty Hall problem, which I urge you to investigate. Only by this means (enlightening ourselves over thorny logical points) can we come to fully understand probability, which is a logic not "hard wired" in our brains.

David, great response, appreciated. I think this is the crux of the issue: I interpreted the question as a description of a real event. The lecturer simply wanted me to think within the walls of the lectures. This is a classic pedagogical mistake in my opinion, but common nonetheless.

Quite simply, if one treats the word "given" in the problem as the equivalent of the pipe symbol in P(A|B), then it's simple - 2/11. If one considers the reality of the situation described, then there is a leap of faith to assume that non-complying scenarios are being disregarded.

Thanks to all those that have responded. Clearly some of you have not read the entire thread, which is understandable since it is so long, but regrettable because there's a lot of repeated material here. I wont address each case individually. Thank you especially to the well reasoned and well thought out comments. I very much enjoy the argument you're offering.

I tried a million iterations of the following program with Matlab and got
>> probparadox
ans =
2.0103 1.0965
>> probparadox
ans =
1.9924 1.0867
The first number is /11 while the second is /6.
Without a doubt, 2/11 wins

n4=0;
n7=0;

for i=1:1000000
x=ceil(6*rand(1));
y=ceil(6*rand(1));
if (x==4 & y~=4)
n4=n4+1;
if x+y==7
n7=n7+1;
end
end
if (y==4 & x~=4)
n4=n4+1;
if x+y==7
n7=n7+1;
end
end
if (x==4 & y==4)
n4=n4+1;
end
end

[(n7/n4)*11 (n7/n4)*6]

Yep, the only ambiguity is in whether the meaning is "only one die shows a 4" (in which case there are 10 possible outcomes, of which 2 are equal to 7, so the probability is 2/10), or "at least one of the dice shows a 4", in which case there are 11 possible outcomes (since 4-4 is now allowed) and the probability is 2/11. Getting angry with the prof is not helping you- I think the emotional frustration has been blocking you from getting the picture.

Now, if the question had been: "I roll two dice. I choose one of the dice at random and look at its value. I find that it's a four. What is the probability that the total of the two dice is 7?", then the answer would indeed by 1/6, as we're asking what the probability is of the other die showing a 3.

I'm sorry to be annoying, but I think you are still missing the point - with the "at least one" there is no ambiguity in the original question! Also I feel that in your post you decide your professor was wrong too quickly. Even now, you are still discrediting your professor by saying that there is a "leap of faith" in the what is meant by the notion of "given". Actually, you do not have to assume some abstract definition of "given" - the mathematical definition corresponds with the natural everyday notion. In fact it is your solution which is quite contrived, as the oracle giving you the information does so in an extremely unusual way.

If you are given something happens in a problem... you take as a _given_ that it happens! You say that the problem doesn't explicitly state "non-complying scenarios are being disregarded". Of course they are, what else would you mean by given? Are you really saying that if the problem was "Given that a single dice roll was a 6, what is the probability it was a 1?" you would answer 1/6 because you can't disregard non-complying scenarios?

You are not describing the solution I posted. Re-read it, and if you still believe I said what you think I said, quote me.

You say "Suppose that this observer made their observation by announcing the value of the die that first stops rolling.", which is immediately wrong because the observer must consider both dice.

A specific number? Like a "4"? Or a "5"?

Yes, a specific number; 1, 2, 3, 4, 5 or 6. But you say that since one of these numbers always appears you could conclude your "chances of guessing the 7 total are better". As Kenny pointed out, the possibilities are not mutually exclusive.

But there was nothing about special 4-producing dice in the question, so I didn't assume that was the case. What makes you think it is?

Rolling 4-producing dice is only one way to think about it, another way is to imagine that two regular dice are rolled and an oracle gives you the specific information and nothing else.

This is similar to a problem I'm having which brought me to this page. Maybe someone can clear up my problem. I'll take a shot at the original.


Original Problem:
Okay, so you have 2 dice. You roll them both and close your eyes. Your buddy looks at the dice and shouts out one of the numbers: "Four!", he yells, melodramatically. Now you have to say what the probability that the two dice together equal 7.


You know that the number of outcomes that include a four (counting that double once, since it's only one outcome, the sneaky devil!) is 11. You know that of those outcomes, only 2 of them add up to 7.


Quick note: your friend is actually Charles Manson and he's told you that if you don't guess the right probability, he's going to cut off your left foot. While working out the math in your head, you realize how lucky you are that he said he only wanted the probablity of the total equalling 7, and didn't say you had to corretly guess what the number was on the second die.


Okay, so you know that the number of outcomes that include a 4 is 11, and you know only two of those outcomes equal 7, so you correctly shout "Please don't hurt me! The probability is 2/11!" and he cackles and runs off to find another victim.


Now, your frustration seems to be rooted in the semantics of the question. You are reading the problem as "Charles Manson wants to cut off your foot, and agrees not to if play a weird math game with him. He makes you roll two dice with your eyes closed and then shouts 'One of them is a four!' and then demands that you give the probability that the other die had while you were rolling them of coming up as the three needed for the dice to total 7." In this scenario, you're totally right: the odds of either die being part of the total of 7 is 1/6. The outcome of either die does not change that probability. Knowing the outcome of one die should not change your expectations of what the second die will be, but does change your expectations of what the total sum will be. If you roll two die, and the first one is a 3, you can't say "well I still have a 1/6 chance for rolling snake eyes!"


So your reading of the question was "what is the probability that the second die will be a 3 thus making the total 7" versus the professor's intention, which was "what is the likelihood that the total sum will be 7 if you know that one of them 4". If I was taking that test, I would have read it the way the professor intended, simply because if I'm given extra information, I tend to assume that it's meant to be used toward the problem, at least when faced with a possible ambiguity.


Part of the problem (I think) is that the sum is supposed to equal 7, and any throw on one die leaves the chance to get a 7 with the other die. Let's say you were playing a modified version of craps where you bet up front what number (total) you are going to throw, you throw the dice under the table. The dealer looks under the table and yells out the outcome of one of the dice. You are allowed to either fold, stand, or double your bet for the second throw. You bet you will throw a total of 5 because 5 is your lucky number. You know that before you throw the first die that your probability of throwing a total of 5 is 4/36 or 1/9. You throw the first die and you get a 5. What is the probability of getting a 5 now? Or, you roll a 3. Are your odds still 1/9, leading you to stand? Or are they 2/11, leading you to double?


I have trouble getting this stuff straight all of the time, my instinct is like yours because it was hammered into my head in grade school that the amazing thing about probability is that each event is exclusive and the probability is the same for each event. Thus it really is possible to throw a coin 100 times and have it be heads each time.


Or, to look at it another way (man, I'm going to be killed for making up math):


If you are throwing the dice in your own scenario, where you say "no matter what the first die is, the odds of the second one are definitely 2/11!" the problem with that situation isn't that your odds are better for getting a 7 no matter what, it's that the probability for the second die DOES change based on the first one, but it doesn't matter since in the case of 7 the odds change for all possible situations. You don't stand a better chance, you just know that the second die has a 2/11 probability based on the first one. If both dice are thrown at the same time, each one has a 1/6 (or 2/12) probability because they are exclusive from each other, but when you look at one before the other, the probability changes from 2/12 to 2/11 because the first die takes away the possibility of the total not including that number.


That last part was totally out of my butt, I admit. Can anyone infer what I'm getting at and set me straight?




Okay, on to my problem, which is (hopefully) much simpler:


When playing backgammon, there are times when you get bumped off the board and there is only one free space for getting back on the board. You have two dice and have to roll the number of that space to get free and back in the game. At least one of the two dice has to be that number, it can't be the total of the dice (so if the number to get free is 6, you're stuck and very pissy if you roll two 3s.) I was trying to figure out the probability of rolling that number in this situation. My instinct was 1/6, but then I started writing out the possibilities to be sure. If there are 11 different combinations with a 6 (and 6 is the number you need to roll), that means that I have a 11/36 chance of getting the number I need. This is all well in good, but that would mean that the probability of rolling and getting a number I don't want (let's say a 2) would be 11/36 and thus, the odds of getting any number would be (11/36)*6, which equals 66/36 (and I would have a really good shot at getting the number, which also seems wrong). That doesn't seem right at all. If I change it up a bit and instead figure out the number of possible 6's total (not counting the double) over the total amount of individual dice possible (11/72) then my odds are closer to fair, but I end up with 66/72 when I multiply the odds out for each number. The only way I can work out the missing 6 is to say those are the double thrown out, but if I account for the doubles for any number I'm not trying to get, I still end up with 11+(12*5) over 72 (71/72). So really, my question is: how do I figure out the odds for throwing the number I need? do I throw out the doubles? Do I throw out all the doubles or just the one for the number I need?

CuBr:

If you are given something happens in a problem... you take as a _given_ that it happens! You say that the problem doesn't explicitly state "non-complying scenarios are being disregarded". Of course they are, what else would you mean by given? Are you really saying that if the problem was "Given that a single dice roll was a 6, what is the probability it was a 1?" you would answer 1/6 because you can't disregard non-complying scenarios?

Re-read the original question. We were told that two dice were rolled. Boom - 36 possible outcomes. Then we are "given" that a 4 appears. I interpret that as being "given" more information, not that the scenario is a given. Subtle, isn't it? As I've said, I could have been "given" the information that a 1 appears, or a 2, 3, 5, or 6 appears. That doesn't change the fact that two dice have been rolled simultaneously. Read the way I contrast this situation with the women with two children problem for more clarification.

You say "Suppose that this observer made their observation by announcing the value of the die that first stops rolling.", which is immediately wrong because the observer must consider both dice.

Yes, a specific number; 1, 2, 3, 4, 5 or 6. But you say that since one of these numbers always appears you could conclude your "chances of guessing the 7 total are better". As Kenny pointed out, the possibilities are not mutually exclusive.

Rolling 4-producing dice is only one way to think about it, another way is to imagine that two regular dice are rolled and an oracle gives you the specific information and nothing else.

Tony:

Knowing the outcome of one die should not change your expectations of what the second die will be, but does change your expectations of what the total sum will be. If you roll two die, and the first one is a 3, you can't say "well I still have a 1/6 chance for rolling snake eyes!"

So your reading of the question was "what is the probability that the second die will be a 3 thus making the total 7" versus the professor's intention, which was "what is the likelihood that the total sum will be 7 if you know that one of them 4".

You bet you will throw a total of 5 because 5 is your lucky number. You know that before you throw the first die that your probability of throwing a total of 5 is 4/36 or 1/9. You throw the first die and you get a 5. What is the probability of getting a 5 now? Or, you roll a 3. Are your odds still 1/9, leading you to stand? Or are they 2/11, leading you to double?

If both dice are thrown at the same time, each one has a 1/6 (or 2/12) probability because they are exclusive from each other, but when you look at one before the other, the probability changes from 2/12 to 2/11 because the first die takes away the possibility of the total not including that number.

Okay, on to my problem, which is (hopefully) much simpler:

P(F) = ( P1(F) x P2(F)' ) + ( P1(F)' x P2(F) ) + ( P1(F) x P2(F) )

In other words, you are successful if the first die is successful only, the second die is successful only, or both dice are successful. Let's stick the numbers in:

P(F) = ( 1/6 x 5/6 ) + ( 5/6 x 1/6 ) + ( 1/6 x 1/6 )
= 5/36 + 5/36 + 1/36
= 11/36.

Same answer. Finally, this can be simplified by considering that a favourable outcome is the conjugate of an unfavourable outcome. That is:

P(F)' = ( P1(F)' x P2(F)' )
= 5/6 x 5/6
= 25/36.

To find the conjugate (the favourable outcomes), we subtract from 1:

P(F) = ( 1 - P(F)' ) = 1 - 25/36 = 11/36.

Right, so that's sorted. Now you wisely tried to double check your answer by considering other scenarios. You are right in assuming the favourable outcomes for any number (not just a '6') are still 11/36. You were wrong in implicitly assuming each case was mutually exclusive. The odds of getting a '6' overlap with the odds of getting a '5'. In other words, the outcomes are not mutually exclusive. To add the probabilities, we would have to remove the overlapping outcomes:

P(5 or 6) = P(5) + P(6 without a 5)
= 11/36 + (11/36 - 2/36)
= 20/36

And so on for the other numbers:

P(any number) = P(6) + P(5 without a 6) + P(4 without a 5 or a 6) + ...
P = 11/36 + (11/36 - 2/36) + (11/36 - 4/36) + (11/36 - 6/36) + (11/36 - 8/36) + (11/36 - 10/36)
= 11/36 + 9/36 + 7/36 + 5/36 + 3/36 + 1/36
= 36/36.

you people have way to much free time

Anonymous, I appreciate you writing in and voicing your insightful thoughts. I'm impressed you took the time to read and comment on something you're clearly not interested in. Perhaps you'd also like to comment on a English language blog somewhere, since you're clearly uninformed in that area as well.

Double check your watch - you'll find that we both have the same amount of time available to us. I choose to host a discussion on a topic I find interesting. You choose to grace strangers with a pathetically formed insult.

Move along. You'll find plenty of other people who believe showing interest in intellectual pursuits is a sign of too much free time. You can help each other achieve nothing together, lest one of you be seen to be pursuing an interest.

My initial take was that the lecturer was wrong to count reflections (if that's what you would call them -- I mean, e.g., both 4-3 and 3-4), because they are really two statements of the same event, i.e., "one die shows a 4." You don't care which die it is; therefore five of the events in the lecturer's sample space are redundant and the chance of a total seven is 1/6.

But you brought a great insight: if the lecturer wants to count reflections, he has to count 4-4 twice. So the sample space is 12 and the chance is 2/12 or 1/6. Bravo.

(I didn't read through all the comments, so I apologize if I'm repeating something that's already been said.)

Interesting perspective mgarelick, thanks for sharing. You've definitely added something new to ponder.

I just re-read Tony's comment and thought of an important clarification: I am not disregarding the new information offered to me about the 4 appearing. It is actually a coincidence that my answer doesn't change! Here's why:

Chance of getting a 7 total from two dice: 6/36 = 1/6.

Chance of getting a 3 when I've already rolled a 4 (and therefore adding to 7): 1/6.

Don't let the fact that they are the same number lead you to believe that I'm disregarding the new information. As Tony cleverly pointed out - if I was rolling for a total of 3, and I roll a 4 first, my chances of getting a 3 total have plummeted! But, if I'm rolling for an 8 total, my initial likelihood is 5/36 - if a 1 comes up I'm buggered, but if a 2, 3, 4, 5 or 6 comes up, I would then have a 1/6 chance of getting my total.

In our Stats lesson today we discussed this "Probability Paradox" and how absolutely absurd it is to have these different logical paths leading to two different, but completely justifiable answers, one of which is conventionally acceptable, and one of which is acceptable to normal human reasoning.

We discussed that the understanding of the complexes of probability do not occur naturally to the human mind. More often than not the "technically correct" answer is not normally logically justifiable.

The amount we choose to spend ruminating on a particular problem is dependent on the motive we have behind studying the subject. If your aim is just to give the technically correct answer, you will not go beyond the 2/11ths answer that you have been taught. But if you choose to think more about this problem and its solution (or lack of it) the more disillusioned you get with the 2/11ths answer that is conventionally acceptable. The mind then veers towards this solution of 1/6th which initially seemed much too obvious to be the correct answer.

This is the problem with such questions, you begin to question the solutions you see and veer on to the question of WHAT exactly is an ACCEPTABLE solution in the first place...

LOL AIDAN WE ALL KNOW THIS IS UR BLOG HAHAHAHAHAHAHAHA

Thanks for writing Rohan, that's an interesting take on things. It just goes to show how sloppy seemingly precise language is when it is subjected to mathematical rigour.

Who's Aidan? This is my blog!

What a lot of useless talk! Talk until the Big Crunch, but the arbiter is the EXPERIMENT. And I am a theorist! - still I worship at the alter of experiment! I care not a whit for all this talk on who is right. Abandon all useless theory, ye who enter here!

Pick up a pair of dice, sit down with a piece of paper and some ink, and start throwing and inking. Throw away all events that are not in the domain (sample space) i.e. no 4 showing. If there is a 4, log the throw as a countable event. Is the total 7? Yes, log it as a winner, else a loser. After some finite time, your winner/event count will converge on a number. What number is it?

Now is the time for explaining the number!

Since you seem to have identified the real problem, whether the chance of 4-4 is 1/36 or 2/36, that is an even easier experiment to do, with a lot more spread between the hypothesized outcomes than between 1/6 or 2/11. Very quickly you see that the chance of 4-4 is 1/36, which is why doubles are so highly valued in dice games.

All this gab reminds me of the time when everyone knew big balls fell faster than small balls. No one would have sullied their hands actually doing it! Really, you people would have saved a lot of time actually doing it instead of debating it.

I am an educator, so this is not a criticism of all of you per se, it is a criticism of our education system that we do not teach problem solving.

There is a very clear theoretical method for solving this problem. Since the dice are independent, and the chance of each number for a given die the same, write a matrix - 1-6 across the top and 1-6 down the left. We all agree that ALL possibilities are covered correctly.

Now fill in the combinations, giving the ALL 36 possible throws on the two independent dice. Note well that there are only 6 doubles, along the diagonal. 4-4 does not appear twice, as do 4-3 and 3-4. Striking all rows and columns without a 4 leaves 11 numbers, of which exactly 2 add to 7. 2/11 is what the theory says, and 2/11 is what the experiment verifies.

It is a mistake to think 4-4' is distinguishable from 4'-4, because the numbers are indistinguishable, ie 4' from 4. We say in Quantum Mechanics that the dice are indistinguishable. What we mean is that the throws are independent, and the results of one dice indistinguishable from the results of the other (the numbers are independent). With dice, there are four independence games that can be played – O (order), D (distinguishable), OD (both), and M, the Matrix game we just did, which has neither O nor D. It turns out, M covers them all, that is, the theory applies to all of them.

Yes, I know, we can make the dice distinguishable, one could be Big RED and the other small blue - but this is precisely game “D”. (R4,b3) is distinguishable from (b4,R3), but (R4,b4) is indistinguishable from (b4,R4). By indistinguishable using distinguishable dice, we mean: The ORDER DOES NOT MATTER. RED first is the same as RED second. 4+3=3+4=7.

Even if we try making ¾ the winner, the analysis does not change.

If you wish to abandon the distinguishable dice, and go to order matters, fine, this is game “O”. (3,4) is different from (4,3), but (4,4) can only occur 1/36 throws.

So clearly, “O” and “D” are the same games as “M”, which I shall simply call “M”.

Now if you want to have your cake and eat it too, order matters and distinguishable dice, game “OD”, fine. We can do that experiment too.

Now (R4,b3) is different from (b3,R4) is different from (R3,b4) is different from (b4,R3), and of course (R4,b4) is different from (b4,R4). But that's it. You got no free lunch. There are now 72 possibilities and 22 of them have a 4, and 4 are winners, and we are back to 4/22, or 2/11.

What you CANNOT do is choose to COUNT the “OD” doubles (R4,b4) and (b4,R4) (odds 2/72) with the “M” winners (3+4) or (4+3) (odds 2/36). You can match odds, but not counts from different games.

So the odds in “M” and “OD” are the same. How do I know? The real answer is the experiment, but all of this arises from dice that are INDEPENDENT. Therefore one die does not know what the other says. Perhaps instead of indistinguishability, we should say independence. The idea is subtle, but from the matrix it jumps out at you.

After all this yaking, no one came even close!

The answer to 6+6? Doesn't this depend on the base? I mean 10 is a perfectly good answer!

Q: How many possible answers are there, given all possible bases?