Basic Probability Paradox
This is actually a continuation of my frustrations with University assessment, masquerading as a mathematical paradox presentation. In actual fact I don't believe there is any paradox. Instead, I believe the University assessors are too interested in teaching people to regurgitate their gospel to worry about issues like the truth. Lets start with the problem description.
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"?
I maintain that the answer is 1 in 6, and yet in a recent quiz I answered 2 in 11, because I knew that's what the lecturer believed. Yep, I actually swallowed my stubbornness, played their game and took the marks. Not mind you, before spending sometime one on one with the lecturer discussing the issue. The only thing I got out of that discussion was that there would be no discussion on the issue. As for the rest of university, the answer is what the lecturer says it is.
Now let me describe the reasoning behind the answers. Firstly, the lecturer's answer of 2 in 11:
Justification of the lecturer's answer
In throwing two dice we have a sample space of 36 outcomes. If one of the dice is a 4, that sample space is restricted to these 11 outcomes (written as first dice-second dice): 1-4, 2-4, 3-4, 4-4, 5-4, 6-4, 4-1, 4-2, 4-3, 4-5, 4-6. If we count the number of those outcomes that have 7 as the total, we find two favourable outcomes (3-4 and 4-3), giving us the probability of 2 in 11. Simple eh? Wrong.
Paradox posed by lecturer's answer
Say I have the two dice in my hand, and I drop one of them (I don't know which one, whichever falls out first). What are the chances it is a 5? 1 in 6 of course. Okay, say I don't make a guess at the first die, but I am told it is a 4. If I then drop the second die, what are the chances it is a 3? Same as dropping any other dice - 1 in 6 of course! A crucial fact in probability theory is that the probability of an event is not effected by history. Rolling a "4" on a fair die has always happened 1 in 6 times, regardless of how many "4"'s (or any such things) have been rolled in the past.
Okay, so lets look at the original question again. Two dice have been rolled. If I were to make a guess that the total of the two dice is 7, I would have a 1 in 6 chance of being right. I don't think there is any controversy here - there are 6 ways out of 36 that the total can be 7, which is equivalent to 1 in 6. Say just before making my guess, I'm given reliable information that one of the dice shows a "4". According to the lecturer, now that I have more information about the situation I can be more confident that the total is 7. That is, my odds of being right about the total being 7 has increased from 1 in 6 to 2 in 11.
Suppose for a moment that I was told one of the dice shows a "1". Again, according to the lecturer, my chances that the total will be 7 have increased to 2 in 11. What if I was told that one of the dice shows a "2". Still 2 in 11? What if I was told a "3", or a "5" or a "6". By the same logic, in each of these scenarios, armed with the information of what appears on one of the dice, I can bet my pretty pennies with extra confidence, because the chances that a total of 7 results have increased from 1 in 6 to 2 in 11. So knowing beforehand that a "1" appears, or a "2" or a "3" or a "4" or a "5" or a "6" appears to increase my chances of guessing the total of 7, compared to not knowing that a "1" or a "2" or a "3" or a "4" or a "5" or a "6" appears. 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. 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 11. I maintain they did not. Rather than leave this proof as the contradiction already laid out (proofs by contradiction always leave me unsatisfied) I'll explain my reasoning as to why the chances of the 7 total are always 1 in 6.
Justification of my answer
Rolling two dice "simultaneously", as explicitly specified in the question, makes no difference to the dice then if they were rolled one after the other. Neither die knows the other one has been rolled. If I were to make a guess on the value of first die to stop rolling, I would have a 1 in 6 chance of getting it right. If I were to make a guess on the value of the second die to stop rolling, I would have a 1 in 6 chance of getting it right. As I mentioned before, the probabilities don't change over time nor due to events happening nearby. Say the first die to stop rolling is the green one and the second the blue one. If after the dice were rolled, I was told that the green one shows a 4, that doesn't change the fact that I have a 1 in 6 chance of guessing the value of the blue one. The dice are independent objects and their outcomes are determined completely without regard of each other.
In fact, even if the dice are not uniquely identifiable, and all I am told is that a "4" appears, I still have a 1 in 6 chance of guessing the value of the other dice. What do I care that the dice I am guessing is blue, or that it was rolled first or that it stopped rolling last? All I care is that a die has been rolled and I'm guessing it's value. But hold on, isn't a double less likely than any other pair? Doesn't that mean that I would be foolish to guess that the other die shows a "4" as well? Well actually, no. I would have just as much chance of being right if I picked 4 than if I picked 5 - that is, 1 chance in 6. The subtle point is that there are actually 5 outcomes which don't produce a double and 1 that does. A double 4 is only more special than a 4 and a 5 because we place special emphasis on it. The chances of a double are still 1 in 6, and a non-double 5 in 6, just as in the case of two unknown dice.
Returning to the question then, if I did not know the value of either die the chances that the total is 7 would be 1 in 6. The favourable outcomes are 1-6, 2-5, 3-4, 4-3, 5-2 and 6-1, that is, 6 favourable outcomes out of 36 possibilities. Look at the favourable outcomes - regardless of the value of the first die, there is only one value of the second die which produces a favourable outcome. Therefore, if I am informed of the value of one of the dice, regardless of which one and of which value, there is a 1 in 6 chance that the other die shows the value I want. So the probability that the total is 7, given the value of one of the dice, remains 1 in 6.
Why the lecturer's answer is wrong
Finally, I want to show where lecturer's assumptions went wrong in coming up with the answer of 2 in 11. Once the sample space is written out, with the 11 combinations of two dice where a "4" appears, and the favourable outcomes identified, it is quite a straightforward deduction to arrive at the answer of 2 in 11. I propose that the error was actually made in the construction of the sample space. The sample space of the two dice is the regular pattern of 36 combinations, each appearing once. In this problem however, it is known that one die shows a "4". Lets imagine how this might occur.
Obviously, there was an observer who provides the extra piece of information. Suppose that this observer made their observation by announcing the value of the die that first stops rolling. In doing so, they have specified a distinction between the dice. This is identical statistically, to following the yellow dice, or the bigger one, or the one that was rolled first. In any case, the decision about which die to call has been made before or regardless of the actual rolling event. 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. Of course, here there is one favourable outcome (4-3) and the probability is simply 1 in 6.
Suppose instead, that no decision is made until the dice are rolled. In other words, there is no distinction between the dice and both dice are considered with equal interest, after they are rolled. Say then, our observer announces the value of the first die half the time and the second die the other half of the time. Say also, that our observer lets us know that a "4" appears. What are the possible outcomes that may have occurred? Well if a "4" and a "3" appeared, then we have a 50% chance that our observer announces the "4" (the other 50% of the time the observer would announce the "3"). The same thing applies to the 10 combinations 1-4, 4-1, 2-4, 4-2, 3-4, 4-3, 5-4, 4-5, 6-4 and 4-6. The exception is if a 4-4 was rolled. In that case, we are guaranteed that our observer will announce a "4". 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.
One last thing...
As an epilogue, there is one question left unanswered - what question has the lecturer answered? There was a lot of merit in the process, and certainly in isolation it seems to make sense, so why doesn't the problem fit? Well I imagine the solution would fit if the following small but critical, extra criteria were adhered to: In each rolling of the two dice, if no "4" appeared, the situation is thrown out, disregarded, and the two dice are rolled again. If we are only asked for the probability of a 7 total once our observer gives us the go-ahead (that is, once at least one "4" appears), we do indeed limit our sample space to that used in the calculation of the 2 in 11 probability. I imagine in this case the answer of 2 in 11 would be accurate.
And there endth this insanely long discussion of a seemingly simple problem. Am I wrong? I'm very interested to hear where my reasoning has missed the boat.