The integers 1 through 6 appear on the six faces of a cube, one on each face. If three such cubes are rolled, what is the probability that the sum of the numbers on the top faces is 17 or 18?
I see a fundamental flaw in this problem. Are the die distinguishable? The question does not specify. The answer key, however, treats the die as distinguishable.
I’ve asked my math teacher, and he pretty much said you should just treat them as distinguishable. This was after I pulled out a set of identical die, rolled them, and asked him to distinguish among the three identical white die :rolleyes:.
I’m saying that there are two ways of looking at this problem:
Way 1:
You can get 17 from:
Rolling a 5, 6, and then a 6
Rolling a 6, 6, and then a 5
Rolling a 6, 5, and then a 6
18 = 6, 6, 6
Way 2:
17 = 5, 5, and 6 (indistinguishable)
18 = 6, 6, 6
The problem is that the problem as stated makes no provisions for distinguishing among the die. If one were to toss all the die at once, is there a difference between a 6, 5, and 6 and a 6, 6, and 5?
You have 2 coins. What are the possible random results when flipped?
Heads-Heads, Tails-Tails, Heads-Tails, Tails-Heads. If I understand you correctly, those last two are what you’re talking about - whether they count as independent events, or if you just consider heads/tails as one possibility.
No, they’re two distinct possibilities. Taking away the idea that Heads-Tails is an equal likely result to heads-heads and tails-tails would suggest that each has a third of a chance of happening, but it’s very easy to demonstrate that each has a quarter, giving 1 heads 1 tails a cumulative 50% of the results, because both heads/tails and tails/heads would have 25% each.
If you took those two coins and flipped them on opposite sides of the table, they’d be distinguishable - your left coin could come up heads or tails and the right coin could come up heads or tails independently. If you shake them up together in a cup, thereby making them indistinguishable, do you think you’ve changed the odds of how they’ll flip?
It’s not about distinguishability vs. indistinguishability. It’s about the assumptions you make. The faces of a die don’t have equal probability of coming up because they’re the 6 distinguishable results of a roll; they have equal probability of coming up because you implicitly assume they have equal probability of coming up. One could, of course, discuss biased dice instead; it’s simply a linguistic convention that one usually isn’t.
In a pair of coin flips, HH, TT, HT, and TH all have equal probability of coming up not because HT and TH are considered distinguishable; you could stop distinguishing between them, and partition outcomes into only “two heads”, “two tails”, and “one of each”, with the latter twice as likely as each of the former. It’s not distinguishability vs. indistinguishability which determines the probabilities; it’s the implicit assumption that individual coin flips have equal heads and tails probabilities, while separate coin flips are probabilistically independent. You could change these assumptions, but the linguistic convention is that they hold unless otherwise noted.
I just find it impossible to tell whether it is him or the the other guy
For the OP:
The outcomes are pretty distinct :-
[QUOTE=IceQube]
Way 1:
You can get 17 from:
Rolling a 5, 6, and then a 6
Rolling a 6, 6, and then a 5
Rolling a 6, 5, and then a 6
18 = 6, 6, 6
[/QUOTE]
The likelyhood of getting 17 is three times that of getting 18.
[QUOTE=IceQube]
Way 2:
17 = 5, 5, and 6 (indistinguishable)
18 = 6, 6, 6
[/QUOTE]
The probability of getting a 17 is the same as getting an 18.
It is pretty simple to demonstrate which is correct - roll 3 dice 500 times and record the 17s and 18s.
I am pretty confident (i.e I’d lay a small bet) that there will be at least twice as many 17s as 18s. The more times you roll the dice, the more distinct the results (i.e I’d put more money on the outcome).
If way 2 is correct, the numbers should be about the same.
Just to note, there are 216 (6[sup]3[/sup]) distinct combinations of 3 dice, so for the stats to look reasonable, you need a large number of trials.
You could also add in the identical probabilities (111=3 equivalent to 18 and 112=4 equivalent to 17) to get a clear result (one probability 3 times that of the other) with fewer trials.
Here is the probability table for the sum of 3 die…
My take on the problem: Are you suggesting that the color of the dice affects the statistics to be expected from the outcomes, such that using three identically colored dice will give different statistics than three differently colored dice?
Are you suggesting that if you couldn’t see the die rolls at all, but only hear the announcement of a friend rolling them in the other room, “Yes, they all came up 6” vs. “No, they didn’t all come up 6”, the odds on those two possibilities would be 50-50?
Etc.
These should help you narrow down what your actual implicit assumptions are and construct an appropriate probabilistic model for the situation. (Hint: The appropriate assumptions are almost certainly that each particular die comes up on each particular side with equal frequency, independently of the other dice). But, again, it’s not about distinguishability. That has nothing to do with it. It’s about your assumptions/probabilistic model.
I believe the difference between distinguishable and indistinguishable affects the possible and total outcomes equally, i.e. 1/3 becomes 3/9. So the final probability remains the same.
Suppose the three dice are three different colors.
And suppose you are asked: How many ways could you roll one 5 and two 6’s?
Red Blue Green
--- ---- -----
5 6 6
6 5 6
6 6 5
In this counting, it matters what numbers come up on which colored cubie. So there are three ways.
But if the dice are all the same color, they all look alike. You MIGHT say that there is only one way.
But you might, instead, distinguish the three dice by “first one”, “second one”, and “third one”. Or “the one on the left”, “the one in the middle”, and “the one on the right”.
Moral of the story: You need to be very sure exactly what the question is!
In general, you always need to know which of these questions you are being asked: If the order of the individual dice matters, or if the order doesn’t matter.
When you are adding up the probabilities, you need to treat them as if the order matters. (Or, as if the dice are different colors. Or as if you distinguish “first rolled”, “second rolled”, “third rolled”.) If the dice are all one color, then the three possibilities in the table above look all alike, but there are still three times as many chances for 17 to come up, just as if they were different colors.
You don’t even need colors. Just imagine that there are three people each rolling one white die. How many ways can they have seventeen? Is there a difference between three people each rolling one die and one person rolling all three dice?
Or just roll the same die three times. This just shows again that distinguishability is irrelevant because the die would be literally identical. All that matters is that the three die-rolling instances are separate.
If I had been your teacher in this situation, I would have put a chalk mark on two of the dice, handed them back to you, and asked you if you thought what I had done would change the probabilities of the outcomes.
That said, this notion of indistinguishability does arise in quantum mechanics, where you literally cannot tell the difference between “Particle A is in State #1 and Particle B is in State #2” versus “Particle B is in State #1 and Particle A is in State #2”. So if they were bosonic quantum-mechanical dice, and you randomly picked a state of the three dice, your second method would be correct; {6, 6, 6} would be just as likely as {6, 6, 5}. (If they were fermionic dice, neither outcome would be allowed. But that’s another story.)
I see. Getting 17 or 18 doesn’t matter whether the dice are the same or not. The probability of getting 17 is always greater than the probability of getting 18.
You can have indistinguishablity in non-quantum systems, but it requires a bit of effort. Imagine that you roll the three dice behind a divider (so you can’t see all three dice at once). To find the result you reach behind the divider, pull out one die, look at the top to see what the answer is, and return the die to behind the divider (without changing what number is on top), then do the same thing twice more. Because you can’t tell whether you’ve pulled out all three dice, or inadvertently pulled out one dice twice (or three times), the statistics for the result become the “indistinguishable” stats that are see in quantum mechanics.
