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I throw two coins (simultaneously). A student (very much a beginner in both math and probability theory) thought that the following 3 outcomes are equally likely: "two heads", "two tails", "a head and a tail".

However much I tried, I couldn't find a clear and obvious explanation of why it's not the case.

Of course, I could say the coins are distinct, so we need to look at how each individual coin falls (leading to 4 equally likely outcomes: HH, HT, TH, TT). But I couldn't clearly explain the concept of "distinct" and why it's important for the probability calculation.

Anyone can help with a simple, precise and very intuitive explanation?

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Do you really want to wreck their intuition about bosons? :) – wnoise Apr 18 '11 at 23:33
I actually thought about it... How to explain to someone the "normal" probability and yet leave open the possibility that it could be different (as in the case of quantum physics)? – max Apr 25 '11 at 0:21
up vote 7 down vote accepted

Try painting one of the coins red. Paint can't possibly affect the probabilities of the oucomes.

Alternately, get ten coins and ask the student how many outcomes she thinks there are and whether ten heads is more or less likely than five heads and five tails.

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Wow very nice idea about the paint! – max Feb 3 '11 at 9:35

Arrange with the student that you'll give them 1 dollar for each HH, and they'll give you 1 dollar for each HT. Then start tossing. Either they'll get the point quite fast, or you'll make a nice profit, so you win either way.

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+1: Although I would change this slightly and offer to give them 2 for each HH and they give you 1.5 for each HT (or TH), so that to their way of thinking they will come out ahead - until the penny drops :-) – Derek Jennings Feb 1 '11 at 9:54

Try to draw the following diagram for him.

enter image description here

-Let him explain why every arrow represents the probability $\frac{1}{2}$.

-Let him find all paths from Start to the result he wants (TH, or HT).

-Let him calculate the total probability.

These steps should be elementary enough for him to get the correct answer.

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There are 4 outcomes, not 3: "HH", "TH", "HT" and "TT". And they are all equally likely, $p = \frac{1}{4}$. So $P(HT) = \frac{1}{4}$. If you don't distinguish between "HT" and "TH", then $P("HT or TH") = P(HT) + P(TH) = \frac{1}{2}$.

I wouldn't write $P(HT) = 2 P(HH)$. This is just my opinion but expressing something in terms of something else that only works by coincidence in this particular case is very confusing, especially to a learner. When your coins are biased, it's broken.

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+1 for the second paragraph. – zyx Aug 31 '11 at 23:17

To distinguish "T" and "H" would be our task. Why not transfer the problem into other settings. For example. We call two people walking together as "TT", while they riding two bycycles as "HH". Then when A is riding one bycicle and B is sitting at the backseat, we denote it as "TH". Finally, "HT" is also similiarly defined.

Such explanations can be various.

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