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Let's say I have a bag of coins, which contains $1$ quarter, $2$ dimes, $3$ nickles and $4$ pennies. If I were to randomly pull out $3$ coins, on average, how much money would I get?

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    Looks like homework...2010-10-10
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    Sorry, not homework (although tangentially, I guess). I'm doing some research into cutsets in graph theory and my stats is pretty dull, but I'd like to know things like the average weight cut on certain sized cutsets. This is a toy problem I was hoping to see an answer to, to better understand the stats2010-10-10
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    Okay, assume you have a graph where the cost of removing a node is equal to it's level (so the root is cost 1, its children cost 2, its children's children cost 3, etc..). One thing I'd like to know is what the average cost of removing three nodes would be. You can't pick and choose which, just three random nodes. That seems analogous to the toy problem presented, to me2010-10-10
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    How you draw the coins might be relevant. Is it one after the other? Or all at the same time (i.e. every set of 3 coins is equally likely)?2010-10-10
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    Not sure I understand the question. Are you asking if it's 'with replacement'? If so, no. Once a coin is selected, it is removed (for that 3-coin selection). If I were to repeat the 3-coin selection process, I would start again from the initial state.2010-10-10
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    I meant do you pick three coins in a bunch or pick them separately without replacement. In one case the order matters, in the other it does not. For instance, if you pick one after other: penny1, penny2, quarter and quarter, penny2, penny1 are different outcomes. Also, are the coins of same type distinguishable? So we can talk of penny1, penny2 etc.2010-10-10
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    Makes sense! Thanks for the clarification. Picking them all at once is fine, the order doesn't matter. Let's also say that coins of the same type are indistinguishable from one another.2010-10-10
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    I suggest you edit your question to state all those assumptions.2010-10-10
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    @Moron: the answer to the original question doesn't depend on whether you pick them simultaneously or separately without replacement.2010-10-10
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    @Qiaochu: I suppose the expectation remains the same. But making the coins of same type indistinguishable matters, I believe.2010-10-10
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    @Moron: not without contradicting the implicit setup of the problem. As long as each individual coin is drawn with equal probability it does not matter whether you consider them indistinguishable or distinguishable.2010-10-10
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    @Qiaochu: From comment by coffee: "Picking them all at once is fine, the order doesn't matter. Let's also say that coins of the same type are indistinguishable from one another." So I am interpreting the outcomes as u quarters, v nickels, w dimes, x pennies where u+v+w+x = 3. The fact that you speak about 'implicit setup' means we might be making different assumptions. But I think I agree with your interpretation though. It might be irrelevant too, as you say in your answer. I apologize for creating confusion.2010-10-10
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    @Moron: if you do things that way you need to be careful to assign each outcome the correct probability. It's more work than necessary; just think of the coins as distinguishable and you can save yourself a lot of grief. I do not see how we could possibly be disagreeing about the implicit setup; we are drawing coins out of a bag and this is unambiguous.2010-10-10
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    @Qiaochu: Yes, and if we interpret that way, it is not obvious to me that the expectation will be the same. I do agree with you about the implicit setup (I said that in my previous comment). When it comes to probability, I prefer clarifying the exact setup. In this case it is just noise I suppose.2010-10-10
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    @Moron: I do not understand. If the expectation is not the same then you have not computed the probabilities correctly.2010-10-10
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    @Qiaochu: Consider this problem: you uniformly select a tuple (u,v,w,x) from tuples such u+v+w+x=3 (and 0<=u<=1 etc) and compute the expected value of 25u + 5v + 10w + x. Are you claiming this will be the same as 19.2? Maybe I am missing something, but it is not obvious to me.2010-10-10
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    @Moron: no, I am claiming this is the wrong problem unless you select the tuples with the correct probability. We agree about this, right? Nothing in the wording of the problem suggests to me that that is a valid interpretation of it.2010-10-10
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    @Qiaochu: If someone says they consider the coins indistinguishable and order does not matter, the above _could_ be a valid interpretation (perhap only in my own mind), but lets not continue this discussion further, as I do agree with you!2010-10-10
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    @coffee: My apologies for confusing matters.2010-10-10
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    No worries, this kind of discussion is very enlightening and is more than I could have asked for with my question :)2010-10-10

3 Answers 3

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Each coin is pulled 30% of the time. So on average you will get 30% of 64 cents=19.2 cents

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    Sorry, each coin is pulled 30% of the time? But there are more pennies than quarters. Surely, you would pull pennies more often than you would pull a quarter.2010-10-10
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    @coffee: Ross is referring to each individual coin, without regard for its monetary value. This solution is correct.2010-10-10
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Ross Millikan's answer is entirely correct, but perhaps it deserves slightly more elaboration.

The expected amount of money you receive is the sum of the expected amounts of money that each coin gives you, which is the value of the coin times the probability that you draw it. (Linearity of expectation!) The probability that you draw each coin is the same: it is one minus the probability that you do not draw it, or $1 - \frac{ {9 \choose 3} }{ {10 \choose 3} } = \frac{3}{10}$. So the expected amount of money you receive is $\frac{3}{10}$ of the total value of the coins.

Note that I am basically treating coins with the same value as distinguishable (pretend they have the same value but look different; it can't affect the answer to the question). The question of distinguishability is a red herring.

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Ross Millikan has already given a very elegant solution, but if you're still in doubt, this problem is small enough that you can easily verify it by letting a computer enumerate all the possibilities. Here's a Python program to do it:

coins = [25, 10, 10, 5, 5, 5, 1, 1, 1, 1]
total = count = 0
for k in range(10):
    for l in range(10):
        for m in range(10):
            if k!=l and k!=m and l!=m:
                total += coins[k] + coins[l] + coins[m]
                count += 1
print "Got", total, "cents in", count, "draws; on average", total*1.0/count, "cents."

The answer comes instantly: "Got 13824 cents in 720 draws; on average 19.2 cents."

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    And of course you can do this computation by hand as well: the total amount of money you draw is the sum of (the total number of times you draw each coin) * (the value of each coin). This is essentially the same computation as the one I did in my answer.2010-10-10
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    @Qiaochu Yuan: Sure, but then you have to do some *thinking*! Whether this requires more or less effort than writing the computer program depends on who you ask, I guess. :-)2010-10-10
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    If you have SAS here's another code example:data _null_; cnt=0; tot=0; avg=0; seed=48743; array coins(10) c1-c10 (25, 10, 10, 5, 5, 5, 1, 1, 1, 1); do n=1 to 100000; call ranperk(seed, 3, of c1-c10); cnt+1; tot+sum(of c1-c3); end; avg=tot/cnt; put avg= 5.1; run;2010-10-21