flipping two coins when compared to a ball and bin situation?

Question

so as i have understood the sample space for the toss of two coins is

TT  
HH   
TH  
HT

even if the two coins are absolutely the same we can color each one different colors thus our sample must contain both TH and HT . so the possibility of getting exactly one H and one T is 1/2.

what about this problem : we have two indistinguishable balls and two different bins . what is the probability of each bin having exactly one ball. we show this by bars and stars like :

  1     * | *    
  2       | * *  
  3   * * |      

and here we say that we have 3 possible out come ?! so the probability is 1/3

but is not it the same as the problem of tossing two indistinguishable coins ?!

in the case of balls and bins cannot we again paint the balls ? thus we have actually two out comes in number 1 .

Answer 1

For the balls in bins problem, there are three distinct events as counted by the "stars and bars" method. However, each such outcome is not equally probable. Their frequencies of occurrence carry a different weight.

In statistics you describe these events as the macrostates of the system. They are composed of numerous (preferably) equiprobable microstates. Basically, microstates are outcomes counted by considering distinctions which are ignored when identifying microstates. (In this case, the distinctness of the balls.)

$$\boxed{\begin{array}{c|cc|c}\text{Macrostate} & \text{Microstates} & & \mathsf P(\text{macrostate}) \\ **| & AB| & & 1/4 \\ *|* & A|B & B|A & 1/2 \\ |** & |AB & & 1/4 \end{array}}$$

Note: In this model we don't find a need to consider the arrangement of the balls inside each bin; we are only concerned with the probability of balls entering bins, as only those outcomes contribute to the weight of the event.

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In point of fact, this is exactly the same process involved in the coin scenario. If we take the macrostate as the count of heads showing, and the microstates are the results of each individual coin toss.

$$\boxed{\begin{array}{c|cc|c}\text{Macrostate} & \text{Microstates} & & \mathsf P(\text{macrostate}) \\ 0 & \mathsf {TT} & & 1/4 \\ 1 & {\sf HT} & {\sf TH} & 1/2 \\ 2 & {\sf HH} & & 1/4 \end{array}}$$

Note The microstates need not be equiprobable; but their probability weight needs to be accountable. Suppose one of these coins was biased. Then while these microstates would not be equi-probable, that bias could be accounted for.

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In summary: When measuring probability of events, it is not sufficient to merely count distinct outcomes, you must also account for their individual probability weights.

Answer 2

As André Nicolas succinctly explains, we would need to establish how the balls are distributed in the bins. In other words, we need to clarify the probability distribution on the set of elementary outcomes. If we consider the balls indistinguishable, that implies that what we are interested in are the resulting configurations of balls in bins, irrespective of how they were placed in the bins, in which case, there are three elementary outcomes: $$(**,\;), (*,*), (\;,**).$$ If, however, we think of the placement of balls in the boxes as being independent events, in which each ball has an equal likelihood of being placed in either bin, then we must take into account the fact that the second elementary outcome described above is twice as likely to occur as either the first or the third. While the balls are in fact indistinguishable, a convenient mental abstraction that helps us conceptualize this unequal probability is to color the balls or make them distinguishable.

Answer 3

Practically speaking a. if you "throw" 2 balls (same or different color) into 2 bins that is equivalent to tossing 2 coins; b. if you take a bin with 2 balls in a row and "throw" in a separator (which is forced to uniformly land left, centre, right) that is the bars+stars model.