Question 15.9: Suppose you toss 100 coins starting with 60 heads and 40 tai...

The change in entropy is

\Delta S=S_{ f }-S_{ i }=k \ln W_{ f }-k \ln W_{ i },                         (15.69)

where the subscript i stands for the initial 60 heads and 40 tails state, and the subscript f for the final 50 heads and 50 tails state. Substituting the values for W from Table 15.4

gives

\Delta S=\left(1.38 \times 10^{-23} J / K \right)\left[\ln \left(1.0 \times 10^{29}\right)-\ln \left(1.4 \times 10^{28}\right)\right]                  (15.70)

Discussion
This increase in entropy means we have moved to a less orderly situation. It is not impossible for further tosses to produce the initial state of 60 heads and 40 tails, but it is less likely. There is about a 1 in 90 chance for that decrease in entropy (-2.7 \times 10^{-23} J / K) to occur. If we calculate the decrease in entropy to move to the most orderly state, we get \Delta S=-92 \times 10^{-23} J / K. There is about a 1 in 10^{30} chance of this change occurring. So while very small decreases in entropy are unlikely, slightly greater decreases are impossibly unlikely. These probabilities imply, again, that for a macroscopic system, a decrease in entropy is impossible. For example, for heat transfer to occur spontaneously from 1.00 kg of 0ºC ice to its 0ºC environment, there would be a decrease in entropy of 1.22 \times 10^{3} J / K. Given that a ΔS of 10^{-21} J / K corresponds to about a 1 in 10^{30} chance, a decrease of this size (10^{30} J / K) is an utter impossibility. Even for a milligram of melted ice to spontaneously refreeze is impossible.