Probability problem - binomial distibution Please help me out, I need to solve this problem, unfortunately I don't have any good way to approach this task 
The keeper of a certain king’s treasure receives the task of filling each of 100 urns
with 100 gold coins. While fulfilling this task, he substitutes one lead coin for one
gold coin in each urn. The king suspects deceit on the part of the sentry and has
two methods at his disposal of auditing the contents of the urns. The first method
consists of randomly choosing one coin from each of the 100 urns. The second
method consists of randomly choosing four coins from each one of 25 of the 100
urns. Which method provides the largest probability of uncovering the deceit?
The answers is "the second method" was solved by binomial distribution. The probabilities are 0.634 and 0.640 accordingly to the textbook "Understanding Probability"
Thanks!
 A: Hint: call $p_1$ the probability that a given urn reveals the deceit when one draws one coin from it and $p_4$ when one draws four coins. Your first step would be to compute $p_1$ and $p_4$. The text of the exercise probably means that the urn reveals the deceit if and only if the lead coin is the coin drawn, respectively is among the four coins drawn. Then $p_1$ and $p_4$ are very simple functions of $n$ the number of coins in each urn. Your second step would be to write the global probabilities $P_1$ and $P_4$ of uncovering the deceit as functions of $p_1$ and $p_4$ respectively. Sub-hint: consider the probabilities of not uncovering the deceit.
A: It is not actually necessary to calculate the numbers to reach the conclusion, though you may need to satisfy your teacher.
The expected number of lead coins found is $\frac{1}{100}$ times the number of coins inspected. 
If you only looked at four coins, either drawn from four different urns or from the same urn, then you would be more likely to find two or more lead coins in the first case, and so also more likely to find zero lead coins in the first case (you might work out these probabilities of finding zero).  So the probability of finding no lead coins when repeating these 25 times is higher in the first case (again you can work this out). So in the original problem the probability of finding at least one lead coin is higher in the second case (easily calculated from the previous numbers).     
A: thanks for help
regarding Didier Piau hint
$$p1=\frac{99}{100} $$ - uncovering deceit by the first method ,so
$$P1 = ( \frac{99}{100} ) ^ {100} $$
probability of uncovering the deceit for the second method might be looking like that
$$p2 = ( 1 - \binom{100}{4} ( \frac{1}{100} ) ^ 4  ( \frac {99}{100} ) ^ {96} ) $$
and
$$P2 = p2 ^ {100} $$
I am not sure if this's right , but at least it makes sense  for me :)
A: I am solving the same question and puzzled by the choosing four coins from each of the 25 urns as well.
However, I think I solved this problem and I would like to shed some light to this problem to people who are puzzled by this problem like I were.
First, we need to solve for the probability for not finding the lead coin for 4 draws out of 100 coins at each of the 25 urn. The solution should be 99/100*98/99*97/98*96/97=96/100.
So we have the probability for not finding the lead coin for each of the 25 independent trial.
Second, we apply binomial formula.
The probability of finding at least 1 lead coin out of 25 independent trial of the 25 urn is 1-(96/100)^25=0.6396 which is very close to the 0.640 indicated by the solution of the book.
Hope this helps.
