Let's imagine you throw five dices. After the throw, you can choose the amount of dices that you want to rethrow (can be a number between 0 and 5).
What is the probability to do a five of a kind with this configuration using the following strategy :
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if you make a five of a kind at first throw, gg wp ;
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if you make a four of a kind, you pick the last dice and rethrow them ;
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if you make a three of a kind, you pick the two dices and rethrow them ;
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if you make a two of a kind, you pick the three dices and rethrow them :
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if you have all the dices different, you just rethrow them all ;
I've coded (hoping to not have failed on that) a small program that simulate this kind of strategy, then I made a huge sample and divided by the number of attempt to have an idea. I obtained $P_{\text{code}} \approx 0,0058$
I wanted to see if i can find this theoretically knowing the probability of five of a kind (6/7776), four of a kind (150/7776), three of a kind (1500 / 7776), two of a kind (5400 / 7776) and one of a kind (720 / 7776)
I've found that $$ P=\frac{6}{7776}+\frac{150}{7776}\times\frac{1}{6}+\frac{1500}{7776}\times\frac{1}{36}+\frac{5400}{7776}\times\frac{1}{216}+\frac{720}{7776}\times\frac{6}{7776} \approx 0,0126 $$
where am I wrong ?