There are 10 children and 10 presents prepared for them. There is one present for each child. Each present has a child's unique name. The children are asked to take turns to take their presents. Out of excitement, the first child picks a present at random and runs away with it. The other 9 children are reminded to carefully check their names. If they can't find their present, they are allowed to take any one at random. They all follow the instruction. What is the probability that the child to pick the last present will pick a present which has her name on it?
Here is my attempt :
For simplicity’s sake let us say there are only 3 kids and 3 presents :
P( k3=p3)= p(k1≠p3)*p(k2≠p3 | k1≠p2)
Where P( k3=p3) is probability of 3rd kid getting the third present which is meant for her. And similarly the inequality would mean not getting it.
If the above approach is right the answer to the original question would be ~0.39.
But I am not sure if this is the right way to solve this problem .
