I am struck with following problem in my research even though it looks very simple. Kindly help me with that. I have modified the question as a real life question.
A shop keeper, say outside a school, conducting a gambling. You need a pay a chocolate to play the game. First he writes a three digit number (000 considered to be a three digit number) in his hand and he will let you choose a three digit number. If both the numbers share a common unit digit he will give you back two chocolates. If unit place and tenth place also matches then he will give you 10 chocolates and if all the three digits are same then he will give you 100 chocolates. I have the following questions in this scenario:
Does every number has equal probability and how to find them?
Is this similar to any well-known probability models?
How badly kids were cheated by the shopkeeper and how to quantify the value?
Sorry for such a simple question. I am working on algebra and my probability understanding is very poor. Thanks for your help.
The probability that shop keeper wins is $\frac{900}{1000}$ (as all not but only unit place number number is to be different)
Probability that Unit digit is not equal = $\frac{9}{10}$. Probability that tenth digit is not equaL = $\frac{9}{10}$. Probability that hundredth digit is not equal=$\frac{9}{10}$.
The probability a child wins 2 chocolates= $\frac{90}{1000}$(unit place equal, tenth different, no condition on 3rd}
The probability a child wins 10 chocolates=$\frac{9}{1000}$
The probability a child wins 100 chocoaltes =$\frac{1}{1000}$
Expected value for a child is $\frac{900}{1000}\times(-1)\ +\ \frac{90}{1000}\times(1)\ +\ \frac{9}{1000}\times(9)\ +\ \frac{1}{1000}\times(99)\ =\ \frac{-630}{1000}\ =\ =\ (-0.63)$
So, shopkeeper expect to earn 0.63 chocoaltes per round.
I don't know about the 2nd question.