# Expected value for choosing number twice

I have a problem:

I can choose twice randomly numbers from 1 to 3 and probabilities to choose each numbers can be adjusted for best end result. Choosing a number is independent. But not equally likely. I can define probability for each number.

If I choose twice same number I get 0 points otherwise I get sum of two choosen numbers. How to calculate best, optimal probability for each number to get max possible sum? And how to calculate expected value after choosing twice?

I tried like this: - Choosing numbers twice from 1 to 3 is actually getting me sum of 2 to 6 - so I got sum matrix: \begin{matrix} 0 & 3 & 4 \\ 3 & 0 & 5 \\ 4 & 5 & 0 \\ \end{matrix}

I think then it would be probabilities for numbers that represent sum like for 3,4 and 5 probability is 2/9, but that looks awfuly wrong.

• Please include your thoughts and efforts in this and future posts. Formatting tips here. – Em. Jan 16 '16 at 23:25
• What do you mean by the best, optimal probability for each number to get max possible sum? – Jendrik Stelzner Jan 16 '16 at 23:30
• probablyme Sorry for posting badly formatted and without my toughts / tries, will try to do better in next posts. @JendrikStelzner best in way that I get highest possible expected value, but so I don't get high probability of getting same number twice coz then I get 0 as end result. – Cazper Jan 16 '16 at 23:34

If choosing a number is independent and equally likely, then if $S$ is the number of points you get, then using your table, the distribution of $S$ is \begin{array}{r|ccc} s&0&3&4&5\\\hline P(S=s)&\frac{3}{9}&\frac{2}{9}&\frac{2}{9}&\frac{2}{9} \end{array} So, to calculate the expectation of $S$, we have $$E[S] = 0(3/9)+3(2/9)+4(2/9)+5(2/9)=\frac{8}{3}$$
• @Cazper You should include in your post that the numbers are not equally likely that you are defining the distribution on the numbers $\{1,2,3\}$. At least, that's what it sounds like you are saying. – Em. Jan 16 '16 at 23:52