Question

you play a game in which two dice are rolled if a sum of 7appears, you win 10 otherwise you lose $ 2.00. if you intend to play this game for a long time should you expect to play this game for a lone time should you expeet to make money. lose money or come out about even?

Answer 1

take the amount you win times the probability you win it and add it up

Answer 2

probability you roll a seven is \(\frac{1}{6}\) and so the probabablity you do not roll a seven is evidently \(1-\frac{1}{6}+\frac{5}{6}\) so you multiply and add
\[10\times \frac{1}{6}-2\times \frac{5}{6}\] and see what you get for your expected value

Answer 3

another damned typo i means \(1-\frac{1}{6}=\frac{5}{6}\)

Answer 4

There are 36 possible outcomes
there are 6 favorable outcomes

\[\langle x \rangle = \sum x P(x)\]
\[=\frac{1}{N} \sum xN(x)\]
\[=\frac{1}{36} ($10\times 6 -$2\times30)\]
\[=\frac{$0}{36}\]
\[\langle x \rangle =$0\]

you can expect to break even over a long period