Suppose I model 2 dice throws by the uniform probability distribution on $\{1,2,3,4,5,6\}^2$ and the following game takes place:
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Alice throws two dice and tells Bob the sum on the two
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Bob tries to guess a value that appears on at least one dice by using this information.
The problem asks this (my question is in the next paragraph): Assuming Bob plays the game optimally (by choosing a value with the highest probability of appearing given a certain sum), what is the overall chance of winning?
I cannot follow how the event 'Bob wins using an optimal strategy' looks like (in term of a subset of some sample space). It seems like I must have a new sample space where each sample should contain the guess as long as the outcome of the 2 throws.
NOTE This event (Bob winning using an optimal strategy) is denoted by C in example 8.12 (page 215) from Victor Shoup's free e-book: "A Computational Introduction to Number Theory and Algebra". The probability of this event is computed using the law of total probability by conditioning C on the events $D_l$ (event that the sum on the two dice is $l$), but I do not see how this is an 'event'.
Best Answer
The particular optimal strategy that is proposed in the book is to choose $1$ if the sum $\ell$ is $7$ or less, and choose $6$ if the sum $\ell$ is greater than $7.$ So the event $C$ is simply this:
$$ (\text{$\ell \leq 7$ and at least one die shows a $1$})\ \mathbf{or} \ (\text{$\ell > 7$ and at least one die shows a $6$}). $$
This is a subset of the sample space $\{1,2,3,4,5,6\}\times\{1,2,3,4,5,6\}.$ You don't need to include Bob's guess in the sample space, because Bob's guess is a deterministic function of the outcomes already listed in the sample space.
For example, if the roll of the dice is $(2,4)$ then Bob will guess $1$ and lose. Therefore $(2,4) \not \in C.$ But $(1,3) \in C$ because $1+3 \leq 7,$ therefore Bob guesses $1$ and wins. Likewise $(3,6) \in C$ because $3+6 > 7,$ therefore Bob guesses $6$ and wins, but $(3,5) \not\in C$ because Bob guesses $6$ and loses.