Guess value on one die by knowing sum of 2 dice

Question

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:

1. Alice throws two dice and tells Bob the sum on the two

2. 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'.

Answer 1

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.

Answer 2

For example, if Alice tells Bob she threw a total of $3$, Bob knows that the dice were $1$ and $2$ in some order. His optimal strategy is to guess either $1$ or $2$, which will be correct all of the time. If he guesses any other number, he is right $0$ of the time. You need to do the same for the other $10$ possible totals, find the chance of each total and Bob's chance of winning with that total, and add them up. You should see a pattern so you don't have to do them all individually.

$C_i$ is the probability Bob wins given that he guesses $i$ given the sum Alice rolled. So if Alice rolls $3$, $C_1=C_2=1, C_3$ through $C_6$ are all $0$.

Answer 3

If Alice says $A$ with $2 \le A \le 12$ then an optimal strategy for Bob is to say $B$ where

• $B \in \{1,2,3,4,5,6\}$
• $A-6 \le B \le A-1$
• Unless $A \in \{2,12\}$, have $B \not= \frac A2$ because other possibilities are twice as likely

We can then make a table

 A   prob(A)    B from     prob(Bob wins | A)   prob(A & Bob wins)   

 2    1/36        1                1/1               1/36  
 3    2/36       1,2               2/2               2/36
 4    3/36       1,3               2/3               2/36
 5    4/36     1,2,3,4             2/4               2/36
 6    5/36     1,2,4,5             2/5               2/36
 7    6/36   1,2,3,4,5,6           2/6               2/36 
 8    5/36     2,3,5,6             2/5               2/36 
 9    4/36     3,4,5,6             2/4               2/36 
10    3/36       4,6               2/3               2/36
11    2/36       5,6               2/2               2/36 
12    1/36        6                1/1               1/36 

So the overall probability that Bob wins is $\frac{20}{36}=\frac{5}{9}$

Answer 4

From the comments under the question and the existing answers, it seems that the problem may lie somewhere else. The book says “$\mathcal C$ is the event that Bob wins following this strategy”. This is very slightly sloppy language (but of a kind that we all use all the time). The idea here is not that “following this strategy” is somehow an aspect of the sample space and we want to find the probability for Bob following this strategy, or condition on Bob following this strategy. It means merely that we assume that Bob follows this strategy, and then we consider the event $\mathcal C$ that he wins. Of course we could also model Bob’s choice of strategy probabilistically, and then the sample space would consist of pairs of strategies and die results, and then we could condition on Bob following this strategy – but that would be rather complicated and not what this formulation is meant to suggest.