After lunch one day, Alice suggests to Bob the following...

After lunch one day, Alice suggests to Bob the following method to determine who pays. Alice pulls three six-sided dice from her pocket. These dice are not the standard dice, but have the following numbers on their faces: - $\operatorname{die} A-1,1,6,6,8,8 ;$ - die B $-2,2,4,4,9,9$ - die C $-3,3,5,5,7,7$. The dice are fair, so each side comes up with equal probability. Alice explains that Alice and Bob will each pick up one of the dice. They will each roll their die, and the one who rolls the lowest number loses and will buy lunch. So as to take no advantage, Alice offers Bob the first choice of the dice. (a) Suppose that Bob chooses die A and Alice chooses die B. Write out all of the possible events and their probabilities, and show that the probability that Alice wins is greater than $1 / 2$. (b) Suppose that Bob chooses die B and Alice chooses die C. Write out all of the possible events and their probabilities, and show that the probability that Alice wins is greater than $1 / 2$. (c) Since die A and die B lead to situations in Alice's favor, it would seem that Bob should choose die C. Suppose that Bob does choose die C and Alice chooses die A. Write out all of the possible events and their probabilities, and show that the probability that Alice wins is still greater than $1 / 2$.

After lunch one day, Alice suggests to Bob the following method to determine who pays. Alice pulls three six-sided dice from her pocket. These dice are not the standard dice, but have the following numbers on their faces:
- $\operatorname{die} A-1,1,6,6,8,8 ;$
- die B $-2,2,4,4,9,9$
- die C $-3,3,5,5,7,7$.
The dice are fair, so each side comes up with equal probability. Alice explains that Alice and Bob will each pick up one of the dice. They will each roll their die, and the one who rolls the lowest number loses and will buy lunch. So as to take no advantage, Alice offers Bob the first choice of the dice.
(a) Suppose that Bob chooses die A and Alice chooses die B. Write out all of the possible events and their probabilities, and show that the probability that Alice wins is greater than $1 / 2$.
(b) Suppose that Bob chooses die B and Alice chooses die C. Write out all of the possible events and their probabilities, and show that the probability that Alice wins is greater than $1 / 2$.
(c) Since die A and die B lead to situations in Alice's favor, it would seem that Bob should choose die C. Suppose that Bob does choose die C and Alice chooses die A. Write out all of the possible events and their probabilities, and show that the probability that Alice wins is still greater than $1 / 2$.

Key Concepts

Strategic Decision Making in Game Theory

The choice of dice in this setting introduces an element of strategy: selecting a die based on known probabilities to maximize the chance of winning. Understanding the interplay between the dice's probability distributions involves game theory concepts, where players make strategic decisions to counteract their opponent's choices.

Pairwise Comparison of Dice Outcomes

When comparing dice with nonstandard numbering, pairwise comparisons are used to calculate the probability that one die will roll a higher number than the other. This involves comparing each outcome from one die with every outcome from the other and aggregating the probability of the favorable events.

Outcome Enumeration and Combinatorial Analysis

Outcome enumeration involves listing each possible result from the dice rolls, while combinatorial analysis counts the number of ways events can occur. This systematic approach is essential when evaluating probabilities in complex scenarios, as it ensures all possibilities are considered and accurately weighted.

Nontransitive Dice

Nontransitive dice are a set of dice for which the relation 'is more likely to roll a higher number than' is not transitive. This means that even if one die tends to beat a second and the second beats a third, the third may still beat the first. This counterintuitive setup is central to understanding cyclic advantages and is key in game theory and probability puzzles.

Discrete Probability and Sample Space

In problems involving dice, each roll produces outcomes that form a discrete sample space. Calculating the likelihood of various events requires enumerating all possible outcomes and determining probabilities for each. A clear comprehension of this concept is crucial in constructing the complete probability model for dice rolling events.

Transcript

00:01 Three unusual dyes, a, and c are constructed such that die a has numbers as given and di b has numbers as given and then down in c.

00:11 First question is if di a and di b are road find the probability that b defeats a.

00:19 That is the number that appears on di b is greater than the number that appears on di a.

00:25 Okay so basically let me just put it in a rectangular format.

00:36 So die a is given as 33448.

00:42 So it's 3 ,4 and 8.

00:47 Otherwise the others just repetition.

00:50 And die b is 1 5 and 9.

00:55 So this is b.

00:57 So 1, 5 and 9.

01:01 Now the outcomes are starting with the horizontal is the sort of x xx is format so this one will be 3 -1, 3 -5 -3 -9 and then this one is 4 -1 -4 -5 and 4 -9 and the next one is 8 -1 8 5 and 8 9.

01:54 So we can see that the one that has got more than half the outcome is the winner in any case.

02:05 So basically, die a.

02:16 Let's see how many times that die a win.

02:23 It wins one.

02:25 Okay, let me use another ink here.

02:29 Let me use red.

02:31 It won here.

02:32 And it won here, it also won here, and where else did it win, it also won here.

02:50 So, die a 1 4 out of 9.

02:58 So die b is 1 5 out of 9.

03:03 So the question says, find the probability that die b, b, beats a.

03:10 So the probability that die b gets so probability that b is greater than a is equals to 5 over 9.

03:28 Okay.

03:29 And then question two says if die b and c are all defined the probability that c beats b.

03:40 All right so basically now we are looking at the other one die b and c so b is still 1 5 and 9 and then c is given as 267 c is 267 okay so we can do the same process there it's 2 1 2 1 2 1 6 1 6 1 6 71 then it's 2565 75 then it's 2 .9 6 .5...

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