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March 12th, 2019 (Permalink)

Three-Dice Monty

Last month, "Three-Card" Monty*, the carnival con artist, shamelessly attempted to swindle a young couple on Valentine's Day using a pair of tricky dice! Now, he's back in his booth on the midway, but this time he has three strange dice.

Imagine that you approach Monty's booth and see on the counter before him three cubical dice: one colored red, the second white, and the third blue.

"As you can plainly see," Monty says to you, "these are three unusual dice. As you probably know, a standard die has one through six dots on its six faces. The number of dots on these dice range from one through nine, instead. Moreover, each of these nine numbers appears twice, both times on the same die. The numbers are distributed in such a way that each die has different numbers on its faces than the other two, so no number is shared among them. This means that when two of the dice are rolled against one another, they never tie. In every other way these dice are the same as standard dice: they're not loaded or gimmicked in any way.

"The first die has the numbers two, four, and nine, twice each on its six red faces. The second, white die has, instead, the numbers one, six, and eight. Finally, the last, blue die has the numbers three, five, and seven. The numbers on each die total thirty.

"Wouldn't you agree that they are evenly matched? I offer you an even money bet that the die I select will beat yours. That is, each of us will put up a dollar, each will roll his die, and the die that comes up with the highest number wins. Moreover, to be perfectly fair to you I offer you the first choice: you may select any one of the three dice and I will take one of the remaining two. What could be fairer than that?

"So, my friend," Monty finished, "which die do you choose?"

Is the bet that Monty is offering you fair? Should you accept or reject his wager? Assuming that you accept it, which die should you choose?


* In case you don't know Monty: he is a trickster, but he always speaks the exact truth. However, this does not mean that he necessarily tells the whole truth. Also, while he is a sharpster, he prides himself in not doing any sleight-of-hand. For previous puzzles involving Monty, see:


March 8th, 2019 (Permalink)

Rule of Argumentation 41: Be as definite as possible!

Before proceeding to this month's rule, I want to mention something that I should have explained earlier in this series, probably in the introduction: Each of these rules is a heuristic or "rule of thumb". In other words, there are exceptions to all of them, that is, situations in which you should not follow them, but such situations are exceptional.

Previous rules in this series were rules that governed the overall process of arguing: appealing to reason, acknowledging one's own fallibility, and focusing on arguments themselves rather than arguers. This is the first rule that deals with the content of argumentation. It says that the claims and arguments you make should be as definite as possible.

In order to be as definite as possible myself, I will explain what I mean by both "definite" and "possible":

Next Month: Rule 5


Notes:

  1. Previous entries in this series:
  2. Unfortunately, there doesn't appear to be a synonym for "unambiguous" lacking a negative prefix, so I choose the ambiguous "definite" so as to remain as affirmative as possible in this rule.
  3. For more on the types of ambiguity, including subtypes and examples, see the entry for the fallacy of Ambiguity.

You chose the red die. Unfortunately for you, Monty chose the blue one. The following table shows all 36 equally probable match-ups between the two dice. The letters in the table indicate which of the dice wins in that match-up:

Blue
335577
Red2BBBBBB
2BBBBBB
4RRBBBB
4RRBBBB
9RRRRRR
9RRRRRR

As you can see, the red die wins against the blue only sixteen times out of the total of thirty-six, with the blue die winning the remaining twenty times. So, the red die will win only 4/9ths of the time and the blue 5/9ths. Thus, an even money bet is not a fair one, and Monty stands to win more often that you.

For what you should have done instead, see below.


You chose the white die. Unfortunately for you, Monty chose the red one. The following table shows all 36 equally probable match-ups between the two dice. The letters in the table indicate which of the dice wins in that match-up:

Red
224499
White1RRRRRR
1RRRRRR
6WWWWRR
6WWWWRR
8WWWWRR
8WWWWRR

As you can see, the white die wins against the red only sixteen times out of the total of thirty-six, with the red die winning the remaining twenty times. So, the white die will win only 4/9ths of the time and the red wins 5/9ths. Thus, an even money bet is not a fair one, and Monty stands to win more often that you.

For what you should have done instead, see below.


You chose the blue die. Unfortunately for you, Monty chose the white one. The following table shows all 36 equally probable match-ups between the two dice. The letters in the table indicate which of the dice wins in that match-up:

White
116688
Blue3BBWWWW
3BBWWWW
5BBWWWW
5BBWWWW
7BBBBWW
7BBBBWW

As you can see, the blue die wins against the white only sixteen times out of the total of thirty-six, with the white die winning the remaining twenty times. So, the blue die will win only 4/9ths of the time and the white wins 5/9ths. Thus, an even money bet is not a fair one, and Monty stands to win more often that you.

For what you should have done instead, see below:


You should have declined Monty's offer, or insisted that Monty pick a die first. Then, no matter which die Monty chose, you could choose one of the remaining two that would beat his die more often than not: see the three tables, above. Very few bets are a sure thing, but you can bet your last dollar that Monty would not have gone for it.

The reason that the person who chooses second can always pick a winning die is because the dice described in this puzzle are "non-transitive" dice*. A relation, R, between two things is called "transitive" if from the fact that x bears R to y and y bears R to z it follows that x bears R to z. For instance, the relation of "taller than" between people is transitive: if Adam is taller than Eve and Eve is taller than Abel, then Adam is taller than Abel.

Now, you would probably expect that the relation of "wins more often than" between dice would be transitive. In other words, if the red die beats the white die most of time, and the white die beats the blue die most of the time, then you'd expect the red die to beat the blue one most of the time. But it doesn't! As you can see above, the blue die beats the red one most of the time.

This is surprising and counter-intuitive, but it may seem less so if you reflect upon the relations "all of" and "most of" between types of thing: the relation "all of" between types is transitive, for if all A are B and all B are C then all A are C. In contrast, if most A are B and most B are C it does not follow that most A are C. For instance, most soldiers are men and most men are civilians, but it certainly isn't the case that most soldiers are civilians―in fact, none are.

Similarly, if one die beat a second all the time and the second beat a third all the time, then the first would beat the third all the time. So, the relation of "beats all the time" between dice is transitive. But "beats most of the time", like "most of", is not transitive.

Warning: Do not try this trick at home. Monty is an expert. Besides, he's a fictional character and cannot get his nose broken or be arrested.


*See:

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