Puzzle: a doubles probability quickie
July 6, 2007 pm31 4:17 pm
Which is more likely, rolling doubles with two regular, 6-sided dice, or rolling *doubles with one 6-sided die and one 4-sided die. Explain why your answer makes sense.
* The dice are numbered 1,2,3,4,5,6 and 1,2,3,4. Doubles is when the numbers on both dice match.
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JD2718 
The possible outcomes for two six sided die:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
….
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
for a total of 36 possible outcomes, all equally likely.
There are six possible doubles.
The possibility of rolling doubles is then 6/36 = 1/6.
The possible outcomes for one four sided die and one six sided die
(1,1), (1,2), (1,3), (1,4)
(2,1), (2,2), (2,3), (2,4)
…
(6,1), (6,2), (6,3), (6,4)
for a total of 24 possible outcomes, all equally likely.
There are four possible doubles.
The possibility of rolling doubles is then 4/24 = 1/6
So, they are equally likely.
interesting. never noticed this.
here’s a good problem from
“mathematics under the microscope”.
the absolute value can be defined
in terms of the maximum function
quite easily: abs(x) := max{x, -x}.
but to define “max” in terms of “abs”
takes a moment’s thought (or, in my case,
several moments). give it a try!
I too determined equally likely! Not what my first instinct was though, interesting. Thanks, more extensions to add to the probability unit next fall.
Given one n-sided die and one m-sided die, where the values of the n-sided die are a superset of the values of the m-sided die, the probability of rolling doubles is always 1/n (in the example above, 1/6) since the probability that the roll of the m-sided die will produce a value that can be matched by the n-sided die is 1. This probability holds, interestingly, even if the m-sided die is unfair (loaded).
in other words:
throw the D4 (4-sided die) first.
regardless of the outcome,
one has a 1/6 chance
of matching the result
upon throwing the D6.
that’s what i came back to mention
when i found that leigh hunt
had pre-empted me … good job!
i probably would have overlooked
the result for a “loaded” Dm …
Nice job all!
Math is a lovely break from union politics and that sort of stuff, isn’t it? There’s nothing like a little arithmetic to steady the nerves.
I’m highlighting Vlorbik’s question, but directing solutions back to this comment-space.
max() in terms of abs():
max(x, y) = (abs(x – y) + x – y) / 2) + y
The idea here is to add the difference between x and y to y if x is larger, and to add zero to y otherwise. Nice problem.
Isn’t that equal to
?
Yes indeed. Your simplification is more elegant.
I remember that max expression from my freshman year of college and Spivak’s calculus. Good times, good times…