Two (Three?) Games
Philip Rash, NCSSM
2016 Teaching Contemporary Mathematics
Conference
Two (Three?) Games
Overview:
•
Game #1 (Bigger Exponent)
•
Game #2 (Biggest Product)
Game #1
“Bigger Exponent”
Two players simultaneously hold out one to five fingers (like rockpaper-scissors, but now you have 5 “things” to choose from
instead of 3). The number of fingers on a player’s hand is the
base, and the number on the opponent’s hand is the exponent.
The player with the higher value wins. For example, suppose
player A shows 5 fingers and B shows two. Then player B wins
with a score of 25 , which of course is larger than player A’s score
of 52 . What is the best strategy?
Expanded version: Allow each player to use 2 hands (i.e. each
picks an integer from 1 to 10).
Game #1
“Bigger Exponent” – What’s the best strategy?
Game #1
“Bigger Exponent” – What’s the best strategy?
Game #2
“Biggest Product”
Start with a natural number, 𝑛. Pick a set of natural numbers that
sum to 𝑛 (repeats allowed). Your score is the product of the
numbers in your set. What is the highest possible score? For
example, if 𝑛 = 10, you could pick {5, 2, 2, 1} with product 20,
but this would be “beat” by {7, 3} with product 21. (But is there
another partition of 10 that would yield a higher product than
21?)
Game #2
“Biggest Product” – What’s the best strategy?
Some rules:
• Do not use any 1’s
• Choosing {2,2} is the same as choosing {4}.
• Do not use any numbers bigger than 4.
• Why? Observe that for 𝑥 > 4, 2 𝑥 − 2 > 𝑥, so anywhere
we have {𝑥} as a subset of our partition (with 𝑥 > 4),
it’s better to use {2, 𝑥 − 2} instead.
Game #2
“Biggest Product” – What’s the best strategy?
Some rules:
• So, basically we’re down to deciding between using 2’s and
3’s. We observe that when partitioning a sum of 6, it’s
better to use {3,3} (product 9) rather than {2,2,2} (product
8).
• Thus, an optimal strategy is to partition into all 3’s, except
use one or two 2’s appropriately if the original number is
not divisible by 3.
Two (Three?) Games
So, it seems the answer is always “three”!
But what’s so special about 3?
Game #2 – Continuous Version
“Biggest Product”
Start with a positive real number, 𝑛. Pick a set of 𝑘 numbers that
sum to 𝑛 so that the product of the chosen numbers is as large
as possible. What’s the best strategy? Why?
Game #2 – Continuous Version
“Biggest Product”
An observation:
When partitioning our integer 𝑛 into 𝑘 terms, the largest product
is obtained when all terms (thus factors) are equal.
For example, with 𝑛 = 10 in the discrete case, the maximal
product was 3,3,2,2 → 36. But we yield a larger product (𝑘 = 4)
with 2.5,2.5,2.5,2.5 → 39.0625.
Game #2 – Continuous Version
“Biggest Product”
When partitioning our integer 𝑛 into 𝑘 terms, the largest product
is obtained when all terms (thus factors) are equal.
Why? In the case where 𝑘 = 2, this is identical to a classical maxmin problem: What values of length and width maximize the
area of a rectangle of fixed perimeter? (When the sides are
equal.)
In the cases with 𝑘 > 2, if we have 2 unequal factors, replace
them with two copies of their mean. Thus the maximal product
has all factors equal.
Game #2 – Continuous Version
“Biggest Product”
When partitioning our integer 𝑛 into 𝑘 terms, the largest product
is obtained when all terms (thus factors) are equal.
But how many terms (i.e. what value of 𝑘) should we use?
(Excel sheet…)
Game #2 – Continuous Version
“Biggest Product”
When partitioning our integer 𝑛 into 𝑘 terms, the largest product
is obtained when all terms (thus factors) are equal.
But how many terms (i.e. what value of 𝑘) should we use?
Maximize 𝑃 𝑘 = (𝑛/𝑘)^𝑘 :
Log of both sides: ln 𝑃 𝑘 = 𝑘 ⋅ ln(𝑛/𝑘)
Differentiating: 𝑃′(𝑘)/𝑃(𝑘) = ln 𝑛 − ln 𝑘 − 1
With 𝑃′ 𝑘 = 0, we have ln(𝑛/𝑘) = 1, and
𝑛
=𝑒
𝑘
Game #2 – Continuous Version
“Biggest Product”
When partitioning our integer 𝑛 into 𝑘 terms, the largest product
is obtained when all terms (thus factors) are equal.
But how many terms (i.e. what value of 𝑘) should we use?
So, it seems we should pick 𝑘 such that 𝑛/𝑘 is as close as possible
to 𝑒.
(See also Geogebra graph of 𝑃 𝑘 =
𝑛 𝑘
𝑘
)
Game #1 – Continuous Version
“Bigger Exponent”
Two players simultaneously write down two positive real
numbers. The number a player writes down is the base, and the
opponent’s number is the exponent. The player with the higher
value wins. What is the best strategy? Why?
Game #1 – Continuous Version
“Bigger Exponent”
Two players simultaneously write down two positive real
numbers. The number a player writes down is the base, and the
opponent’s number is the exponent. The player with the higher
value wins. What is the best strategy? Why?
Suppose Player A picks 𝑘 and Player B picks 𝑥. Then Player A
wins if 𝑘 𝑥 − 𝑥 𝑘 > 0.
Thus, Player A wants to maximize 𝑘 𝑥 − 𝑥 𝑘 .
Game #1 – Continuous Version
“Bigger Exponent”
Thus, Player A wants to maximize 𝑘 𝑥 − 𝑥 𝑘 .
I.e., can Player A pick a value of 𝑘 so that ℎ 𝑥 = 𝑘 𝑥 − 𝑥 𝑘 , 𝑥 > 0 is
as large as possible – or even better, always positive? If so, then
𝑘 is the value that A should play, forcing B to never win.
Let’s look at a graph of ℎ(𝑥)…
(Geogebra…)
Game #1 – Continuous Version
“Bigger Exponent”
Thus, Player A wants to maximize 𝑘 𝑥 − 𝑥 𝑘 .
Let’s look at a graph of ℎ 𝑥 = 𝑘 𝑥 − 𝑥 𝑘 (for 𝑥 > 0)…
So we suspect ℎ(𝑥) has a minimum (and sole root) when 𝑥 = 𝑒.
Indeed with 𝑘 = 𝑒, ℎ′ 𝑥 = 𝑒 𝑥 − 𝑒 ⋅ 𝑥 𝑒−1 , which equals zero when
𝑥 = 𝑒. Furthermore, ℎ 𝑒 = 0 as well, so 𝑒 is the sole root of ℎ(𝑥).
Game #1 – Continuous Version
“Bigger Exponent”
Let’s also look at a graph of the implicit curve 𝑥 𝑦 = 𝑦 𝑥 .
𝑥 𝑦 > 𝑦 𝑥 (X wins)
𝑦 𝑥 > 𝑥 𝑦 (Y wins)
𝑦𝑥 > 𝑥𝑦
𝑥 𝑦 > 𝑦 𝑥 (X wins)
The implicit curve has
2 “branches”.
Obviously one is the
graph of 𝑦 = 𝑥, but
what is the other one?
Game #1
See any
resemblance?
The nonlinear branch is indeed nontrivial.
Daniel Bernoulli, Christian Goldbach, and
Leonhard Euler all worked on it; Goldbach
used a parameterization “trick” and
showed this curve is traced by the
parametric curve
𝑡
1
𝑥 𝑡 = 1+
𝑡
1
𝑦 𝑡 = 1+
𝑡
Where 𝑡 > 0 or 𝑡 < −1.
(See Geogebra graph…)
𝑡+1
Two (Three?) Games
So, it seems the answer is always “three”!
Two (Three?) Games
So, it seems the answer is always “three”?
No, it seems the answer is always “𝑒”!
Two (Three?) Games
Philip Rash, rash@ncssm.edu
http://www.ncssm.edu/tcmconference
What’s So Special about 3?, Debra K.
Borkovitz, Mathematics Teacher, Vol 108 No 1,
August 2014, pp.71-75.