Puzzles and Games
Herein are a small collection of diversions that
I find interesting, usually because their solutions are creatively
simple. Some of them are
not mine (and I have been quite relaxed about providing the sources),
so if you would like further references simply e-mail me.
Please contemplate...and feel free to e-mail your thoughts to me.
-MMI.
You can find some of the solutions here.
Fun
Games
Bridge
Technical
Rob:
"I just picked two integers greater than one and their sum is less than 100."Adam:
"Adam, their sum is ..." (he whispers it to Adam).
"Rohit, their product is..." (he whispers it to Rohit).
"Rohit, we don't know the numbers."Rohit:
"Now I do."Adam:
"Me too".
What were the numbers?
If instead of $1 I said if you fail you owe me $X, what is the maximum value of X for which you will take on this bet?
What if only one person (either a perpetual liar or perpetual truth teller) was standing at the fork. What question would you ask.
What about if player A has N>1 more coins than B?
What is the value of the sum |a1 - b1| + |a2 - b2| + ... + |an - bn|?
Two concentric circles are as pictured. The chord in the larger one which is tangent to the smaller one has length 100. What is the area of the ring between the two circles? |
Bidding: W N E S -- -- -- 1D *majors 2D* 2H# 3H 3NT #heart stopper, diamond support, good hand 4H X -- 5D Opening Lead HQ: Plan the Play. NORTH: 86 AK6 Q97 AT984 SOUTH: K74 5 AKJT86 J76
973 973 A973 973 A86542 QJT T864 2 2 JT82 J2 Q8654 K AKQJ5 KQ65 AKTSouth is declaring in 6H. West to lead. What do you lead? What leads allow South to make the contract-and how? What leads set the contract-and why?
AKQJ2 - A 8765432 - T87654 T8542 9763 J64 532 KQJT9 - 93 AKQJ KQT987 ASouth is declaring 6NT and the lead is the Club K.
P[|X-Y|<=2]<= 3 P[|X-Y|<=1]
I have managed to show P[|X-Y|<=2]<= 5 P[|X-Y|<=1], but the factor of 3 seems to have eluded me. This is problem number 2 in chapter 2 of the text "The Probabilistic Method" by N. Alon and J. H. Spencer, Wiley, 2000.
What is N(p) as a function of p?
Here is a naive computer scientists argument that the answer to this question depends on the continuum hypothesis. Let a0 be aleph 0 and a1 be aleph 1. Treating a0 and a1 as algebraic quantities, suppose that a0 samples are drawn, then the probability that at least one is rational is
1-(1-a0/a1)^(a0) - > 1-((1-a0/a1)^(a1/a0))^(a0^2/a1).
Now suppose that a1 samples are drawn, then the probability that at least one is rational is
1-(1-a0/a1)^(a1) - > 1-((1-a0/a1)^(a1/a0))^(a0).
Since a0/a1 - > 0 lets postulate that (1-a0/a1)^(a1/a0) - > 1/e. Since a0^2/a1=0, the top expression is 0 and the bottom expression is 1. Thus, if N < = aleph 0 then p=0; if N > = aleph 1 then p=1. The only way to obtain 0 < p < 1 is if there is some cardinality between aleph 0 and aleph 1, i.e. the continuum hypothesis.
So can someone put some beef into my argument or are my heuristics just plain false?
Question: For a large enough lattice (m,n), no matter how you color the vertices, there is a set of four monochromatic vertices at the corners of a square (the square may be of any size)?
The original motivation for this problem is to construct a misere game for two players who alternately color the vertices on a large enough grid. The loser is the first player to create such a monochromatic quadruplet.
If you have puzzles that can be solved with some ingenious and creative trick, I would like to know of it!