The worlds first riddle!

CARDS
The second player should always win.
Winning first moves are:
1-3
2-3
3-4
4-3

For the last two cases, the rest of the game will look like:
N-(5-N)

The other two cases are a little more complicated.
If possible, you want to hit (and keep the opponent from hitting) 5X+2 if there are at least 4-X of each rank left.

Mark, to err is human. QED you are human. However, I am more human than you are, as I err more often.

Cards: Humm, I need to think on this awhile.


A ball of radius 1 is in a corner touching all three walls. What is the radius of the largest ball that can be fit into the corner behind the given ball


If we cut a pie 100 times, what is the maximum number of pieces we will end up with

The minimum is obviously 101.

BALL
2-sqrt(3)

PIE
5051 (N^2 + N)/2 + 1

CARDS
I'll look into producing a complete list of responses for the second player.

the 1st player will win

by playing ace at first
then if the 2nd player played 2 or 3 or 4 he would play 6-n to force the sum to 7
and then he plays 5-n to reach 22

if the second player played ace to block the 6-n rule
the 1st player would play 4
if the 2nd player played 2 or 3 or 4 he would lose for the above reason(6-n then 5-n)
if the 2nd player played ace again
the 1st player would play 4 again and so

if the 2nd player continued to play ace the 4 aces will end at 21 and he will play 2 or 3 or 4 to lose
if he played 2 or 3 or 4 he would lose due to 6-n and 5-n rule






again if the 1st player started with 3 or 4 he would lose due to 7-n then 5-n rule that player 2 would apply

if player 1 started with 2 the 2nd player would keep on playing 3 to force him to end all 2 cards or let him to apply 5-n rule so player 1 would also lose

i hope my bad language to be easy for you.

usamashaker:
Let's play. You go first. I'll make your first two moves for you based on the strategy you described.

1-3
3-2
?

i will play 3

1-3 (4)
3-2 (9)
3-2 (14)
?

1-3
3 (7)
2
3 (12)
2
3 (17)
?

1-3 (4)
3-2 (9)
3-2 (14)
3-2 (19)

i have 3 aces
4 cards 2
one card 3
4 cards 4

i will play 3 to win

I interpreted the instructions to mean there is a common set of 16 cards (4 of each rank) that we alternate turning over. That would leave:
3 aces
1 two
0 threes
4 fours

Try: Please clarify the instructions to the game.

Actually, this example from Try's original post clarifies it:

"Again, supposing the play was 3-1, 1-2, 3-3, 1-2, 1-4, scoring 21, the second player would win again, because there is no 1 left and his opponent must go beyond 22."

mmmmmmmmm
they are only 16 cards

ok

again

the player who reach 7 with 3 pairs remaining the sum of each of them =5 wins
reach 12 with 2 pairs of the above mentioned win
17 with one pair wins

the player who reach 7 without 3 pairs remaining lose
12 without 2 pairs lose
17 without one pair lose

so player 1 can win if he stsrted with the ace
if player 2 replyed with 2 or 4
player 1 would reply with 4 or 2 to reach 7 with 3 pairs the sum of each is 5

also if player 2 played (3)
1--3
1--2--------7 without 3 pairs left

1--3
1--3or4
4or3--12 with 2 pairs left

1--3
1--1
4--2------12 without 2 pairs left

1--3
1--1
4--3or4
4or3-------17 with one pair left


1--3
1--1
4--1
4--3or4
4or3-----22

1--3
1--1
4--1
4--2----17 with out any pairs left

also if player 2 played (1)

1--1
3--continue as above

I was wrong! In my haste, I made waste. The first player wins by leading with a 1. Here's a complete tree of moves. Note that the strategy is the same for 1-4 2-X and 1-2 4-X. The same is true for 1-1 3-X and 1-3 1-X.

Cards:
How fantastic. Online gaming.

Usamashaker writes, "i hope my bad language to be easy for you."

Your answer is so clear; you could have written it in hieroglyphs and the world would still understand.


You wrote:
the 1st player will win by playing ace at first.
the player who reach 7 with 3 pairs remaining the sum of each of them =5 wins
reach 12 with 2 pairs of the above mentioned win
17 with one pair wins


Mark agrees.


I say:

Apart from the exhaustion of cards, the winning series is 7, 12, 17, 22.

If you can score 17 and leave at least one 5-pair of both kinds (4-1, 3-2), you must win.

If you can score 12 and leave two 5-pairs of both kinds, you must win.

If you can score 7 and leave three 5-pairs of both kinds, you must win.

Thus, if the first player plays a 3 or 4, you play a 4 or 3, as the case may be, and score 7. Nothing can now prevent the second player from scoring 12, 17, and 22. The lead of 2 can also always be defeated if you reply with a 3 or a 2. Thus 2-3, 2-3, 2-3, 2-3 (20), and, as there is no remaining 2, second player wins.

Again, 2-3, 1-3, 3-2, 3-2 (19), and second player wins. Again, 2-3, 3-4 (12), or 2-3, 4-3 (12), also win for second player.

The intricacies of the defence 2-2 I leave to the reader. The best second play of a first player is always 1.

The first player can always win if he plays 1, and in no other way.
Here are specimen games: 1-1, 4-1, 4-1, 4 (16) wins. 1-3, 1-2, 4-1, 4-1, 4 (21) wins. 1-4, 2 (7) wins. 1-2, 4 (7) wins.

I apologise for the somewhat ambiguous instructions.


A ball of radius 1 is in a corner touching all three walls. What is the radius of the largest ball that can be fit into the corner behind the given ball?

May I at this point clearly state, "Whilst I appreciate many think the question impossible as no detail sizes are given and therefore impossible to answer. It is easy if you are a genius.

Mark:
BALL
2-sqrt(3)


The large ball will be touching the three walls, which can be thought of as half an enclosing cube. If we draw a diagonal from the bottom (where the other sphere is to be placed) to the opposite end of this imaginary cube, then the centre of both spheres will always lay somewhere along this line.

We can imagine taking a very small sphere touching the three walls in that small area and increasing it's size until it hits the other sphere. It will necessarily happen along the diagonal we have constructed. So, we need only find the length of that diagonal, and that will be the diameter of the smaller sphere.

The length of the entire constructed diagonal is 2\sqrt{3}. The part of it that goes from the end of the sphere to the corner of the room is \sqrt{3}-1, since 2x + diameter = 2\sqrt{3}. If we construct a similar cube around the smaller sphere, then the ratio of the small diagonal length to cube side length will be the same.

For the larger sphere the diagonal length is \sqrt{3}-1 and the cube length is 2, and for the smaller sphere the diagonal side length is \sqrt{3}-1-2r and the the cube length is 2r. Setting these ratios equal yields 1/(\sqrt{3}-1) = r/(\sqrt{3}-1-2r). This yields the answer, r = 1/4*(2-\sqrt{3}).



If we cut a pie 100 times, what is the maximum number of pieces we will end up with?

Dang, he is good, very good.
Mark
PIE
5051 (N^2 + N)/2 + 1


A straightforward math problem. Just searching for a pattern, list the first few terms:
1 cutting - 2
2 cuttings - 4
3 cuttings - 7
4 cuttings - 11

n cuttings - ?

The formula is (can be found, by counting line intersections):
MP = 1 + 1 + 2 + 3 + ... + n
= 1 + n(n + 1)/2
= (n^2 + n + 2) / 2

MP for 100 cuttings = (1002 + 100 + 2) / 2
= 10102 / 2
= 5051

Well done Mark.





One of the American states runs a lottery in which players select five numbers from between 1 and 32 (inclusive). Each player may choose the numbers of their own liking or allow the computer that prints out the ticket to randomly select the five numbers.

At a convenience store which sold the tickets, and where I had stopped for gas, the clerk had just bought one of these randomly generated tickets. She took a look at it and, seeing that her ticket contained a consecutive pair of numbers, reasoned that it would never be a winning ticket and discarded it.

Leaving aside that fact that any ticket could be a winner, was she right in doing so

More precisely, what is the probability that five numbers selected at random, uniformly and without replacement, from between 1 and 32 will contain a consecutive pair

Note: 1 and 32 are not a consecutive pair.

Lottery:

No, she was not right in doing so.

Example: The odds of the numbers being 1 2 3 4 5 (drawn in any order) are equal to the odds of any other combination of numbers being drawn.

I'll get right on that probability question...

LOTTERY
103096/201376 = .512

*****************WARNING - PET FREE ZONE********************


DrewDad wrote, "No, she was not right in doing so."

Mark, so glad to see you are ok. If you have not answered all the forum questions by 9am, I get worried. I was going to ask one of the guys to call round and ask your mom if you could come out to play. :wink:

Mark
LOTTERY
103096/201376 = .512

Although I no longer have the proof, I can assure any reader the answer is, correct; that approximately 51.2% of the drawings will contain an adjacent pair of numbers.


An ordinary die is rolled until the running total of the rolls first exceeds 12. What is the most likely final total that will be obtained

Idea/Clue
The possible outcomes are 13 through 18. However, there are different probabilities for each. For example, for 18 to be the final total, the previous total must have been 12, and the last roll must have been 6.


And now - Riddle favourites.


Until I am measured
I am not known,
Yet how you miss me
When I have flown.

I am


Never was, am always to be,
No one ever saw me, nor ever will
And yet I am the confidence of all
To live and breathe on this terrestrial ball.

I am


Alive without breath,
As cold as death,
Never thirsty, Ever drinking,
Clad in mail, Never clinking,
Drowns on dry land,
Thinks an island, Is a mountain,
Thinks a fountain, Is a puff of air.

I am

die:

13 28.5%

DIE
I get:

13: 0.279263
14: 0.236996
15: 0.192313
16: 0.145585
17: 0.097371
18: 0.048472