3,5

But since every face of a dice is different, maybe it will change the probability (the face with 6 holes is lighter than the one with 1 hole).

The probability of each outcome is a constant at 1/6 and hence the expected value is:

1/6*(1+2+3+4+5+6) = 3.5

7

38.5

if c is the side of the small square c^2 is the surface of the small square. If a is the longer side of the triangle and b its short side the 4 triangles have a surface of 4*1/2*a*b=2ab.The whole big square surface is then 2ab+C^2 (4 triangles+ 1 small square)

(a+b)^2 is the surface of the large square

(a+b)^2 =a^2+b^2+2ab

Combining the 2 equality:

a^2+b^2+2ab= 2ab+c^2

or

a^2+b^2=C^2

Jean

an infinite number, so long as both hemispheres are in play

None. Places can't walk !

Assuming it is a standard die, then you keep rolling until you get a 6

The expectation is also 6

The expectation are 3,5

So even if you still have a turn take the 4 and go home

I defenitely want the other envellope.

If 10=X then the other envellope contain 20$
If 10=2X then the other envellope contain 5$

Since you gave me the envellope randomly, there is a 50/50 chance for the two outcome.

So 25/2 = 12,5$

I'm richer with 12,5 than 10$

You Swap

The probabilty of winning the fisrt time is 1/3.

When you know that one door does not hide the price the probability for the door you have not choosen is now 1/2, which is better that you first choice of 1/3

Jean

all the yellows on one jar, one orange alone, and 4 oranges in with the yellows.

You have 50 pairs of 101 (1+100, 2+99, 3+98, etc.) so the answer is 50*101 = 5050

sum = n(n+1)/2..here n=100
so sum = 100(100+1)/2 =100*101/2= 5050...

watch die hard (3)

You fill the 3 pints jug and poor it in the 5 pints jug twice.
The 5 pints is now full and you have 1 pint remaining in the 3 pints.
Trow the 5 pints away.
Put the remaining pint in the 5 pints.
Fill the 3 pints again and put it in the 5 pints. Here you are with 4 pints

Fill up the 3 pints, empty the 3 pints into the 5 pints. Fill up the 3 pints again, and top off the 5 pints, leaving 1 pint in the 3 pint jug.
Empty the 5 pint jug, pour the remainder 1 pint in the 3 pint jug, into the 5 pint jug (this leaves 1 pint in the 5 pint jug).
Fill the 3 pint jug and pour it into the 5 pint jug.

Now you have 4 pints in the 5 pint jug

Two, because you know that the one awry marble is heavier.

Divide the 9 balls into 3 sets of 3. Weigh three balls against three balls. If it tips either way, you know which side is heavier, or if they balance, the wrong ball must be in the third set of 3. Either way, it results in you having it narrowed down to three.

Just do that again, applying the same logic to 1 ball vs 1 ball.

Yes, you remove all seven matchsticks in the fourth column and you will always win.

Given the complexity of this game, the way to deduce the answer is to reduce it to its simplest form and then add more complexity. So, starting with a game with one matchstick and one column it is clear that you do not want to start first, as doing so always results in loss.

Moving onto the two column game, it is clear playing first carries an advantage, you just remove the second column of three matchsticks on your first go and you win.

The three column game still has few enough possibilities that you can work through all of them, if you do so you discover that you will always lose if the second player plays an optimal strategy.

This then brings us to the answer, since you can force your opponent to play the three column game simply by removing the fourth column on your first go.

First person always loses for sure. Optimal play is to start by picking a single one from any of the rows. To jump ahead, the next optimal play for your opponent (we are P1) would be to pick another single from any column not already picked from, until you got to a (0,2,4,6) arrangement. At that point, any move you make will give your opponent a winning move, which is explained below.

It's a really long explanation, so I'll be intentionally "brief", but there are a few endgames to achieve. Using form (1,3,5,7) to designate 4 columns, the two true endgames are if, after you move, you can get to (2,2) or (1,2,3).

Given this, there are 2 ways to get there.

1. matching - if, after your move, everything is paired up i.e. (3,3) or (1,3,3,1) or (4,4), the you win. You can, from this point on, just copy your opponent until either you can get to a (2,2) or make a winning move.
2. (1,2k,2k+1) - i.e. creating a (1,4,5) in this case. You can copy moves until you get to a (1,2,3), or, if the opponent crosses out the single, drop into the matching strategy.

I left out a whole lot of logic, but its worth a thought.

And to the admin, a decent game is with two players, to have one player draw the game, and the other gets to pick who goes first. Especially at a pub.

that's a tough one! i needed to print out a pretty substantial list of winning/losing boards to see the pattern. i hope i have it right, now, though. seems you ought to look at the base 2 representation, no? if the bit-sum of each bit is 0 (assign each column the number of sticks in base-2), you have a losing board - otherwise you have a winning board. not too hard to show that any move changes this property, and that given any configuration without this property it seems you can make it so with a single move to hand the other player back a losing board, which gives you a winning strategy (as long as you can turn numbers into base-2 reasonably well in your head!).

that said, i guess that's a losing board?

this deserves a 10, I've no real idea of the answer

The set of solutions evolve from infinity to zero.

When radius = 0 (very small) any point is the desired point, as radius increases, the locus of desired points moves towards the poles. When radius = infinity, poles dont exist. Just flat surface, so 0 desired points.

1) cut the first rope in half.
2) burn the second rope from both ends, at the point where the fire meets 30 mins would have past.
3) once the 30mins have past, burn the second half rope from both ends, at the point where the fire meets 15 mins would have past.

?

Person 1 and person 2 cross: 2 minutes
Person 1 goes back: 1 minute
Person 5 and 10 cross: 10 minutes
Person 2 goes back: 2minutes
Person 2 and 1 cross: 2minutes
= 17 minutes

Bob always drinks half of what you drink. Let amount Bob drinks = B, Let amount you drink = Y:
B + Y = 1
1/2B = Y

So B gets 1/3 and you get 2/3

This one look hard but is easy

If you drink 500ml and he drink 250ml, then that's 2/3 for you 1/3 for him.

This ratio won't change for the rest of the liter

125/62.5
31.5/15.625

But hey!! What's the point of drinking water in a bar!

X = Bob Volume

2x+X=1

3X=1

X=0.33

You is 0.67

Jean

4 weighings

1st weighing:
split the marbles into 3 groups of 4. put any two groups on scales

2nd weighing:
- if the original weighing is balanced, take one group off, put 3rd group on, which contains the bad marble. note whether or not the marble is heavier or lighter based on whether the group is lighter or heavier.
- if the original weighing is unbalanced, take the "heavier" group off, put 3rd group on. if balanced now, the group you took off contains the bad marble, and it's heavier. if still unbalanced, the original "lighter" group from the first weighing contains the bad marble, and it's lighter.

3rd weighing:
split the bad group into pairs and weigh. find out which pair has the bad marble.

4th weighing:
split the bad pair into single marbles and weigh. locate the bad marble.

3 weighings
key
l = light (or a potentially light set)
h = heavy
u = unknown

12u marbles

weighing 1 4u v 4u reserve 4u

1.1 level
8g 4u marbles

1.2 tipped
4l, 4h, 4g marbles

weighing 2
1.1 weigh 3u v 3g

1.1.1 level 11g 1u
1.1.2 tipped unknown on light side 9g 3l
1.1.3 tipped unknown on heavy side 9g 3h

1.2 split 4l = 1l + 3l, split 4h = 1h + 3h, split 4g = 1g + 3g
weigh 1l + 3h v 1h + 3g
outcomes
1.2.1 level 3l 9g
1.2.2 same direction, 1l, 1h, 10g
1.2.3 switched 3h 9g

weighing 3
1.1.1 weigh 1u v 1g
1.1.1.1 1l 11g
1.1.1.2 ih 11g

1.1.2 weigh 1l v 1l
1.1.2.1 level have 11g and 1l
1.1.2.2 tips so have 1l 11g

1.1.3 weigh 1h v 1h - same as for the light weighings we
have
1.1.3.1 level have 11g and 1h
1.1.3.2 tips so have 1h 11g

1.2.1 1l v 1l
1.2.1.1 level have 11g and 1l
1.2.1.2 tips so have 1l 11g

1.2.2 1l v 1g
1.2.2.1 level 1h 11g
1.2.2.2 tips 1l 11g

1.2.3 1h v 1h
1.2.3.1 level 1h 11g
1.2.3.2 tips 1h 11g

in each case we know the false and if it is light or heavy

360/12=30
So there is 30 degree per hour (or per 5 minutes)

Between 45 and 2 there is 5 hours, so 5*30=150 degree

or 360-150=210 degree by the other side

 172,5

you go first. place the coin in the middle. From then on, mimic the second players move on the opposite side of the table.

No.
The opposite corners are the same color (either both black or both white). So, removing them leaves you with more squares of one color. A domino always covers black and white squares in pairs. Therefore, no arrangement of dominos can cover the entire board.

23 and 24 have same answer, ha ha.
6x64 = 384
because in a rubik's cube the interior cubes do not exist

10x10x10 becomes 8x8x8 = 512

23 and 24 have same answer, ha ha.
6x9 = 54

I will increase its difficulty. What is the probability to make an obtuse triangle?

1/4

It depends on how good you are at scoring baskets. If we assign a probability p, to your chance of scoring with any given shot, then clearly there are three distinct cases where p<0.5, p=0.5 and p>0.5.

This makes sense also intuitively; if you are a poor shot, you would take one go and hope you scored, if you were a good shot you would feel more confident being allowed one slip up.

If you work through the maths, you see for the two from three game:

s=score, m=miss

mmm (1-p)^3 lose
mms p(1-p)^2 lose
msm p(1-p)^2 lose
smm p(1-p)^2 lose
mss p^2(1-p) win
sms p^2(1-p) win
ssm p^2(1-p) win
sss p^3 win

we see that we need 3*p^2(1-p) + p^3 > p in order for it to be better to play the two out of three game. This is satisfied for p>0.5. For p=0.5, it makes no difference, both have an equal chance of winning. p<0.5 you are better off taking a single shot.

Here is a hint:  how can you simulate a fair dice using a fair coin?

To simulate a dice with a good coin

You need to throw the coins three times.

If first time is Tail

1,3,5

If Firts time is head

2,4,6

Out of the two next throws, one needs to get three equaly probable events . we can select Tail, Tail and  Head, Head  and Tail , Head  (if we get Head, Tail we ignore and replay)

THen we can say Tail,Tail is the lower of the three (either 1 or 2)

Head, HEad is the Higher: either 5 or 6

andTail ,H ead is the middle: either 3 or 4

Know we have to find a way to aplly this system to a biased dice.

One possibiilty is to toss the dice as many time as it takes to get all numbers 1,2,3,4,5,and 6 and record the number of trials N1

THen the operation is repeated if the number of time N2 is <N1 we consider we get a  tail if N2>N1 it is a head

if N1=N2 we ignore the results

THe operation is repeated until we get two other numbers

Jean

Roll it six times. If the nth throw shows the highest number, n is our answer. If there is a tie, repeat.

while (true){
roll the dice six times
if you get a permutation of (1,2,3,4,5,6) (i.e. each throw results in a different number), then return the first throw
}

not an expert in stats, but I think the proof is as follows:

the chance you get any one permutation of (1,2,3,4,5,6) is p1*p2*p3*p4*p5*p6 (where p1 - prob you get 1 in one toss, etc.., and p1+p2+p3+p4+p5+p6 = 1)
for each n in [1:6], there are 5! permutations of (1,2,3,4,5,6) with n being first.
Thus the probability of getting n from the algorithm above is (5!) * p1*p2*p3*p4*p5*p6, equal for all n in [1:6]