Set 1-8 in these 8 boxese such that consecutive numbers wont intersect vertically/horizontally or diagonally

                                                                   
                                             
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You have a dice which has numbers from 1..6. Another dice is provided to you empty; you need to place numbers on this dice so that whenever both dices are used there is an equal probablity of getting a sum between 1.. 12

Solution::
Totally there are 6*6 possible combinations from two dices. Since we need the sum from 1..12 with equal probability there should be three combinations among these 36 which sum up to each number in 1 to 12 to achieve a equal probability for all sums.

Other Dice Numbers Presently _ _ _ _ _ _

Getting 1 we need 0 on the other dice
Getting 12 we need 6 on the other dice

State of Numbers on other dice now 0 6 _ _ _ _

We need to get SUM=1 in three different combinations and only possibility is 1 and 0 on both dices so put two more 0's on second dice and we have three ways to get 1.

State of numbers on other dice now 0 6 0 0 _ _

Now all the sum's from 7 to 12 can be achieved with one of the numbers be 6 and using the other numbers from first dice. Eg:: 7 can be 1 and 6 from dice 1 and dice2 and if we put two more sixes in dice2 then again 1 and other 6's on dice2 give us 3 ways for seven
Similarly for 8 we can 2 from dice1 and the three sixes one by one we will have 3 ways for 8 and so on for all numbers from 7 to 12. Again we can use the three 0's on dice2 in the similar way we have used three sixes above for getting three ways for sum 1 to 6.

State of numbers on other dice now 0 6 0 0 6 6

In how many ways 3 identical coins can be placed in 5x5 grid so that no two coin come in same row and same column

First coin can be placed in one of 25 places. For second, only 16 places are possible because of the same column and same row condition. Similarly, only 9 places for third coin. So, total ways = 25 * 16 * 9. but since coins are identical you can't distinguish one from another and thus final answer is ( 25*16*9 ) / 3 = 1200.

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Two_envelopes_problem

1. Someone gives you two envelopes with cash inside, and tells you that one of the envelopes has twice as much as the other. You're asked to pick an envelope and open it. You pick Envelope 1 and you open it to reveal $100. Awesome! And now that person asks you if you'd like to switch to Envelope 2, for a cost of merely $5. What should you do ?
2. The basic setup: You are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are offered the possibility to take the other envelope instead.^[3]
3. The switching argument: Now suppose you reason as follows:

1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.
6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
7. So the expected value of the money in the other envelope is
   
   
8. This is greater than A, so I gain on average by swapping.
9. After the switch, I can denote that content by B and reason in exactly the same manner as above.
10. I will conclude that the most rational thing to do is to swap back again.
11. To be rational, I will thus end up swapping envelopes indefinitely.
12. As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction

1. The puzzle: The puzzle is to find the flaw in the very compelling line of reasoning above

1. http://en.wikipedia.org/wiki/Two_envelopes_problem

There are n buses in a city. Each of them carries at most m passengers. Find the probability that at least two of them carry the same number of passen

There are n buses in a city. Each of them carries at most m passengers. Find the probability that at least two of them carry the same number of passengers.



This is equal to 1-(probability that they all carry different number of passengers). That probability in paranthesis is equal to (m+1)!/((m+1-n)! * (m+1)^n). This is given that (m+1) >= n. If (m+1) < n, then the probability is 1.

- m+1 because maximum m passengers mean m+1 different possiblities (a bus could carry 0 passengers).
- The first bus has (m+1) possibilities in terms of number of passengers. The second has m+1, the third has m+1 and so on. Total number of orderings possible is (m+1) ^ n.
- For the number of passengers to be different in each bus, first bus has m+1 possibilities, second has m, third has m-1 nd so on. For n buses, this is equal to (m+1)!/(m+1-n)!. Dividing this by (m+1)/n gives the possibility. Subtract this number from one to find at least two buses with the same amount of passengers.


Solved by James