m00li

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exactly 6 rolls have a result of 5 = ^{50}C_{6}(1/9)^{6}(8/9)^{44}
10 or fewer rolls have a result of 5 = S_{0>10}^{50}C_{k}(1/9)^{k}(8/9)^{50k} , where S_{a}_{>b} denotes sum over a to b

The youngest child among the oldest children in each column is NEVER younger than the oldest child among the youngest children in each row.
He is either older or of the same age.
The same age happens when the youngest in any row is the oldest in that column.
Number of ways in which the ages could be same can be calculated as below:
Choose any place (chosen cell) in the 5x10 matrix. This can be done in 50 ways
The kid in this cell must be the youngest in the row and oldest in the column. All other places can be filled randomly.
So, there are 9 (in row) and 4(in column) "special cells" where combinations must be made. For other cells, the rest can be placed in 36! Ways.
The 9 chosen kids in rows and 4 chosen in column can be palced in 9!*4! Ways.
Now the only thing that remains to do is to choose 1 kid for the chosen cell and 9 kids for the rows and 4 kids for the columns in the special cells.
For any given kid lets say there are x kids younger than him/her and 49x elder kids
Now we need to choose 4 kids out of x younger kids and 9 kids out of 49x elder kids.
Note that x varies from 4 to 40 ONLY.
^{4}C_{4} * ^{45}C_{9} + ^{5}C_{4} * ^{44}C_{9} + .... + ^{40}C_{4} * ^{9}C_{9}
Using excel for this, the total combinations comes out to 9.378E + 11
Now, the total possible ways of placing the kids is 50!
Total ways in which the age is same is = 50 * 36! * 4! * 9! * the above combinations
Relying on excel getting the calculations right, there is 1 in 200 times* that the age is same.
So, the probability we are looking for is 1/200 = 0,5%
* Excel calulated it to be 1 in 200.2 times. I simplified it to make the answer look elegant
Nicely done

you posted the answer too!
realized now that ts an old problem

Yes Bonanova, I did not permit zeros as I thought in that case Rainman would have given a different initial reply (that he can never find out YSan's product)

3*8
you never mentioned all 4 had to be used 
Fifty school children lined up at random in five rows and ten columns.
No two of the children have the same birthday.
What is the probability that
the youngest child among the oldest children in each column is
NOT older than
the oldest child among the youngest children in each row ?

I think there is only one answer. Here are the details of what I did. See if I missed something.
You said, since Rainman knows the product, he doesn't even have to think about the other number.
The problem is that, if there are two possible numbers, he doesn't know the answer.
Consider the two suggested answers: 1114111 and 9869689.
They have sums of 10 and 55 respectively
They have products of 4 and 1679616 respectively.
Initially
The 2200 possible ABCDCBA numbers have 64 different sums, from 0 to 63.
9 of them, 0 1 2 3 4 5 6 62 and 63 correspond to unique products.
They are, respectively, 0 0 0 0 0 0 0 4251528 and 4782969.
Since Ysan can't determine the product, none of these are her sum.
We can eliminate numbers having those sums, leaving 2177 possibilities.
After Ysan's first statement
The 2177 possible numbers have 679 different products.
327 of them, that do not include 4 or 1679616, correspond to unique sums.
Since Rainman can't determine the sum, none of them are his product.
We can eliminate numbers having those "unique" products, leaving 1844 possible numbers.
Since 4 and 1679616 were not eliminated, 1114111 and 9869689 are still possible.
After Rainman's first statement
The 1844 possible numbers have 52 different sums.
5 of them, 7 8 56 57 58, correspond to unique products.
The products are, respectively, 0 0 1679161 2125764 and 2125764.
Since Ysan still can't determine the product, none of these are her sum.
We can eliminate numbers having those sums, leaving 1825 possible numbers.
After Ysan's second statement
The 1825 possible numbers have 351 different products
At this point, only 1 of them, 1679616, corresponds to a distinct sum, 55.
Rainman says I know kman's number
So Rainman must have been given the product of 1679616.
So (product=1679616) points to only one number
9869689 sum=55.
While (product=10) points to two different numbers that have different sums.
1114111 sum=10
2111112 sum=9
==========
Looking closer,
Before Ysan's second statement, (product=1679616) pointed to two different numbers.
(1679616=TABLES1[;7])/[1]TABLES1 < this asks for rows where col7 (prod)=1679616
9 8 6 9 9869689 55 1679616
9 9 8 4 9984899 56 1679616 < so the number 9984899 was possible at this point.
Then Ysan's second statement eliminated (sum=56), (see red above) making 9869689 unique.
Also, before Ysan's second statement, (product=4) pointed to two different numbers.
(4=TABLES2[;7])/[1]TABLES2 < this asks for rows where col7 (prod)=4
1 1 1 4 1114111 10 4
2 1 1 1 2111112 9 4
Ysan's second statement did not eliminate (sum=9) or (sum=10).
This leaves (product=4) ambiguous. So Rainman's product could not be 4.
In any event, I'm marking the puzzle solved. Good work.
First of all, thanks for the star. Its been decades since I got one
Now,..
"The 1844 possible numbers have 52 different sums.
5 of them, 7 8 56 57 58, correspond to unique products."
Actually, 7 and 8 are eliminated earlier as both correspond to unique numbers 1111111 and 1112111. So rainman would have guessed them straight away as he would have got the product 1 or 2, respectively. Your statement should read:
"4 of them, 9 56 57 58, correspond to unique products."
The sum 9 gets eliminated here. 9 gives two products3 and 4. If rainman has 3, he will tell the sum, immediately after ysan's first statement.
ALTERNATIVELY  you state "327 of them, that do not include 4 or 1679616, correspond to unique sums". 3 happens to get eliminated here as it has a unique sum 9.
Therefore after Rainman's 1st statement the sum 9 corresponds to only 1 unique product 4. It has to be eliminated otherwise YSan will know rainman's product (and the number). Thus "4 of them, 9 56 57 58, correspond to unique products."

Nice work.
But since Rainman knows the product, he knows at least one of these is not the right number.
Since, Rainman knows the product, he doesn't even have to think about the other number. This doesn't change the fact that there are indeed two answers to your problem.
Or, am I wrong in stating that there are two possible answers to your question?

1114111
9869689 
a 3 legged stool with vertically straight legs of heights 10 meter,1 meter and 1/2 meter will be very unstable. do u mean to say all three legs are equal? if so, then i do not understand the question

15:27:34
 1

I didn't do it, I just remembered it

has to be Mary
as no one else is present in the living room according to the finding in my first post. she has to be there and be the killer 
four,
number of letters used

thats why i wrote that it needs some refining as theoretically speaking you cannot apply the same strategy any more but have to modify it now,

Player 2 can guarantee at least a tie: Put a mirror in the center of the paper. Anytime player 1 places a point, place yours at its mirror image's location. On player 2's final move, since player 1 cannot move again, player 2 is free to look for whatever spot on the paper will give him the most area and he might be able to get a win, but will at the very least be able to get a tie by copying player 1's move.
Finding a strategy for player 1 that forces that to become a tie rather than an outright win for player 2 when he makes his last move is not easy (for me).
How can you mirror player 1's first move in the center of paper?
If player 1 places their point at (0, 0), place yours at (0, 1/infinity).
Assume player 1 knows your strategy
Assuming you place the your first point at (0,1/x), where x is very very large but finite, player 1 keeps it at (0,1/x+1/y),where y is very very large but finite. By your strategy, you will place your 2nd point at (0,1/x1/y). Player 1 now keeps his 2nd point at (0,1/x).
Now, where do you keep your 3rd point, as (0,1/x) is already occupied?

question looks incorrect.
the informants are providing info only about who was in which room, so we can at best deduce who was where. The question assumes that by knowing who was where, we will know the murder. The only relationship between a murderer and location information is that the murderer was alone in his room. This must imply that the murderer is the ONLY person who is alone in a room, otherwise we cannot deduce the murderer from the location information alone.
Having said that, additional information on why I think so:
Assume: Pat is not in 2nd bedroom. Implies, John, Mary, Tim and Don are liars. Implies, Pat and Kelly speak truth. But they cannot be as they do not concur on Don's location. Hence our assumption is wrong and Pat is in 2nd bedroom.
Fact established: Pat in 2nd bedroom. Pat and Kelly are liars
Assume: John speaks truth. Implies, Tim is in bathroom and John in 2nd bedroom. Implies Mary and Tim are liars as they imply John is not in 2nd bedroom. Also implies, Don lies, as he says Tim is in kitchen. Hence, there are 3 new liars, totalling 5 liars, which is a contradiction.
Fact established: John is a liar
Assume: Mary speaks truth. Implies, John is in living room and Tim is in Kitchen. Implies, Tim and Don are liars. Contradiction as there are again 5 liars (including above 3 established liars)
Fact established: Mary is a liar.
Fact established: Tim and Don speak truth
Fact established: Kelly is in bathroom. John is in 1st bedroom. Tim is in kitchen. Don is in 2nd bedroom with Pat.
Mary may have or might not have been in kitchen. In any case it will leave at least two people (among Kelly, Tim and John) alone. But to infer the murderer we need only one person alone. Hence I think information is insufficient. 
The first player can not lose this game if he plays it logically. consider this:
wherever the 2nd player puts his 'current/last' point, the first player should put his point such that:
his point lies on a line from the center to the 2nd player's 'current/last' point and immediately behind it (i.e. a ray starting from the center will hit 2nd player's 'current/last' point and then immediately hit first player's 'current/last' point)
If I'm the second player and I know you'll use that strategy, then after you place your first point, I will draw two lines going through your first point and running parallel to the edges of the paper, and I'll place my four points where those lines reach each edge of the page. Well, maybe just a tiny bit away from the edge so you could still place a point just a little bit farther away from the center than mine. I would then win with 75% of the area.
and I concede defeat

after each drawing the number of white balls in the urn can decrease only by 2 or not decrease at all.
After each drawing the number of black balls can either go down by 1 or go up by 1
Hence, at the outset (assuming there is at least one ball in the urn to begin with):
if the number of white balls is even (or 0), ultimately all white balls will finish and only 1 black ball can remain.
if not, ultimately only one white can remain 
More recursions:
Let f(t,m) denote the number of ways of arranging m colored balls in t slots, where t >= m and supply of each m coloured ball is infinite. Then final answer is ^{n}C_{m}f(t,m) / n^{t}
Here are the different recursions I have come up with for f(t,m) (alas, no direct formula in m,n)
1) m^{t}  ^{m}C_{1}f(t,1)  ^{m}C_{2}f(t,2)  ^{m}C_{3}f(t,3) ....  ^{m}C_{m1}f(t,m1)
2) ^{t}C_{1}f(t1,m1) + ^{t}C_{2}f(t2,m1) + ^{t}C_{3}f(t3,m1) + ... + ^{t}C_{tm+1}f(m1,m1)
3) mf(t1,m1) + m^{2}f(t2,m1) + m^{3}f(t3,m1) + ... + m^{tm+1}f(m1,m1)
4) m( f(t1,m) + f(t1,m1) )
where f(t,1) = 1, f(t,2) = 2^{t}2, and f(t,t) = t!
And finally one kind of non recursive form:
5) f(t,m) = m^{t}  ^{m}C_{1}(m1)^{t} + ^{m}C_{2}(m2)^{t}  ^{m}C_{3}(m3)^{t} + ....(1)^{m1}(^{m}C_{m1}) (This one gives the AHA insight into the solution)
 1

The first player can not lose this game if he plays it logically. consider this:
wherever the 2nd player puts his 'current/last' point, the first player should put his point such that:
his point lies on a line from the center to the 2nd player's 'current/last' point and immediately behind it (i.e. a ray starting from the center will hit 2nd player's 'current/last' point and then immediately hit first player's 'current/last' point) 
understood the exact question now

More recursions:
Let f(t,m) denote the number of ways of arranging m colored balls in t slots, where t >= m and supply of each m coloured ball is infinite. Then final answer is ^{n}C_{m}f(t,m) / n^{t}
Here are the different recursions I have come up with for f(t,m) (alas, no direct formula in m,n)
1) m^{t}  ^{m}C_{1}f(t,1)  ^{m}C_{2}f(t,2)  ^{m}C_{3}f(t,3) ....  ^{m}C_{m1}f(t,m1)
2) ^{t}C_{1}f(t1,m1) + ^{t}C_{2}f(t2,m1) + ^{t}C_{3}f(t3,m1) + ... + ^{t}C_{tm+1}f(m1,m1)
3) mf(t1,m1) + m^{2}f(t2,m1) + m^{3}f(t3,m1) + ... + m^{tm+1}f(m1,m1)
4) m( f(t1,m) + f(t1,m1) )
where f(t,1) = 1, f(t,2) = 2^{t}2, and f(t,t) = t!

There is no information given in the question on consensus building. If I understand your assumption correctly, if 500 (captain) proposes a distribution in which no. 499 to 201 don't get anything, they will still agree to it. This implies that they fear, that killing no.500, will start an avalanche of killings. Why?Lets see this in more detail.I am assuming that it is established that if there are 202 pirates. No 202, will distribute 1 coin to no. 200, 198,.....,6,4,2, and they will all live.Now, if there are 203 pirates(p=203). Suppose 203 again proposes 200,198,196,...,4,2 to get 1 coin each. No one will agree, because they know they can make him walk the plank, and still get the same distribution. So, no. 203 has to propose a distribution among 202,201,199,197,195,...5.3.1 but he can pacify at most 100 and thus can never get the majority. So no matter what, in this scenario, no. 203 always dies.Next, assume there are 204 pirates. No matter what 204 proposes, no. 202 to 1 will have at most 100 consents (as they know, that none of them from 1202 can die). If 203 doesnt agree, he dies next, so 204 will get 101 votes and including his own, he can survive. So, if there are 204 pirates they all live but 201,202,203,204 do not get anything.p=205: captain can never survive as the remaining 204 know that, when 205 dies, they will survive. (this is exactly like p=203 in the sense that, the captain dies no matter what)p=206 to 207: captain proposes 1 coin distribution to 2,4,6,8,...,200 and gets 100 votes. no. 201204 will never agree. so 104 dissenters. hence the captain dies in p=206207p=208: if captain dies, so do 207 to 205, so 205208 will agree to what 208 proposes (distribute 1 coin to 2,4,6,..,200) and there will be 104 votes. so no. 208 survivesp=209 to 215: they can propose to distribute to 1,3,5,7..,199 but will have at least 108 dissenters, so the captains will always diep=216: proposes to distribute 1 coin each to 2,4,6,8,...200 and survivesHence, the final solution:(1) n<=200:n keeps 100  [(n1)/2] coins for himself and gives 1 coin each to no's n2,n4,n6,...(2) for n>200(2.1) Let n = 200 + 2^{k} (k=1,3,5,7...) i.e. n=202,208,232,...captain proposes 1 coin each for 200,198,196,..,6,4,2 and there are exactly 100 + 2^(k1) dissentors so the proposal is accepted(2.2) for n=200 + 2^{k} (k=0,2,4,6,8..) i.e. n=204,216,264,..captain proposes 1 coin each for 199,197,195,..,5,3,1 and there are exactly 100 + 2^(k1) dissentors so the proposal is accepted(2.3) for n = 200 + m (m is not a power of 2 and m>2)captain dies no matter what he proposes, next captain dies no matter what, and so on, till n reaches (2.1) or (2.2) above and follows that case respectively.
Hard number sequence: 4, 8, 16, 26, 41, 57, 79, 104, 138, 184, 241, ?
in New Logic/Math Puzzles
Posted
Let S_{1}=(n2)(n3)(n4)(n5)(n6)(n7)(n8)(n9)(n10)(n11)
S_{2}=(n1)(n3)(n4)(n5)(n6)(n7)(n8)(n9)(n10)(n11)
S_{3}=(n1)(n2)(n4)(n5)(n6)(n7)(n8)(n9)(n10)(n11)
so on till
S_{11=}(n1)(n2)(n3)(n4)(n5)(n6)(n7)(n8)(n9)(n10)
where n is any natural number
then the sequence is given by