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#26 Re: Help Me ! » Guess the coins » 2016-04-09 13:29:03
then he would have said 2,2 because both numbers (9,10) are in both subsets.
What happens if Alexander says (1,1) when he has picked 9,10?
#27 Re: Help Me ! » Guess the coins » 2016-04-09 09:23:16
OK, so:
Let's name the coins 1 to 10. We have 9 possible coin sets to guess: 1-2, 2-3, 3-4, 4-5, 5-6, 6-7, 7-8, 8-9 and 9-10.
All possible answers that Alexander can give are 0, 1 and 2 (coins) for each subset, thus 9 possible combinations.
Let's take these two subsets for example: A: {3,4,5,9,10} and B: {5,7,8,9,10}.
Below we list Alexander's replies for each subset A - B and the set of numbers that Byron guesses:
A B
0 0 ----->1,2
0 1 ----->6,7
0 2 ----->7,8
1 0 ----->2,3
1 1 ----->5,6
1 2 ----->8,9
2 0 ----->3,4
2 1 ----->4,5
2 2 ----->9,10
So this way Byron can always win, no matter which pair of number Alexander has put in mind.
#28 Re: Help Me ! » Guess the coins » 2016-04-09 03:48:12
This is just my guess, maybe we must wait for Salem to confirm 
But it doesn't say "two different subsets"!
That changes things a lot.
#29 Re: Help Me ! » Guess the coins » 2016-04-09 03:30:53
I mean that the two subsets may also have some common numbers.
Overlapping? I do not understand.
#30 Re: Help Me ! » 15 people in a row » 2016-04-09 01:31:06
Number of combinations with 3 adjacent males to 15 seats: 1586131 (pattern is 3^13 - 2^13)
Number of combinations with 2 adjacent females to 15 seats: 1586131 (pattern is 3^13 - 2^13)
Number of combinations with both 3 adjacent males and 2 adjacent females to 15 seats: 523250 (pattern is 3^(13-1) - 2*2^(13-1) + 1)
Now what?
#31 Re: Help Me ! » Guess the coins » 2016-04-09 00:34:35
Bobby, now that I think of it, the subsets can also be overlapping. I think that I am close to something...
If I understand correctly choosing two subsets will split the 10 numbers into 3 pieces. Is that correct?
#32 Re: Puzzles and Games » Day-off » 2016-04-07 22:12:46
I think it must be something related to some unique property - maybe related to the number of employees (20). Otherwise the number is not used!
#33 Re: Help Me ! » 15 people in a row » 2016-04-07 06:25:52
So we can have 10 men at maximum and 8 women at maximum, if this is of any help
#34 Re: Help Me ! » Guess the coins » 2016-04-06 04:15:27
Very intriguing puzzle though!
#35 Re: Help Me ! » Guess the coins » 2016-04-05 13:34:46
The two subsets plus the ones not belonging to any of the two subsets. Right.
#36 Re: Help Me ! » Guess the coins » 2016-04-05 09:10:29
Cannot even imagine any solution!
#37 Re: Puzzles and Games » Perfect Squares » 2016-04-04 21:27:55
Wow! Nehushtan, I like your solution!!
#38 Re: Puzzles and Games » Day-off » 2016-03-28 05:59:46
Nice observation, but how would he determine the "first"? For example, what if 5 students didn't show up at school on the next day? Who would be the first?
Hi, I guess he said in his announcement that "the first employee who accept his advice will be charged with the day off" so since no one of them would like to be charged with the leave day, no one will accept the advice first and all of them will come to work the next day hence the manager achieves his objective. (This is a kind of "foxy" advice)...
#39 Re: Puzzles and Games » Day-off » 2016-03-26 23:44:52
Hi Samuel,
So I guess it would not be a specific employee, but someone who would meet some condition, right?
#40 Re: Help Me ! » Integers arrangements » 2016-03-21 03:06:18
Wow!! Heap sort!
#41 Re: Help Me ! » Train to Dawanglu » 2016-03-20 01:11:39
Bobby, sorry to sound a bit stupid; what is "simulation"?
Just did a rough simulation. I think the exact answer is 301 / 625
#42 Re: Help Me ! » Train to Dawanglu » 2016-03-15 21:26:00
but "in mathematics you don't understand things..." - right?
I think I understand now.
#43 Re: Help Me ! » Train to Dawanglu » 2016-03-15 15:07:46
My understanding is that we have at maximum one train per minute: Since the probability for the train to come is 70% within each minute, this means it may come but there is also a chance it may not come (with 30% probability) within the first minute. Same for the second minute: again the probability for the train to come is 70% and so on.
#44 Re: Help Me ! » Train to Dawanglu » 2016-03-15 08:49:27
Yes, but with probability 70% in each minute. I am not sure what exactly this means. If, for example, the first train comes exactly at 8:01, and they miss it, then within the second minute, a second train may come, but with probability 70% (and they may also miss it).
For the first minute, the overall probability is 0.7 x 0.4, but how about in the 5 minutes?
That seems to mean there are 5 trains coming, one for each minute.
#45 Re: Help Me ! » Train to Dawanglu » 2016-03-15 07:56:22
I don't think it is like this. It says that every minute there is 70% probability that the train comes (within this minute). Right?
That sort of assumes that the trains come exactly every minute?
#46 Re: Help Me ! » Train to Dawanglu » 2016-03-15 07:22:07
So, would it be [(7/10)x(4/10)]^5?
Hi;
I would think so but the problem is unclear to me.
#47 Re: Help Me ! » Train to Dawanglu » 2016-03-15 02:38:43
@bobbym: is it correct to assume that the probability (for the two guys to catch the train) within EVERY minute from 8:00 to 8:05 is independent, since it is uniformly distributed in the course of each minute?
#48 Re: Help Me ! » Open polo » 2016-03-08 10:14:02
So now you can focus on my rotten apples
Absorbing Markov Chains are a good way to describe random walks like this. They only look tough and they are a breeze to compute.
#49 Re: Help Me ! » Open polo » 2016-03-08 07:50:41
That's very hard for me to understand, but the result is the same as the one I got with my simplistic approach
Many thanks!!
Hi;
The Markov Chain of this process is,
We can use an initial state vector of
orto compute the chance that the team starts at home versus on the road. Getting the mean of the first passage time to the appropriate absorbing state (one of the last 4 rows) we get a expected number of games of 16 / 3
#50 Re: Help Me ! » Open polo » 2016-03-04 21:32:42
Good morning to all of you
I believe the answer is 5,3333 and I think that Nehushtan's post #3 is very close to the solution. Indeed we have an even number of games, and we can examine two different cases, one is WW (2 wins, or LL, symmetrically) and all others that start with an even number of games where nobody wins and then we have a winning streak of 2 more games (for example, WHWHWH plus HH at the end). The first case is with probability 2 . 3/4 . 1/4 = 3/8. Thus the other cases (those with more than 2 games in total) occur with probability 5/8. But then what?
If I am to put this into a Markov chain then I must assign probabilities to who goes first. I can assign .5 to each one. Before I start, do you have the answer to this problem so we can tell if I get there?