In 1896 lord Coin has decided to play a game. From the January 1 till December 31 every day he chooses among two match boxes an arbitrary one and placed a match from it to another box (if the chosen box was not empty). If the chosen box was empty then he placed a match from
the other box to the chosen one. What is the probability that after the December 31 the both boxes will have an equal number of matches if at the beginning each box had a) n = 400 b) n = 200 c) n = 100 matches?
Complete bouncer .... just waiting for the solution
ReplyDeletejust got close to ... there are 365 days ... arbitary box can be selected with 0.5 probability ... different 'n' and 365 days will have some effect on probability ..
ReplyDelete:)
waitin for the soln too .. 1896 is a leap year as its a multiple of 4 . so its 366 days not 365.
ReplyDeleteWhen n is 400 is it 0.5 * 366 for the boxes to be equal ?
explain pls
In the first case you can never get an empty box, while in the second case, if at any moment you exhaust one of the boxes you can never get back to boxes with the same number of matches, so in both cases the probability is equal to the number of sequences of length 366 of 1's and -1's summing 0 divided by the total number of such sequences, or binom(366,183) / 2^366, approximately 0.04. Case c must be more complicated.
ReplyDeletehow about the case in which the same box is chosen in every alternate turn to end up with the same no. of matches (duh...)
ReplyDeleteABABABABABAB... 366 times ... so the prob wud be (1/2)^366
case a) 400 matches, 366 days so no way to ever empty a matchbox
ReplyDeleteways in which we can get equal matches = ways of selecting matchbox A 366/2 times = 366 C 183
total no of ways of selecting matchboxes = 2^366
therefore prob = 366!/(183!^2)*2^366 (phew!)
case b) 200 matches - caseb.1 no matchbox emptied
ReplyDeletecase b.2 matchbox A emptied
b.1 = same as above
b.2 = 0 as emptied matchbox cannot be filled up in time
case c) 100 matches case c.1 same as above
case c.2 matchbox a emptied...and refilled
A selected B+200 times
AAABAA... and A+B = 366, so 2b+200 = 366, b = 83,a = 283,
ans = (366 C 83)/ 2^366.... multiply by 2 to get probability for b emptied and refilled as well