1but celestine in ur 2nd eqn........u need 2 cases when d flipped coin shows head and when it shows tail.............uve taken only when it shows head..............check tht part............
Someone walks into your room and dumps a huge bag of coins all over the floor so that no coins are on top of any other .A robot then comes into the room and is programmed such that if it sees a head, it flips it to tails. if it sees a tail, it throws it in the air. the robot moves around randomly forever. will there be a convergence in distribution of heads vs. tails?
If yes, what?
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26 Answers
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9yes i got that
only their relative ratio can be const not the tot no
62okie.. celestine.. i will check that.. but i hope u got what i am trying to do ;)
the post above Highlight Replied#18 Posted 00:54am 14-12-08
had a mistake i am hiding that.. i realised it as soon as i posted that one ;)
9but nishant acc to ques wen tail comes head doesnt inc by 1 only tail dec by 1
so ur logic isn correct acc to ur q
please verify post 14 ;)
62it is the same i guess...
at infinty time the distribution will remain same if you can find a number for which the next time the probability will remain the same!
This could be a bit ahead of time for u to udnerstand.. but this is not
"purely" a question fo JEE level.
1Doubt............why at infinte time shud d distribution be same........shudnt v find tht??????
62see at the end of the long time the distribution of the Head and tail shoud remain same..
so the value of H that is shoudl be the same.. (There is something alled the expected value.. but we will not go into it)
If we get a head the no of heads will decrease...
if we get a tail no of heads will increase by a probability 1/2
There was a typo above...
H = (H-1) (H/H+T) +1/2(H+1)(T/H+T)
Solve this to get the answer :)
1And if theres a flip wont d no. of heads dec by 1 and d no. of tails becums T+1.............rite??
62Dont read.. Nishant went mad for a bit! :D
Assume that no of heads is H
and tails is T
then if a flip occurs the number of heads should remain the same
"There are flaws in this proof" pls find them.. but the answer remains the same :)
H = (H-1) (H/H+T) + (H+1)(T/H+T)
Solve H/T you will get the answer :)
1i mean ...........d second part..........when it is thrown into air.........for tht ull hav 2 cases rite..........one head and one tail.........i meant tht not flip sorry.......
9no akand i had by mistake typed the opp and uve justified it too;)
ive edited it now
let at steady state H=xT
H/H+T → H-1 , T +1 ( wen it picks head)
T/H+T → H ,T-1 ( wen it picks tail)
jus eq prob of loosing H = X times prob loosing T
youll get X=2
1wel i think..........lets assume there r 50 heads and 50 tails initially....
so when d robot sees a head it makes into tail....and when it sees d tail it flips..........after flipping we may either get a head or a tail....if we get head it becomes tail...........so we hav another tail............but if v get a tail it again flips and again v hav 2 cases..........it continuos forever..........
so i think wat celestine has written is correct........no. of tails=2no of heads
9got it
no of heads = 2 X no of tails
cos only then
prob of loosing head = 2 prob of loosing tail
11This'll go on forever and ever and robot's battery life'll get over! That's the best ans i've got fr this question.
P.S : Pls don't b offended by it! :)
33The no of heads and tails converges to or say no of heads and tails after infinite time become...
11What do you mean by "a convergence in distribution of heads vs. tails"?