subject:"Re\: \[algogeeks\] Re\: puzzle"

Re: [algogeeks] Re: puzzle

2012-08-12 Thread Amitesh Singh

Does the pattern comes in this way? HT,TH,TT or HT(X)TH(X)TT ??

Let me know.

--
Amitesh

On Sat, Aug 11, 2012 at 11:24 PM, Piyush pkjee2...@gmail.com wrote:

How can I find the expected number of tosses , required to obtain a
{HT,TH,TT} , by using random variables??

On Friday, December 31, 2010 8:27:46 PM UTC+5:30, Dave wrote:

@Anuj and Bittu: It is not necessary to know the bias. You can
simulate the flip of an unbiased coin with multiple flips of a biased
coin: Flip it twice. If the result is HT, consider it a Head. If the
result is TH, consider it a Tail. If the result is HH or TT, repeat
the process. It terminates with probability 1. Now use the resulting
Head or Tail in the procecure for deciding with a biased coin.

Dave

On Dec 31, 7:07 am, Anuj Kumar anuj.bhambh...@gmail.com wrote:
in case the coin is not biased, we can flip the coin twice and define
the
rules as if {H,H} comes then ignore it i.e. dont take it as a flip and
the 3
other events would be valid onces and could occur with equal
probabilities.

In case of a biased coin please specify the probability of getting
heads and
that of getting tails.

On Fri, Dec 31, 2010 at 4:11 PM, bittu shashank7andr...@gmail.com
wrote:
At a restaurant, how can Veronica choose one out of three desserts
with equal probability with the help of a coin? What if the coin is
biased and the bias is unknown?

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Anuj Kumar
Third Year Undergraduate,
Dept. of Computer Science and Engineering
NIT Durgapur- Hide quoted text -

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Re: [algogeeks] Re: puzzle

2012-08-12 Thread Amitesh Singh

if you meant to calculate the E[x] for [HT,TH,TT]. It can be solvable but
very lengthy/boring.

I shall give you an example which should help you.

Let E[X] = x be the expected no. of coin flips to get [HT]

1) if first flip is a tail, we have wasted one flip hence. E[X1] = 1/2*(1+x)
2) if first flip is a head, and second flip is a head, hence E[X2] =
1/4*(1+1+x)
3) if first flip is a head and second flip is a tail, we are done then.
hence E[X3] = 1/4*(1+1)
We have,

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