probability and statistic question?

Question

consider flipping an unfair coin. the probability of getting a head is 6/10. the probability of getting a tail is 4/10. determine how many times we need to flip the coin such that 50% are heads.
for example, how many times you need to flip the coin so that out of all total number of flips, you'll end up with half of them heads.

determine the expected number of flips that need to be flipped to find out if we can obtain 50% heads.
note that the probability of getting head is 60% and tail 40%

Answer 1

I don't think you worded this question properly.

I can flip a coin twice and get 1 heads and 1 tails.

I doubt that's what you're looking for.

Answer 2

this is an example of a random walk.

define the mass function f such that

f(x) = p for x = 1 (coin toss is a head)
f(x) = 1- p for x = -1 (coin toss is a tail)
f(x) = 0 other wise

Let Sn be the partial sum of n coin flips, i.e., Sn = X1 + X2 + ... + Xn.

To have the same number of heads as tails you will have S_2n = 0. not that 2n is always an even integer, since you cannot have an odd number of tosses and a sum of zero.

P( S_2n = 0) = 2n C n * (p) ^ n * (1-p) ^ n

where p = probability of a head.

In this case, here is a list of the probability that on the 2n th toss you will have S_2n = 0

P(S_ 2 = 0) = 0.48
P(S_ 4 = 0) = 0.3456
P(S_ 6 = 0) = 0.27648
P(S_ 8 = 0) = 0.2322432
P(S_ 10 = 0) = 0.2006581
P(S_ 12 = 0) = 0.1765791
P(S_ 14 = 0) = 0.1574077
P(S_ 16 = 0) = 0.1416669
P(S_ 18 = 0) = 0.1284447
P(S_ 20 = 0) = 0.1171416
P(S_ 22 = 0) = 0.1073443
P(S_ 24 = 0) = 0.09875672
P(S_ 26 = 0) = 0.09116005
P(S_ 28 = 0) = 0.08438816
P(S_ 30 = 0) = 0.07831221
P(S_ 32 = 0) = 0.07283036
P(S_ 34 = 0) = 0.06786075
P(S_ 36 = 0) = 0.0633367
P(S_ 38 = 0) = 0.05920315
P(S_ 40 = 0) = 0.05541415
P(S_ 42 = 0) = 0.05193097
P(S_ 44 = 0) = 0.0487207
P(S_ 46 = 0) = 0.04575509
P(S_ 48 = 0) = 0.04300978
P(S_ 50 = 0) = 0.0404636
P(S_ 52 = 0) = 0.03809804
P(S_ 54 = 0) = 0.03589682
P(S_ 56 = 0) = 0.03384557
P(S_ 58 = 0) = 0.03193155
P(S_ 60 = 0) = 0.03014338
P(S_ 62 = 0) = 0.02847091
P(S_ 64 = 0) = 0.02690501
P(S_ 66 = 0) = 0.02543746
P(S_ 68 = 0) = 0.02406085
P(S_ 70 = 0) = 0.02276844
P(S_ 72 = 0) = 0.02155412
P(S_ 74 = 0) = 0.02041233
P(S_ 76 = 0) = 0.019338
P(S_ 78 = 0) = 0.01832647
P(S_ 80 = 0) = 0.01737350
P(S_ 82 = 0) = 0.01647516
P(S_ 84 = 0) = 0.01562787
P(S_ 86 = 0) = 0.0148283
P(S_ 88 = 0) = 0.01407341
P(S_ 90 = 0) = 0.01336035
P(S_ 92 = 0) = 0.01268653
P(S_ 94 = 0) = 0.01204950
P(S_ 96 = 0) = 0.01144703
P(S_ 98 = 0) = 0.01087701
P(S_ 100 = 0) = 0.01033751

if you are looking for the expectation of 2n, well you have an infinite sum that does not converge and ergo, there is no expectation. this is because in the experiment you can fall below 0 and above 0 as many times as you like.