A coin is tossed 6 times. What is the probability that the number of heads in the first 3 throws is the same in the last 3 throws?

In 3 tosses of coin
you have 8 possible results
all are equally likely
TTT (0 heads)
TTH (1 head)
THT (1 head)
THH (2 heads)
HTT (1 head)
HTH (2 heads)
HHT (2 heads)
HHH (3 heads)
P(h =0) = 1/8
P(h = 1 ) = 3/8
P(h =2) = 3/8
P(h =3) = 1/8

so for 1st 3 rolls number of heads = 2nd 3 rolls number of heads
means either both =0 , both = 1 , both = 2 , both = 3
P((both = 0) = (1/8) * (1/8) = 1/64
P(both = 1 ) = (3/8)*(3/8) = 9/64
P( both = 2) = 3/8* 3/8 = 9/64
P(both =3 ) = 1/8*1/8 = 1/64
P( number of heads in 1st 3 rolls equals the # in the last 3 rolls)
= 1/64 + 9/64 + 9/64 + 1/64 = 20/64 = 10/32 = 5/16
P( number of heads in 1st 3 rolls equals the # in the last 3 rolls) = 5/16

6

Chances of heads in the first three...and last three...tosses:

0 = 1/8
1 = 3/8
2 = 3/8
3 = 1/8

(1/8)² + (3/8)² + (3/8)² + (1/8)²

= 1/64 + 9/64 + 9/64 + 1/64

= 20/64

= 5/16

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Prob = 1/2

is that including the chances of landing on the side and if so what is the current probability