Quote:
Originally Posted by **TheBramble** Dcraig kind of hit on what I’m driving at here.
In any series of 4 coin flips, while the probability of getting 2 consecutive heads is indeed 1 in 4 and the probability of getting 2 consecutive tails is 1 in 4, statistically, the probability of getting two consecutive tails OR two consecutive heads is 1 in 2.
But of the 16 possible permutations from a 4 coin flip, there are 12 in 16 occurrences of two consecutive Heads, and 12 in 16 occurrences of two consecutive Tails. On that basis, the probability of two consecutive Heads in a 4 coin toss exercise is 3 in 4. Same for two consecutive Tails.
So if two consecutive tails has a 3 in 4 and two consecutive heads has a 3 in 4, what it the probability of two consecutive heads OR two consecutive tails? |
ok. still on first page, and I am now confused.
if you flip a coin 4 times, the permutations of consecutive Heads positions are: (1,2), (2,3) or (3,4).
the 4 heads in a row option gives you the above all in one go, so, say 3 again,
total 6.
but, strictly speaking, only 4 of those throws results in a 2-consec result.
dont know where you get the 12 possibilities from.
from dcraigs .75; you ask how many times would you need to throw to get a Head. If you get a Head on the first throw, you stop throwing.
You only flip the coin again if the first wasnt a Head. this negates one of dcraigs options, so we are back to 0.5.
EDIT: writing before thinking. yes, the possibility of (1,2) and (2,3) also exists, so HHHT shows 2 hits. so, yes maybe 12. havent checked it.
I was assuming finishing once you got your consec.
EDIT2: yes, there are 12 instances of consecs. (Pascals Triangle)
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Last edited by trendie; Mar 27, 2009 at 2:03pm.
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