True. I guessed student A because I guess I didnt read it real good. I didn't see the expect part. I was taking it for the whole instead of the next flip.
Student A also has a good response. If you say it is a 10:10 average then that would be right. Student B is saying this particular flip will be a 12:8 = 3:2 So for every 5 flips 3 will be heads. That is off setting the odds. The odds go up with each flip staying the same. So it is a 1 out of 2(2^1) chance to be heads then the second time is 1 out of 4(2^2) and the 3rd time is 1 out of 8(2^3) and 4th time is 1 out of 16 (2^4). Within the bounds of this single flip, of course the odds are 1/2. But within the context of the 5th flip, the odds 1/32 (2^5). Therefore if you look at it through an average of all flips then Student A is right. If you are looking at it from a perspective where you examine the context of the next flip only then Student B is right. For all we know the next 4 flips will be all tails, giving us our 1:1 odds in which case it will be Student A.
Which is right for homework though? I am now guessing student B.
A person sets out to toss a coin 20 times. On average, they expect to get 10 heads and 10 tails.
Their first four tosses hapen to be heads. What do you expect to happen in the next 16 tosses?
Student A: I still expect 10 heads and 10 tails, and since I've already got 4 heads, I now expect 10 tails and 6 heads from the remaining 16 tosses. So in the next few tosses I expect more tails than heads.
Student B: There are 16 tosses to go. For these 16 tosses, I expect 8 head and 8 tails. This means I now expect 12 heads and 8 tails from my original 20 throws.
Which argument is right, and why?