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I am studying expected values in probability and was wondering if/what the difference is between these two problems?

Problem 1a: What is the probability distribution of the number of heads you can get from flipping 4 coins once?
Problem 1b: What is the probability distribution of the number of heads you can get from flipping 1 coin 4 times?

Problem 2a: What is the probability of getting 4 heads from flipping 4 coins once?
Problem 2a: What is the probability of getting 4 heads from flipping 1 coin 4 times?

This seems to me the same type of problems within each set. Is it?

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    $\begingroup$ Yep. Just write a number on every coins and write the sequence starting from coin 1 to coin 4. $\endgroup$ – Gâteau-Gallois Nov 8 '18 at 19:03
  • $\begingroup$ If I wanted to find the probability of getting heads 100% of the time in 4 flips, then I would simply do P = (1/2)^4? Getting heads X number of times in a finite number of flips means that the flips can no longer be treated independently because they are conditional on getting heads multiple times in a row? $\endgroup$ – Evan Kim Nov 8 '18 at 19:09
  • $\begingroup$ For your first question, yes. For the second one, in the general case you need a combinatorial factor to "set" the position of the heads in the series of outcomes. For instance if you want to compute the probability to get 2 heads, there exists $\binom{4}{2} = 6$ ways to do so (for instance HHTT, HTHT, HTTH, and so on)... So you probability will be $(1/2)^2 (1/2)^2 \binom{4}{2} = 6 (1/2)^4$. Check the binomial law for further details. Note that this is consistent with your answer for 100% of heads, since $\binom{4}{4} = 1$. $\endgroup$ – Gâteau-Gallois Nov 9 '18 at 9:22

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