Three fair coins are tossed and D is the positive difference between the number of heads and the number of tails obtained, so D takes the values 1 and 3. Tabulate the probability distribution of D and calculate E(D).

I have made groups like

with difference 1:


with difference 3: (HHH) (TTT)

later I have done

$6 (1/2)^3$ (for the condition where difference is 1) AND $2 (1/2)^3$ (for the condition where difference is 3)

The answers I am getting are 3/4 for the first condition and 1/4 for the second condition but I am not sure if this is the right way

  • 3
    $\begingroup$ What have you tried, and where are you having problems? This is not a good site to just have people do your homework for you. $\endgroup$ Commented Apr 1, 2015 at 21:09
  • 1
    $\begingroup$ I would never trust coins I got from a fair. They're bound to be weighted, or have two heads, or split into two mid-air. $\endgroup$
    – Joffan
    Commented Apr 1, 2015 at 21:21
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    $\begingroup$ Your answers seem correct. Just the difference there - it's not 0, it's 3. And also you stopped before calculating the expectation, it seems. You only calculated the probabilities. $\endgroup$ Commented Apr 1, 2015 at 21:31
  • $\begingroup$ oh yes sorry for the mistake $\endgroup$
    – Hamza750
    Commented Apr 1, 2015 at 21:32
  • $\begingroup$ Because there are only 8 outcomes in terms of H's and T's when three coins are tossed, I suppose you were intended to get the distribution of D by direct enumeration--the way you began. Make a distribution table, telling the probability of each value of D. Then use it to find E(D). $\endgroup$
    – BruceET
    Commented Apr 2, 2015 at 0:21

1 Answer 1


Your calculation of the probability distribution of the absolute difference is correct:

  • $\Pr(D=1) = \frac34$
  • $\Pr(D=3) = \frac14$

You could have used these to find the expected absolute difference:

  • $E[D]=1\times \frac34 +3 \times \frac14 = \frac32 .$

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