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The question is: "Maria flip a coin for $6$ times while Davide for $7$ times. What is the probability that Davide obtains more heads than Maria?"

I solved this problem analysing $7$ cases:

$1)$ Maria doesn't obtain a head, therefore Davide has to obtain at least a head. The probability is: $$P=\frac{2^7-1}{2^7}$$ $2)$ Maria obtains a head, therefore Davide has obtain at least two heads. The probability is: $$P=\frac {2^7-(1+7)}{2^7}$$ (indeed there are $7$ possible dispositions with a only head for example a possible disposition is $TTHTTTT$)

$3)$ Maria obtains two heads, therefore Davide has to obtain at least three heads. The probability is: $$P=\frac {2^7-(1+7+21)}{2^7}$$ (indeed there are $21$ possible dispsitions with two heads and five tails, e.g. $TTTTTHH$

$4)$ Maria obtains three heads, therefore Davide has to obtain at least $4$ heads. The probability is: $$P=\frac {2^7-(1+7+21+35)}{2^7}$$ (in this case there are $35$ possible dispsositions with $3$ heads and $4$ tails).

$5)$Maria obtains $4$ heads, Davide has to obtains at least $5$ heads. The probability is: $$P=\frac {2^7-(1+7+21+35+35)}{2^7}$$

$6)$ maria obtains $5$ heads, Davide has to obtain at least $6$ heads. The probability is: $$P=\frac {2^7-(1+7+21+35+35+21)}{2^7}$$

$7)$ Maria obtains $6$ heads, Davide has to obtain $7$ heads and the probability is: $$P=\frac {2^7-(1+7+21+3535+21+7)}{2^7}=\frac {1}{2^7}$$ (indeed there is an only case:$HHHHHHH$)

Is there a simplest method ?

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We say that Maria wins if Davide does not get more heads than Maria. We will show that the probability Maria wins is $\frac{1}{2}$.

Maria wins if (i) she is leading after each has tossed $6$ times, or (ii) if they are tied and then Davide tosses a tail. Let $p$ be the probability Maria is leading after $6$ tosses. Then the probability they are tied is by symmetry $1-2p$. It follows that the probability Maria wins is $$p+\frac{1}{2}(1-2p).$$

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Another trick is since you know the coin is fair, it suffices to show that the number of outcomes in each scenario is equal, as all outcomes are equally likely.

Consider a given outcome where David wins. If you take all of his coins and flip them, this gives an outcome where David does not win (as he has an odd number of coins). This is a bijection between our two sets of outcomes. As all outcomes are equally likely and the two sets of outcomes are the same cardinality, we conclude our desired probability is $\dfrac{1}{2}$.

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The answer is $${{\sum_{i=0}^6}[{6\choose i}{\sum_{j=i+1}^7}{7\choose j}]}\over {2^{13}}$$

It is just summation of all possible combinations of their coin tosses where David has more heads over the total possibilities. $13$ coin tosses $=2^{13}$. The combos are counting the number of possible places for the H coin toss/which toss it is. E.g., one of $6\choose 2$ is $THHTTT$ where heads was gotten on second and third coin tosses.

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  • $\begingroup$ But... why do you think so? $\endgroup$ – user147263 May 31 '15 at 1:30
  • $\begingroup$ it is just summation of all possible combinations of their coin tosses where David has more heads over the total possibilities. $13$ coin tosses $=2^{13}$. the combos are counting the # of possible places for the H coin toss/which toss it is. e.g., one of $6\choose 2$ is $THHTTT$ where heads was gotten on second and third coin tosses. is this correct? not sure. $\endgroup$ – miniparser May 31 '15 at 21:47
  • $\begingroup$ @Yes: is it correct, though? thanks $\endgroup$ – miniparser May 31 '15 at 22:25
  • $\begingroup$ Yes, it evaluates to $1/2$. $\endgroup$ – user147263 May 31 '15 at 22:31

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