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Me and friend flip a coin 4 times, whoever gets most heads wins. Each coin flip is independent of each other and the coins are fair. (So the probability of a flip is 1/2)

Show that the probability of a tie (we get the same number of heads) is the same as getting exactly 4 heads and 4 tails on 8 coin flips. Use this answer to calculate the probability of someone winning (getting more heads than the other person).

Also, If I toss the coin 5 times, while my friend only tosses hers 4 times, calculate the probability that I will get strictly more heads than my friend.


So the possible outcomes for the first question is ((HHHH),(HHHH)),((HTHH)(HTHH), ((HHTH),(HHTH)), ((HHHT),(HHHT)),..., ((TTTT),(TTTT)). Am I thinking of this right? So I am trying to prove that the probaility of the above outcomes is equal to the probability of getting (HHHHHHHH)? Still not sure how to figure out the second and third question.

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  • $\begingroup$ You cannot answer like that. Start computing player A gets 4 heads independently , player B the same, and then player A and player B. Do the same with 3 , 2 , 1 and 0 heads, alone and both. And then, you wil get elements to answer. $\endgroup$ – user354674 Aug 14 '16 at 22:56
  • $\begingroup$ You can prove that all ties (as well as 4 heads and 4 tails, which is essentially the same thing as a tie, where you play with 1 coin) have a 1/2 probability with induction. For the second and third question, you should do a binomial distribution. $\endgroup$ – The Monkey Aug 14 '16 at 23:55
  • $\begingroup$ @The monkey - I'm missing what the link is between the two probabilities, other than they both work out at the same value - having worked that out, the chance of a winner is surely 1 - p, the chance of an individual winner (1 - p) / 2 - I'd tackle the last one by working out p(1),p(2) ... for each of them, then I can work out p(n>m and m) for each value of m $\endgroup$ – Cato Aug 15 '16 at 9:29
  • $\begingroup$ 'A friend and I' you wouldn't say 'me flip a coin 4 times' - did a prof write it? Tell him off. $\endgroup$ – Cato Aug 15 '16 at 10:37
  • $\begingroup$ @Andrew yup, ill also tell him i got the answers from stackexchange lol $\endgroup$ – Deegeeek Aug 15 '16 at 21:12
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"Show that the probability of a tie (we get the same number of heads) is the same as getting exactly 4 heads and 4 tails on 8 coin flips."

Let $H,T$ denote the number of heads and tails repectively flipped by you in $4$ flips .

Let $H',T'$ denote the number of heads and tails respectively flipped by your friend in $4$ flips.

Then the probability of a tie is $\Pr\left(H=H'\right)$.

But $T'$has the same distribution as $H'$ and there is independence, so we observe:

$\Pr\left(H=H'\right)=\Pr\left(H=T'\right)=\Pr\left(H=4-H'\right)=\Pr\left(H+H'=4\right)$

The last probability can be recognized as the probability of $4$ heads by $8$ flips (the flips of you and your friend taken together).

"Use this answer to calculate the probability of someone winning (getting more heads than the other person)."

Now we have:

  • $\Pr\left(\text{tie}\right)=\Pr\left(H+H'=4\right)=2^{-8}\binom{8}{4}$.
  • $1=\Pr\left(\text{you win}\right)+\Pr\left(\text{tie}\right)+\Pr\left(\text{friend wins}\right)$
  • $\Pr\left(\text{you win}\right)=\Pr\left(\text{friend wins}\right)$

leading to: $$\Pr\left(\text{you win}\right)=\frac{1}{2}\left(1-2^{-8}\binom{8}{4}\right)$$

"Also, If I toss the coin 5 times, while my friend only tosses hers 4 times, calculate the probability that I will get strictly more heads than my friend."

If you toss $5$ times then think of it as a match1 as described above that is followed by an extra toss of you, and call the whole thing match2.

Now apply that:

$$\Pr\left(\text{you win match2}\right)=$$$$\Pr\left(\text{you win match1}\right)+\Pr\left(\text{match1 ends in a tie}\wedge\text{extra toss is a head}\right)=\Pr\left(\text{you win match1}\right)+\Pr\left(\text{match1 ends in a tie}\right)\Pr\left(\text{extra toss is a head}\right)=$$$$\frac{1}{2}\left(1-2^{-8}\binom{8}{4}\right)+2^{-8}\binom{8}{4}\times\frac12=\frac12$$

There is another (more elegant) route to this result.

Let $H,T$ denote the number of heads and tails repectively flipped by you in $5$ flips .

Let $H',T'$ denote the number of heads and tails respectively flipped by your friend in $4$ flips.

The probability of winning for you is $\Pr(H>H')$ and just as above we find:

$\Pr\left(H>H'\right)=\Pr\left(H>T'\right)=\Pr\left(H>4-H'\right)=\Pr\left(H+H'>4\right)$

The RHS is the probability that by $9$ flips there are more heads than tails. Symmetry then tells us that this equals $\frac12$.

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  • $\begingroup$ How does Pr(match1 ends in a tie∧extra toss is a head) = Pr(extra toss is a head)Pr(extra toss is a head) ? Is this because the events are independent? $\endgroup$ – Deegeeek Aug 15 '16 at 21:52
  • $\begingroup$ And does Pr(extra toss is a head) just equal 1/2? So in general, Pr(you win match2) = winning match 1 multiplied by 1/4? $\endgroup$ – Deegeeek Aug 15 '16 at 21:54
  • $\begingroup$ Sorry, but the last line in my answer was not correct (caused by copying, pasting and no checking). I have repaired now. On your comments: yes, because the events are independent, and yes P(extra toss is head)=1/2. $\endgroup$ – drhab Aug 16 '16 at 7:36
  • $\begingroup$ What you say about "multiplied by 1/4" is unclear to me. $\endgroup$ – drhab Aug 16 '16 at 8:18

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