Suppose kitty tosses a coin 6 times. What is the probability that she counts more "heads" than tails?

This is a question from the test and I got only partial marks so I am wondering what is the solution to this.

What I did was I assume the number of coin outcomes,for example: hhhhhh,tttttt,and so on and I use the formula P(E) = n(E)/n(S) where n(E) is the number of events n that has heads more than tails.

  • $\begingroup$ Are you familiar with the binomial distribution ? $\endgroup$ – callculus Nov 25 '15 at 7:12
  • $\begingroup$ no I didn't learn that. $\endgroup$ – guest11 Nov 25 '15 at 7:13
  • $\begingroup$ But I think it is necessary. Try to understand the link. If you have any question feel free to ask: mathsisfun.com/data/binomial-distribution.html $\endgroup$ – callculus Nov 25 '15 at 7:22
  • $\begingroup$ my teacher never mentions this in class but thank you so much. $\endgroup$ – guest11 Nov 25 '15 at 7:23

Add up the following:

  • The probability of getting $4$ heads and $2$ tails is $\dfrac{\binom64}{2^6}=\dfrac{15}{64}$
  • The probability of getting $5$ heads and $1$ tails is $\dfrac{\binom65}{2^6}=\dfrac{6}{64}$
  • The probability of getting $6$ heads and $0$ tails is $\dfrac{\binom66}{2^6}=\dfrac{1}{64}$

Hence the probability of getting more heads than tails is $\dfrac{15}{64}+\dfrac{6}{64}+\dfrac{1}{64}=\dfrac{22}{64}$

  • $\begingroup$ hello I am not familiar with that notation (6 over 4)...how it is calculated? $\endgroup$ – guest11 Nov 25 '15 at 7:20
  • $\begingroup$ @guest11: Then you're in a bit of trouble there, as in other questions of yours (some of which I have answered myself). "$6$ over $4$" is a method for counting the number of ways to choose $4$ items out of $6$ unique items without replacement (i.e., you cannot choose an item and put it "back in the box"). In short, $\binom64=\frac{6!}{4!\cdot(6-4)!}$. I hope you are familiar with the factorial operator "!"... :) $\endgroup$ – barak manos Nov 25 '15 at 7:24
  • $\begingroup$ yes thank you! because there are many ways to write this out so I am not used to writing this way but I know how it works now thanks to you! Im voting now thank you. $\endgroup$ – guest11 Nov 25 '15 at 7:37
  • $\begingroup$ It's a bit shorter to use the logic from down below in this case, only done properly....6 choose 3 is 20, so 20 times you have equality. Of the remaining 44, the situation is symmetric so half the time you have more heads, half the time more tails. Thus you get to 22/64 in a tiny bit less work :) $\endgroup$ – Alan Nov 25 '15 at 7:48
  • $\begingroup$ @guest11 $\binom{n}{r} = {^n{\rm C}_r}$ You may be more familiar with this notation for the binomial coefficient. $\endgroup$ – Graham Kemp Nov 25 '15 at 7:55

The probability of getting the exact same number of heads and tails is ${^6{\rm C}_3}\cdot\tfrac 1{2^6} = \frac 5{16}$

Due to symmetry then, the probability of seeing more heads than tails is $\tfrac 12\big(1-\tfrac 5{16}\big)$, that's $\color{green}{\tfrac {11}{32}}$.   This is also the probability of seeing less heads than tails.

Assuming, that is, that the coins are unbiased.

$$\dfrac {11}{32}$$

  • $\begingroup$ +1 simple , short and can be used for arbitrary number of tosses without increasing the number of calculations $\endgroup$ – miracle173 Nov 25 '15 at 8:19
  • $\begingroup$ do i use this formula and apply for a similar question? $\endgroup$ – guest11 Nov 25 '15 at 18:25

The first step is to think of the different ways that it is possible to have Kitty count more heads (H) than tails (T). So the only way that that can occur is

  1. Kitty gets 4 heads and 2 tails
  2. Kitty gets 5 heads and 1 tail
  3. Kitty gets 6 heads and 0 tails

Now, any other scenario that occurs would not allow Kitty to have more heads than tails.

The next step is then to calculate the probability of 1., 2., and 3. occurring.

Naively, you could try to enumerate and count all of the different possibilities, but that would be a very long list to have to keep track of.

Instead, let's consider the following: Let's define some notation. Let $X$ be the number of heads that Kitty gets when tossing 6 coins. And so based on 1., 2., and 3. we are interested in the calculating the following three probabilities:

$Pr(X=4)$, $Pr(X=5)$ and $Pr(X=6)$. Why are these relevant? Think about it this way: We want to calculate

\begin{align*} Pr(\text{Kitty gets more heads than tails})&=Pr(X>3)\\ &=Pr(X=4)+Pr(X=5)+Pr(X=6)\\ &={6\choose 4}(0.5)^4(0.5)^2+{6\choose 5}(0.5)^5(0.5)^1+{6\choose 6}(0.5)^6(0.5)^0\\ &=15\times(0.5)^6+6\times(0.5)^6+1\times(0.5)^6\\ &=(0.5)^6\times(15+6+1)\\ &=0.34375 \end{align*}

The above solution makes use of the fact that if a random variable $X$ follows the Binomial distribution with proportion $p$ and number of trials $n$, then the probability of seeing $x$ "successes" in $n$ trials is

$$Pr(X=x)={n\choose x}p^x(1-p)^{n-x}$$

@guest11, I just realized you said you haven't learned the binomial formula which is what the above answer depends on. Let me try to explain it with out it.

Imagine the same setup as above. So let $X$ be the number of heads that Kitty gets when tossing 6 coins.

Without using the binomial formula (or distribution) we can say that for Kitty to get more heads than tails we need to calculate the following:

\begin{align*} Pr(\text{Kitty gets more heads than tails})&=Pr(X>3)\\ &=Pr(X=4)+Pr(X=5)+Pr(X=6)\\ \end{align*}

But so without using the binomial formula, how can we solve this? Well let's try to calculate each probability one at a time.

So, what the probability that Kitty gets 6 heads, i.e., $Pr(X=6)$. Well that can only happen one way. Kitty has to get the following sequence of coin flips HHHHHH. And we know (or assume) that the coin is fair and so the probability of getting a head on any one flip is $\frac{1}{2}$ or 0.5. But we also know that flipping coins are independent events and so we can say that $$Pr(X=6)=Pr(\text{Kitty gets } HHHHHH)=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=(0.5)^6$$

That's the simple one. But how can we calculate the probability that Kitty gets 5 heads? Well, there are 6 ways for that to happen: HHHHHT, HHHHTH, HHHTHH, HHTHHH, HTHHHH, THHHHH. And so we can say that the probability that Kitty gets 5 heads is \begin{align*} Pr(X=5)&=Pr(\text{Kitty gets } HHHHHT \text{ or } HHHHTH \text{ or }HHHTHH\\ &\text{ or }HHTHHH \text{ or }HTHHHH \text{ or }THHHHH )\\ &=Pr(\text{Kitty gets } HHHHHT) + Pr(\text{Kitty gets } HHHHTH) + Pr(\text{Kitty gets } HHHTHH)\\ &+Pr(\text{Kitty gets } HHTHHH )+Pr(\text{Kitty gets } HTHHHH)+ Pr(\text{Kitty gets } THHHHH )\\ &=\left(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\right) + \left(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\right) \\ &+\left(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\right) + \left(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\right)\\ &+\left(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\right)+\left(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\right)\\ &=6\times\left(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\right)\\ &=6\times(0.5)^6 \end{align*}

And so you see that we have

\begin{align*} Pr(\text{Kitty gets more heads than tails})&=Pr(X>3)\\ &=Pr(X=4)+Pr(X=5)+Pr(X=6)\\ &=Pr(X=4)+6\times(0.5)^6+6\times(0.5)^6 \end{align*}

Which is matching our answer using the bionmial formula. We could similarly solve for $Pr(X=4)$ but in that case there are 15 ways to write down the number of ways of getting 4 heads and 2 tails (but I think at this point you should be able to do it and convinced that you don't want to!).

And so hopefully you see why the binomial formula is such a useful tool.

  • $\begingroup$ Wow...using decimal numbers is the whole new thing but thank you. $\endgroup$ – guest11 Nov 25 '15 at 7:27
  • $\begingroup$ omg thank you so much ...but how to do ? I already voted ? can I vote for you also? omg thank you so much... my teacher wouldn't have explained that much for me... thank you again $\endgroup$ – guest11 Nov 25 '15 at 7:59
  • $\begingroup$ @guest11 sadly you cannot accept two answers. If you prefer my solution you can just click the green check mark but this would take away the check mark from barak manos $\endgroup$ – RustyStatistician Nov 25 '15 at 8:07
  • $\begingroup$ @guest11: you can accept only one answer but I think you can unaccept an answer an accept another one. But you can upvote as many answers as you like. $\endgroup$ – miracle173 Nov 25 '15 at 8:22

Since head and tails have same probability, what you seek is half the probability of not getting an equal number of head and tails. And you get 3 head and 3 tails in ${6\choose 3}=20$ different ways out of $2^6=64$ possible outcomes.

So, the requested probability is: $$ p = \frac{1 - \frac{20}{64}}{2} = \frac{11}{32}. $$

added. A very intuitive way to understand what is happening. Consider the 6-th power of the binomial $T+H$: $$ (T+H)^6 = T^6 + 6 T^5 H + 15 T^3 H^2 + 20 T^3 H^3 + 15 T^2 H^4 + 6 T H^5 + H^6. $$ If you develop the product $(T+H)^6$ you have to choose T or H for six times. So you get all possible outcomes of a 6-tuple of coin tosses. The binomial formula tells us how we can put together the outcomes if we are not interested in the order. So, out of $2^6=64$ possible outcomes you have one of only tails $T^6$, six of $T^5H$ (i.e. 5 tails and one head) and so on. You see that to have more heads than tails you have to consider $15+6+1$ out of $64$ possibilities. Or, more quickly, you note that this number is half of the total minus 20: $(64-20)/2$ over $64$ possibilities.

  • $\begingroup$ But the questions says "more heads than tails". $\endgroup$ – barak manos Nov 25 '15 at 7:14
  • $\begingroup$ I added the conclusion... $\endgroup$ – Emanuele Paolini Nov 25 '15 at 7:41
  • $\begingroup$ No factor of times 2 in your choose..there's only 20 configurations of 3 heads and three tails. $\endgroup$ – Alan Nov 25 '15 at 7:47
  • $\begingroup$ right... corrected! $\endgroup$ – Emanuele Paolini Nov 25 '15 at 7:52

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