6 Heads Or 60 Heads?

Probability Level 2

Which of the above scenarios is more likely to occur?

Scenario A : 6 fair coins show heads when I flip 10 of them.

Scenario B : 60 fair coins show heads when I flip 100 of them.

Image Credit: Flickr Felicia Keelin .

A B Both are equally likely

3 solutions

Andrew Ellinor
Oct 21, 2015

While a misguided intuition may lead us to think that the outcomes are equally likely given that the proportion of coins showing heads is the same in each scenario, it is actually more likely for 6 coins to show heads when flipping 10 than it is for 60 coins to show heads when flipping 100.

We can see this using the Binomial Theorem , which says the likelihood of scenario A occurring is given by ${10 \choose 6} \left(\frac12\right)^{10} \approx 0.2051 = \color{#3D99F6}{20.51\%}.$

Whereas the likelihood of scenario B occurring is given by

${100 \choose 60} \left(\frac12\right)^{100} \approx 0.0108 = \color{#D61F06}{1.08\%}.$

This stark contrast can be explained in terms of the variance we should expect from variance in the binomial distribution which says that the standard deviation of a random variable such as the one above is equal to $\sqrt{np(1 - p)}$ .

In scenario A, this standard deviation is $\sqrt{10(0.5)(1 - 0.5)} \approx 1.58$ heads. This means that the result of 6 heads deviates by less than one standard deviation from the mean $(6 - 5 < 1.58)$ .
However, in scenario B, the standard deviation is $\sqrt{100(0.5)(1 - 0.5)} = 5$ . The result of 60 heads deviates by two of these standard deviations from the mean $(60 - 50 = 2(5))$ , making scenario B much farther from the average case.

Well, I think that the actual issue is that you're doing a "continuity correction" going from a range of 10% (5.5 to 6.5 out of 10) to 1% (59.5 to 60.5 out of 100).

The issue that you're getting at is:

Which is larger

probability of 6 heads out of 10

10 X probability of 60 heads out of 100.

It is interesting that the first probability is higher when we use 6, but lower when we use 5! This has to do with the variance (and higher order moments).

Calvin Lin Staff - 5 years, 7 months ago

More precisely, this follows directly from the Sanov's Theorem , which is one of the central result of the Large Deviation Theory . In both the cases in this problem, the empirical distribution of the sets considered are the same, viz. $\hat{p}=(0.6,0.4)$ . Sanov's theorem states that the probability of this empirical distribution for $n$ tosses is given by $\approx 2^{-nD(\hat{p}||q)}$ , where $q=(0.5,0.5)$ , the actual distribution of the fair coin. Hence it is clear that as $n$ increases, the probability decays exponentially in $n$ (unless of course, $D(\hat{p}||q)=0$ , i.e. $\hat{p}=q$ , as you have observed).

Abhishek Sinha - 5 years, 7 months ago

one way I thought of it was that the 100 coins has a greater area for failure, and getting 61 or more heads or 59 or lower tails

Tyler Hill - 5 years, 7 months ago

Ariella Lee
Dec 8, 2015

Scenario A: Each of the $10$ coins can either be heads or tails (two options per coin), so there are $2^{10}$ possible outcomes. There are $\binom{10}{6}$ successes, i.e. ways to choose 6 coins to be heads (order in which we choose the coins doesn't matter as long as we choose the same coins, and we can't choose the same coin twice). Probability: $\frac{\binom{10}{6}}{2^{6}}\approx 0,205$

Scenario B: By the same reasoning as Scenario A, there are $2^{100}$ possible outcomes and $\binom{100}{60}$ successes. Probability: $\frac{\binom{100}{60}}{2^{100}}\approx 0,011$

Probability of Scenario A is higher, so it is more likely to occur.

Alphonse Fullo
Nov 4, 2015

the least the possibility, the greater the chances of correct answer

But, both have equal probabilities. 6/10 = 60/ 100. Remeber we have to reduce them to the simplest form.

Ashish Menon - 5 years, 6 months ago

I agree with ashish Siva.there is no point going deep into mathematical analysis

Ràunak Patël - 5 years, 6 months ago

Then consider this, 60 of 100 times head is one combination out of 101 possible combinations whose sum of probabilities of all combinations equals 1. The chance of that specific combination occurring is far smaller than the likeliness of 6 of 10 times heads occurring in 11 possible combinations. The probability of getting 50 heads and 50 tails is not 50% in 100 flips, Take a simple example of flipping a coin 2 times and flipping a coin 4 times. In the 2 times case you have 4 combinations HH,HT,TH,HT. 1H1T does occur 50% of the time. In a four flip scenario you have 16 event specific scenarios HHHH,HHHT,HHTH,HHTT HTHH,HTHT,HTTH,HTTT THHH,THHT,THTH,THTT, TTHH,TTHT,TTTH,TTTT The total probability of specific distributions 4H=1/16, 3H1T=4/16, 2H2T=6/16, 3T1H=4/16, 4T=1/16

Owen Berendes - 5 years, 3 months ago

@Owen Berendes – Exactly, i am not against. But there are different answer for different viewpoints. Thanks for the explanation. Your answer is right. My answer is by elementary viewpoint.

Ashish Menon - 5 years, 3 months ago

They do not have equal probabilities, take a simple analog flipping heads 1 out of 1 time, or flipping heads 2 out of 2 times. The first probability is 50%, the second probability is 25%.

Owen Berendes - 5 years, 3 months ago

If you are seeing by that view then yes. But if you look by elementary view of probability then the answer should be 6/10. But, anyways you're answer is right. I can say that it is wrong☺

Ashish Menon - 5 years, 3 months ago

6 Heads Or 60 Heads?

Image Credit: Flickr Felicia Keelin .

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