A problem involves the flipping of a weighted coin.

Question

Assume that the coin is weighted so that a tail is 6 times as likely as a head. The coin is flipped 6 times. What is the probability that both heads and tails occur?

Answer 1

This is quite a beautiful question.

We wish to find the probability that both heads and tails occur in 6 flips of a weighted coin.

Since we are dealing with a weighted coin, the probability of getting heads in a single flip is not the same as it would be for a fair coin. We have to start by finding the probability of getting heads and the probability of getting tails in a single flip for this weighted coin before we can proceed with the rest of the question.

1. "Assume that a coin is weighted so that a tail is 6 times as likely as a head."

Let us use this sentence to find the probability of getting a heads and that of getting a tails in a single flip of this weighted coin.

Let us label p the probability of getting heads in a single flip of this weighted coin.

Since a tail is 6 times as likely as a head, what is the probability of getting tails (in a single flip) in terms of p?

It is 6 times as large as p, that is 6 * p or simply 6p.

In a single flip of the coin, there are only 2 possible outcomes: getting heads or getting tails.

The probabilities of these two outcomes must add up to 1 (100%, everything that is possible).

In other words, we must have:

p + 6p = 1

Thus

7p = 1

from which we can conclude that

p = 1/7

That is, the probability of getting heads in a single flip of this weighted coin is 1/7

and the probability of getting tails in a single flip of this weighted coin is 6*1/7 = 6/7.

These probabilities do satisfy the condition that "a tail is 6 times as likely as a head"

1/7 + 6/7 = 1

both 1/7 and 6/7 are between 0 and 1

This is a quick check showing us that our work so far is correct.

We now have the necessary knowledge to look at the actual question.

1. The coin is flipped 6 times. What is the probability that both heads and tails occur?

Instead of looking at a single flip, we are looking at 6 successive flips.

We wish the following condition to be true "both heads and tails occur".

There are many different ways in which this can be satisfied...

In 6 flips, we could have:

1. 1 head and 5 tails
2. 2 heads and 4 tails
3. 3 heads and 3 tails
4. 4 heads and 2 tails
5. 5 heads and 1 tail

which would all satisfy the condition "both heads and tails occur".

What is more is that each of these could happen in many different ways!

We can get 5 heads and 1 tail by having the tail on the 1st flip, or the 2nd flip, or the 3rd flip, or the 4th flip, or the 5th flip, or the 6th flip.... Phew!

Let us think about a different approach...

What is it that we do NOT want?

The only outcomes that we do NOT want are getting 6 heads or getting 6 tails.

Those are much easier to work with than the possibilities that we listed above.

What is the probability of getting 6 heads in 6 flips of this weighted coin?

The probability of getting a head in a single flip is 1/7.

Assuming that the flips are independent, the probability of getting 6 heads in 6 flips would be

1/7 * 1/7 * 1/7 * 1/7 * 1/7 * 1/7 = (1/7)⁶ = 1 / 117,649

What is the probability of getting 6 tails in 6 flips of this weighted coin?

The probability of getting a tail in a single flip is 6/7.

Assuming that the flips are independent, the probability of getting 6 tails in 6 flips would be

6/7 * 6/7 * 6/7 * 6/7 * 6/7 * 6/7 = (6/7)⁶ = 46,656 / 117,649

Both heads and tails occur in 6 flips is equivalent to

"we are okay with anything except for 6 heads or 6 tails"

In formal terms, the event "both heads and tails occur in 6 flips" is the complement of

"we get all 6 heads or we get all 6 tails in 6 flips".

Using the rule of complements,

the probability that both heads and tails occur in 6 flips of this weighted coin is:

1 - (1/7)⁶ - (6/7)⁶ = 1 - 1 / 117,649 - 46,656 / 117,649 = 70,992 / 117,649