Coin Flipping with Bias

Question

This is a question on basic probability, and I wonder if my approach is correct. Moreover, if my answer is correct, I wonder if there is another way to do it without using the law of total probability (simply use counting method for example):

Question: A bag contains 6 coins, 5 fair coins, and one which has heads on both sides. Suppose you grab 2 of the 6 coins, and flip them both. What is the probability that you get exactly one head and one tail?

My Attempt so far:

Let $A$ be the event that we get exactly one head and one tail and $B$ be the event that the coin is biased. Then $$ P(A) = P(HT) + P(TH) = P(H)P(T) + P(T) P(H) = 2P(T)P(H) $$ Now $$ P(H) = P(H \cap B) + P(H \cap B^c) = P(H | B) P(B) + P(H|B^c) P(B^c) = 1 \times \frac{1}{6} + \frac{1}{2} \times \frac{5}{6} = \frac{1}{6} + \frac{5}{12} = \frac{7}{12}. $$ Similarly, $$ P(T) = P(T \cap B) + P(T \cap B^c) = P(T| B) P(B) + P(T|B^c) P(B^c) = 0 \times \frac{1}{6} + \frac{1}{2} \times \frac{5}{6} = \frac{5}{12}. $$ Then we have $$ P(A) = 2P(T)P(H) = 2 \times \frac{7}{12} \times \frac{5}{12} = \frac{35}{72} \approx 48.61\%. $$ (This seems like a pretty high probability, which makes me think this reasoning invalid.)

Answer 1

Here is a much more direct way, which works whenever we have $n\geq2$ fair coins, and one biased HH coin.

• If we don't pick the biased coin, clearly the probability of one H one T is $1/2$.
• If we do pick the biased coin, then we are guaranteed one H. So we need the other (fair) coin to land T, which has probability $1/2$.

So in any case the probability of one H one T is $1/2$.

Answer 2

Why is coin 1 lands H and coin 2 lands T not independent? Intuitively, they are not correlated, right?

I will answer that question in the addendum, after attacking the original problem.

Let event $E_1$ denote the event that both coins selected are fair coins.

Let event $E_2$ denote the event that one of the coins selected is the biased coin.

Let $F$ denote the event that whichever coins are selected, you get one Heads and one Tails.

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Then, the desired computation is $S + T$ where

• $S = p(E_1) \times p(F|E_1)$.

• $T = p(E_2) \times p(F|E_2)$.

$\displaystyle p(E_1) = \frac{\binom{5}{2}}{\binom{6}{2}} = \frac{10}{15} = \frac{2}{3} \implies $

$\displaystyle p(E_2) = 1 - p(E_1) = \frac{1}{3}$.

$\displaystyle p(F|E_1) = \left[\frac{1}{2} \times \frac{1}{2}\right] + \left[\frac{1}{2} \times \frac{1}{2}\right] = \frac{1}{2}.$

The above computation represents the combined probabilities of $HT$ and $TH$, where one of the coins is arbitrarily designated as the first coin to be flipped.

$\displaystyle p(F|E_2) = \left[1 \times \frac{1}{2}\right] + \left[0 \times \frac{1}{2}\right] = \frac{1}{2}.$

The above computation represents the combined probabilities of $HT$ and $TH$, where the biased coin is arbitrarily designated as the first coin to be flipped.

So,

$$S = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}, ~~T = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \implies S + T = \frac{1}{2}.$$

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The shortcut is in realizing that $p(F|E_1) = p(F|E_2)$, which implies that the event $F$ is independent of event $E_1$. So, since all that you are actually interested in is $p(F)$, you can use your intuition to reason that $p(F)$ must be $(1/2)$, since it does not matter what the relative probabilities are of $p(E_1)$ vs $p(E_2)$.

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Addendum

Why is coin 1 lands H and coin 2 lands T not independent? Intuitively, they are not correlated, right?

This is a totally different question from the question of whether events $F$ and $E_1$ are independent events. The easiest way to answer this question is with intuition.

If the first coin lands H, then the probability of the first coin being the biased coin is slightly greater than normal. To demonstrate this, use Bayes Theorem.

The chance of the first coin landing Heads is

$\displaystyle \left[\frac{1}{6} \times 1\right] + \left[\frac{5}{6} \times \frac{1}{2}\right] = \frac{7}{12}$.

This implies that the probability that the first coin is biased, given that it lands Heads is

$\displaystyle \frac{(1/6)}{7/12} = \frac{2}{7}$, rather than $\displaystyle \frac{1}{6}.$

So, once you realize that the probability of the first coin being biased is slightly greater than normal, given that the first coin lands Heads, this must imply that the probability that the second coin is biased must be less than normal.

So, if you have the implication that there is a smaller chance than normal that the second coin is biased, this means that there is a greater chance than normal that the second coin comes up Tails.

Thus, the event of the first coin flip being Heads and the second coin flip being Tails are not independent events.

Answer 3

The way I see it is that there are two 'experiments': Firstly which two coins you grab and secondly which sides they land on.

Grabbing two coins:

First coin	p	second coin	p	overall p
normal	5/6	normal	4/5	4/6=2/3
		both heads	1/5	1/6
----------	-	-----------	-	---------
both heads	1/6	normal	1	1/6

So 2/3 of the time we have two normal coins and 1/3 of the time we have one normal coin and one coin with two heads.

Which side they land on:

coins	p	sides	p	overall p
both normal	2/3	hh	1/4	1/6
		ht or th	1/2	1/3
		tt	1/4	1/6
----------	-	--------	-	---------
one biased	1/3	ht or th	1/2	1/6
		hh	1/2	1/6

So the probability that we have one heads and one tails is 1/3+1/6=1/2. I hope that I explained it clearly enough.