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I came across this question recently and can't seem to find the correct approach. Any help would be appreciated!

An experiment consists of first tossing an unbiased coin and then rolling a fair die.

If we perform this experiment successively, what is the probability of obtaining a heads on the coin before a $1$ or $2$ on the die?
$P(\text{Heads}) = \frac{1}{2}$
$P(1,2) = \frac{1}{3}$
If $A_i$ represents the event that a $1$ or a $2$ is rolled on the $i^{th}$ toss, then I have to find the following:
$$\bigcup^{\infty}_{i=1} P(A_i).$$

But I am not sure how to find this and also incorporate the probability of landing on heads before this... Am I approaching this correctly or should I be assigning random variables and working from there?

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Terminology question; why are they referred to as 'unbiased' coins and 'fair' die, why not 'unbiased' die, or 'fair' coins? – Psytronic Mar 24 at 15:56
up vote 8 down vote accepted

This is the probability of no-heads and probability of no-1,2 before than a head appear in $n$ play, where a play is tossing in order first a coin and after a dice.

So a play before the last play (when a head happens) is no-head AND no-1,2. Because the two events are independent one of each other (coin and dice) then we have that the probability for some $n$ that a head happen before a 1 or 2 in the dice is

$$\left(\frac12\cdot\frac46\right)^{n-1}\cdot\frac12$$

because the probability that the dice show something different than one or two is $\frac46$, and the probability than the coin show tail or a head is $\frac12$. Then we have $n-1$ plays where we cant have a head or a 1 or 2, and in the last play we can have in the coin a head (in the dice doesnt matter what we get after we toss the coin).

Then the probability that this happen in any $n$ number of plays is the probability that this happen in one play OR two plays OR three plays OR..., i.e.

$$\sum_{n\ge 1}\left(\frac13\right)^{n-1}\cdot\frac12=\frac12\sum_{n\ge 0}\left(\frac13\right)^n=\frac12\cdot\frac1{1-\frac13}=\frac34$$

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Thank you, this is definitely the most understandable method for me! – Stephf97 Mar 24 at 10:48

A simple way to find the probability is to condition on the result of the first round. It is clear there is some probability $p$ of obtaining a head before (though not necessarily immediately before) a $1$ or $2$. Call that the probability of winning.

We win if (i) we get a head on the first round or (ii) we get a tail, don't roll a $1$ or $2$, but ultimately win.

The probability of (i) is $\frac{1}{2}$.

For (ii), note that the probability of tail and then something other than $1$ or $2$ is $\frac{1}{2}\cdot \frac{4}{6}$. Given this has happened, the probability of ultimately winning is $p$. Thus $$p=\frac{1}{2}+p\cdot \frac{1}{2}\cdot \frac{4}{6}.$$ Solve this linear equation for $p$. We get $p=\frac{3}{4}$.

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Use recursion; Let $p$ be probability of your event. Then, we have $$p= \frac{1}{2} + \frac{1}{2}\times\frac{2}{3}\times p,$$ where first term is probability of having head in first toss and second term results from tail in coin toss and 3-6 in first roll and having head before 1-2 in the next tries.

Thus, $p=0.75$.

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What you are describing is a series.

You could think of this as a game between Alice and Bob, where Alice flips the coin (wins with a head) and Bob rolls the die (wins with 1 or 2). Essentially you are asking what is the probability that Alice wins before Bob $$P(A<B).$$

Well, she could win before Bob in the

  1. First round $(A_1)$ with chance $P(A_1) = 1/2$.
  2. Second round $(A_2)$, which means Alice lost, Bob lost and then Alice flipped a winning Head. This occurs with chance $$P(A_2) = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{2}.$$
  3. Third round $(A_3)$, which means Alice lost, Bob lost, Alice lost, Bob lost, and then finally Alice flips a winning head. This occurs with chance $$P(A_3) = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{2}.$$
  4. Etc.

Since the events are disjoint, we can add up the probabilites. This gives us (the series you want), $$P(A<B) = \sum_{k = 1}^\infty P(A_i) = \sum_{k = 1}^\infty \left(\frac{1}{2}\right)^{k-1}\left(\frac{2}{3}\right)^{k-1}\frac{1}{2} = \frac{3}{4}.$$

The reason why I frame this in terms of a game is because alongside evaluating a series, we can also use craps principle; regarding a particular round, \begin{align*} P(A<B) &= \frac{P(\text{Alice wins})}{1-P(\text{Draw})} = \frac{1/2}{1-(1/2)(2/3)} = \frac{3}{4}\\ &= \frac{P(\text{Alice wins})}{P(\text{Alice wins})+P(\text{Bob wins})} = \frac{1/2}{1/2+(1/2)(1/3)} = \frac{3}{4}. \end{align*}

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