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We have a rigged coin, the probability of obtaining heads is triple that of obtaining tails.

Consider the variable $X$ defined as follows: We flip the coin twice in a row.

  • If Heads are obtained on the first roll and Tails on the second, we take $X = 1$.
  • If you get Tails on the first roll and Heads on the second, we take $X = 2$.
  • Otherwise, the coin is tossed twice (If you get any outcome other than or you repeat the process from the start.)

I want to find the density function of the variable but I'm confusing with that "Otherwise" means, the probability of getting head first and then tail is $\frac{3}{4}*\frac{1}{4} = \frac{3}{16}$, same for the first tail and then head. So $P[X=1] = 3/16$ and $P[X=2] = 3/16$, is this enough or I'm missing something? Any help is appreciated!

duster
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    What do you mean "otherwise the coin is tossed twice"? It was already tossed twice. Are you saying "if you get any outcome other than $HT$ or $TH$ you repeat the process from the start."? – lulu Apr 14 '21 at 17:04
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    Probabilities should add up to 1. The "Otherwise" means that you can take $X=1$ from other events, not only "heads-tails" but also "heads-heads-heads-tails". Try to think how you can account for that. – AnilCh Apr 14 '21 at 17:05
  • @lulu Yes! I will add that clarification – duster Apr 14 '21 at 17:06
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    The way it is phrased grammatically is poor, and it is unclear whether that is the fault of who gave you the problem or how you had copied the problem. The standard problem which is like the one you describe, the "otherwise" here would mean that we try again to flip both coins to try again to get one of the first two scenarios. You should find that $\Pr(X=1)=\Pr(X=2)=\dfrac{1}{2}$. This is the typical way of simulating a fair coin toss using an unfair coin. – JMoravitz Apr 14 '21 at 17:06
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    It should be mentioned that this might possibly require an unbounded number of tosses (as there is a strictly positive probability of having flipped $n$ heads in a row for any finite number $n$) and that you can not simulate a fair coin with a biased coin with a desired maximum number of tosses in the event that the probability of heads on the biased coin was rational and not equal to $\frac{1}{2}$. See Here. – JMoravitz Apr 14 '21 at 17:10

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First, if the probability of heads is three times the probability of tails, let $\Pr[H] = 3\Pr[T] = p$. Then since $\Pr[H] + \Pr[T] = 1$, we obtain $$3(1-p) = p,$$ or $p = 3/4$. Therefore, for a sequence of two independent coin flips, $$\Pr[HH] = (3/4)^2 = 9/16, \\ \Pr[HT] = (3/4)(1/4) = 3/16, \\ \Pr[TH] = (1/4)(3/4) = 3/16, \\ \Pr[TT] = (1/4)^2 = 1/16.$$ If, after two coin flips, the outcome is neither $HT$ nor $TH$, then the coin is flipped two more times--i.e., the process is repeated--then we can see that because $\Pr[HT] = \Pr[TH]$, we must have $$\Pr[X = 1] = \Pr[X = 2] = 1/2.$$ This can also be computed as $$\Pr[X = 1] = \frac{\Pr[HT]}{\Pr[HT] + \Pr[TH]} = \frac{3/16}{6/16} = \frac{1}{2},$$ and similarly for $\Pr[X = 2]$.

The question becomes more interesting (and difficult) if one simply flips until the last two flips is either $HT$ or $TH$; e.g., if the sequence of flips is $HHT$, then $X = 1$ because the last two flips was $HT$. Similarly, $TTTTH$ will yield $X = 2$. These are not allowed in the original phrasing of the question. I leave the exploration of this situation as an exercise for the reader.

heropup
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