This question comes from an exercise in book: If I'm tossing coins. I tossed 2 heads in a row using a coin. What’s the probability now that the next coin will be heads?

Here's what I thought:

Tossing a coin three times are independent events. So,
Probability of getting a head will be 1/2 = 50 percent chance.
Probability of getting a head second time will be = 1/4 = 25 percent chance.
Probability of getting a head third time will be = 1/8 = 12.5 percent chance.

Is this correct? Because the answer says that it would be 50 percent.

  • $\begingroup$ throws are assumed independant of each other, getting $2,4,40$ heads in a row doesn't change the probability of the next throw. $\endgroup$ – Jean-Sébastien Oct 11 '12 at 2:09
  • $\begingroup$ What you compute is the probability of having $3$ Heads in a row, what you really want is to have a Head, given that you just had $2$ $\endgroup$ – Jean-Sébastien Oct 11 '12 at 2:09
  • $\begingroup$ What would be the correct wording for a question according to my answer? $\endgroup$ – user121314 Oct 11 '12 at 2:18
  • $\begingroup$ Of course it is "50 percent" or $\frac{1}{2}$, since previous tosses do not affect current outcome. In case you want $HHH$ out of three tosses, it is $\frac{1}{2^3}$ $\endgroup$ – Alex Oct 11 '12 at 2:31

To answer your question in your comment, If you have seen conditionnal probabilities, denote $A$ the event getting a HeadS on the third throw, and $B$ the event getting HeadS on throw $1$ and $2$. Argue that $A$ and $B$ are independant and so $\Pr(A|B)=\Pr(A)=\frac{1}{2}$.

If you haven't seen conditional probabilities, then you can say that since each throws are independant (I at least assume so), the first run of HeadS does not change the probabilities, so it remains $1/2$


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