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A person decides to flip a coin until it lands on heads.

Part 1:

What is the probability that the first time it lands on heads is on the 4th flip (that is, it lands on 3 tails before it lands on heads for the first time).

I understand this question. Using the geometric distribution, I solve for P(x=$4$):

P(Heads) = $0.5$
P(Tails) = $0.5$

P(X=$4$) = $0.5^3$ x $0.5^1$ = $0.5^4$ = $0.0625$

Part 2:

If their first $3$ flips are tails, what is the probability that the next flip will be heads?

I'm not sure how where to start? To me, this sounds exactly like Part 1. Any help is appreciated.

Thanks

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    $\begingroup$ As long as the tosses are independent of each other no matter if the last $1$ billion tosses were tail when the probability of the next toss being head is asked. It is $1/2$ $\endgroup$ – Seyhmus Güngören Oct 3 '14 at 21:11
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Each toss is independent of the last. If I'm not mistaken, the idea that the previous tosses impart "probabilistic weight" on subsequent tosses is known as the gambler's fallacy.

Even if my last 100000 tosses yielded tails, my next flip still has a 50% probability.

The reason this might not seem intuitive is because in real life, if your last few thousand tosses were tails, the coin probably isn't fair. But with this sort of calculation, we assume that the coin is fair, thus implying that our 100000-tail streak is a lovely coincidence.

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Since any toss does not affect or influence the outcome of any other toss, the probability of getting a heads is the same old $1/2$.

Another way to think of it is as follows: Let's say your friend flipped the coin 5 times and he got a certain heads/tails combination you don't know of. Now if you toss the coin, and get heads, you can't say anything about what your friend got. Basically, his tosses are independent of yours, or in general any toss has no influence on any other toss.

(Can be counterintuitive in the beginning)

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