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If you toss a fair coin four times, the probability of any specific outcome is the same. Consider the following example:

Event A: Heads, Tails, Heads, Tails

Event B: Heads, Heads, Heads, Heads

P(A) = P(B) = 0.5^4 = 6.25%

However, the chance of having any Tails in the sequence is : P(Any Tails) = 1 - P(B) = 1 - 0.0625 = 93.75%

If we tossed a fair coin 100 times, then P(Any Tails) would be very close to 1, which makes me think that a sequence that has only heads is less likely to happen than any sequence that has at least one tails.

Should the probability of P(Any Tails) affect the probability of P(B)? If yes, how should it be computed? Otherwise, why is my intuition wrong ?

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    $\begingroup$ $P(\text{any tails}) = 1 - P(\text{all heads})$. $\endgroup$ – Frans Rodenburg Feb 28 '18 at 4:35
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The problem is this "which makes me think that a sequence that has only heads is less likely to happen than any sequence that has at least one tails."

In reality, all the unique sequences are still equally probable. It isn't that the sequences with at least one tail are each more likely, ONLY that there are so many more of them!

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If we tossed a fair coin 100 times, then P(Any Tails) would be very close to 1, which makes me think that a sequence that has only heads is less likely to happen than any sequence that has at least one tails.

Yes. And difference is due to difference between "single tail" and "at least one tail".

  • In case of single tail, say, in position 2 we have sequence 11111... for all heads and sequence 11011... for tail at position 2. Probability to get such events are the same and equal to (1/2)100.

  • In case of "at least one tail" we are comparing sequence 1111... vs list of sequences 0111..., 1011..., 1101... etc, which obviously, counted as total, shall be 100 times more probable. In other words, it is now one events vs 100 events.

Let's do some math and look at the problem as a case of Binomial Distribution.

Probability of k tails in n trials is given as

P(k|n, p) = C(n, k) * pk * (1-p)n-k, and C(n, k) is binomial coefficient

C(n,k) = n!/(k! (n-k)!)

For the case of fair coin p = 1/2, n = 100, so

P(k|100, 1/2) = C(100, k) * (1/2)100

We could compute ratio of at least one tail to no tails probabilities

P(1|100,1/2) / P(0|100,1/2) = ... = 100!/99! = 100

which is exactly what we surmised earlier.

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