A coin-flipping question

Probability Level 3

Flip a fair coin 5 times and note the results. If the result is either 3 heads/2 tails or 3 tails/2 heads, call the sequence normal . If the result is anything else, a 4-1 or 5-0 split, call the sequence wild .

If you do this procedure 3 times, what are the chances of at least 1 of the sequences being wild?

387/512 407/512 367/512 397/512 377/512

Denton Young by Denton Young , 47, USA

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2 solutions

Denton Young
Aug 2, 2015

The chance of a single sequence being normal is 20/32 = 5/8.

So the chance of all 3 sequences being normal = ( 5 / 8 ) 3 (5/8)^{3} = 125/512.

The chance of at least 1 sequence being wild is thus 1 - (125/512) = 387/512

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Ian McKay
Sep 8, 2015

The odds of a single sequence being all heads is 1 in 32. The odds of a single sequence being all tails is 1 in 32. The odds of a single sequence being 4-1 heads is 5 in 32. The odds of a single sequence being 4-1 tails is 5 in 32. Total those to find the odds of a single wild sequence: 12 in 32, (6 in 16-- 3 in 8).

We then know the odds of a single sequence being normal is 5 in 8 (5/8)

We then take the probability of one sequence being normal , times the probability of the second sequence being normal (nothing changed, so they are the same). 5/8 * 5/8 = 25/64.

Then, we multiply that by the probability of the third sequence being normal (25/64 * 5/8 = 125/512).

The only way we wouldn't get at least one wild sequence is if all three sequences were normal . Therefore, we can take the complement of the normal probability to find the probability that at least one is wild 1 - (125/512) = 387/512

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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