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weehong · #22 08-28-2013, 03:32 AM

Quote:

Originally Posted by yaser

The probability of getting 10 heads for one coin is ${1 \over 2} \times {1 \over 2} \times \cdots \times {1 \over 2}$ (10 times) which is aprroximately ${1 \over 1000}$ .

Therefore, the probability of not getting 10 heads for one coin is approximately $(1-{1 \over 1000})$ .

This means that the probability of not getting 10 heads for any of 1000 coins is this number multiplied by itself 1000 times, once for every coin. This probability is therefore $\approx (1-{1 \over 1000})^{1000}$ .

This is approximately ${1 \over e}$ since $\lim_{n\to\infty} (1-{1\over n})^n = {1\over e}$ . Numerically, ${1 \over e}\approx{1\over 2.718}\approx 0.37$ .

Therefore, the probability of this not happening, namely that at least one coin of the 1000 coins will give 10 heads, is 1 minus that. This gives us the answer of approximately 0.63 or 63% that I mentioned in the lecture.

I captured the above idea, however I am confused why the following approach gives wrong answer:

Probability of all heads in 10 flips by one coin: p = 0.5^10
Probability of all heads in 10 flips by any of 1000 coin: 1000*p = 97.7%

In above I assumed there were 1000 10-flips.