There are two biased coins A and B. We have been given the following:

P(H|Coin A) = 0.9
P(T|Coin A) = 0.1
P(H|Coin B) = 0.1
P(T|Coin B) = 0.9

We are also given that Probability of tossing Coin A or Coin B randomly is 0.5

How could we find the probability of "11th toss = Head", given that we have 10 heads in a row ?

  • $\begingroup$ This depends on whether you select the coin just once, or anew on each trial. $\endgroup$ – Graham Kemp Jun 23 '17 at 8:56
  • $\begingroup$ We are selecting the coin just once. I would love to see how answer changes if we select the coin anew on each trial $\endgroup$ – user3036345 Jun 23 '17 at 9:34
  • $\begingroup$ If anew on each trial then $0.5\cdot0.9+0.5\cdot0.1=0.5$. $\endgroup$ – drhab Jun 23 '17 at 9:58
  • $\begingroup$ @drhab True so biases will be cancelled out of two coins $\endgroup$ – user3036345 Jun 23 '17 at 12:04

Given we have coin A, the probability of tossing 10 Heads in a row is $0.9^{10}$; given we have coin B, the probability of tossing 10 Heads in a row is $0.1^{10}$. Using Bayes rule we have

$$P(A|10H)=\frac{P(A)P(10H|A)}{P(10H)}=\frac{0.5\cdot 0.9^{10}}{0.5\cdot 0.1^{10}+0.5\cdot 0.9^{10}}=\frac{0.9^{10}}{0.1^{10}+0.9^{10}}$$

$$P(B|10H)=\frac{P(B)P(10H|B)}{P(10H)}=\frac{0.5\cdot 0.1^{10}}{0.5\cdot 0.1^{10}+0.5\cdot 0.9^{10}}=\frac{0.1^{10}}{0.1^{10}+0.9^{10}}$$

The probability $P$ that the 11th toss is Heads given 10 previous tosses were Heads is

$$P=P(A|10H)\cdot 0.9 + P(B|10H)\cdot 0.1$$

  • $\begingroup$ wonderful - Thanks for your explanation $\endgroup$ – user3036345 Jun 23 '17 at 6:45

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