Gambler and Four Coins

Question

A Gambler has four coins in his pocket. Two are double-headed, one is double-tailed and one is normal coin. The coins cannot be distinguished until one looks at them. The gambler takes a coin at random, opens his eyes and sees that the upper face of the coin is the head. What is the probability that the lower face is a head?

Given answer is 4/5.

I think it should be 2/3.

Comments

H/H, H/H, T/T, H/T

LH-> lower side is head

UH-> upper side is head

P(LH/UH) = P(LH$\cap$UH)/P(UH) = $\frac{\frac{2}{4}}{\frac{1}{4}*(1+1+0+\frac{1}{2})}$ = 4/5

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see total faces = 8

number of faces = head = 5 so probability of head = 5/8

now second face also head = 4/8 = 1/2

so probability = probability of 2 head / given that 1st is head= (1/2)/(5/8 )= 8/10 = 4/5

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Thanks

@joshi_nitish

@Anu007

For such a great response and good explanation.

Thanks Guys.

Answer 1

This types of questions are always interesting. Initially it seems like that question is asking what is the probability of selecting a two headed coin, as we are given that upper part is Head. The answer that directly jumps is:

$ Prob(Lower\; side\;is\;head) = \frac{2}{4}\; = \frac{1}{2}$

but the questions, in which something is "GIVEN", we need to be careful as this changes the whole "SAMPLE SPACE" and Bayes theorem has to be used.

$ Prob(A|B) = \frac{Prob(A \cap B)}{Prob(B)} $, so here

A: Lower side is Head

B: Upper side is Head

$Prob(B) = \frac{1}{4}(1+1+0+\frac{1}{2}) = \frac{5}{8}$

$Prob(A \cap B) = \frac{1}{2} $. this is selecting a two headed coin.

so answer is:

$ Prob(A|B) = \frac{Prob(A \cap B)}{Prob(B)} = \frac{\frac{1}{2}}{\frac{5}{8}} = \frac{4}{5}$