JEST Exam

Question

Two gamblers have an argument. The first one claims that if a fair coin is tossed repeatedly, getting two consecutive heads is very unlikely. The second, naturally, is denying this.They decide to settle this by an actual trial; if within n coin tosses, no two consecutive heads turn up, the first gambler wins.

(a) What value of n should the second gambler insist on to have more than a 50% chance of winning?

(b) In general, let P(n) denote the probability that two consecutive heads show up within n trials. Write a recurrence relation for P(n).

(c) Implicit in the second gambler’s stand is the claim that for all sufficiently large n, there is a good chance of getting two consecutive heads in n trials; i.e. P(n) > 1/2.In the first part of this question, one such n has been demonstrated. What happens for larger values of n? Is it true that P(n) only increases with n? Justify your answer.

Answer 1

1)

for n=5 we can get probability greater than 50, p(5)=19/32

 

2)

so basically we want to generalize the formula

Considering opposite approach

In N trials no 2 consecutive head appears

n=0 ->1

n=1 ->2

n=2 ->3

n=3 ->5

and so on basically we can formulate it as T(n)=T(n-1)+T(n-2)

so T(n) gives the number of outcomes where no two consecutive head appears

so the probability of 2 consecutive head appear will be

p(n)=1-$\frac{T(n)}{2^{n}}$

 

 

3)

let us demonstrate n=4

T(n)=8

p(n)=50%

n=5

T(n)=13

p(n)=$\frac{19}{32}$=59%

no let us take n=10

T(n)=144

p(n)=$\frac{55}{64}$ =85%

so an we increase n probability of getting 2 consecutive head increases

$\lim_{n\rightarrow \infty }\frac{T(n)}{2^{n}} \rightarrow 0$

p(n)$\rightarrow$1

Comments

@Tesla!, how did you derive the recurrence relation?

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Ohh my mistake it's for 2 heads not 2 consecutive heads, will change answer

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Ya, @Tesla, if possible kindly mention the steps for deriving the recurrence relation

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Check it now @Sukannya