Math, asked by vickysmilevick3069, 1 month ago

A coin toss has possible outcomes H and T with probabilities 3/4 and 1/4 respectively. A gambler observes a sequence of tosses of this coin until H occurs. Let the first H occur on the nth toss. If n is odd, then the gambler’s prize is −2^n , and if n is even, then the gambler’s prize is 2^n . What is the expected value of the gambler’s prize?​

Answers

Answered by amitnrw
7

Given : A coin toss has possible outcomes H and T with probabilities 3/4 and 1/4 respectively.

A gambler observes a sequence of tosses of this coin until H occurs.

Let the first H occur on the nth toss.

If n is odd, then the gambler’s prize is −2^n , and if n is even, then the gambler’s prize is 2^n .

To Find  : What is the expected value of the gambler’s prize

Solution:

H occurs on nth toss

Expected value  in case of odd

= (-2)¹ . (3/4)¹  + (-2)³.(3/4)(1/4)² + _____

= -2 * (3/4) [ 1  +  2²/4²  + 2⁴/4⁴  + ______  ]

= ( -3/2 )[ 1  +  2²/4²  + 2⁴/4⁴  + ______  ]

Expected value  in case of even

=  ( 2)²(3/4)(1/4)  + ( 2)⁴(3/4)(1/4)³ + ____

= 2² *(3/4)(1/4) [ 1 +   2²/4²  + 2⁴/4⁴  + ______ ]

= (3/4) [ 1 +   2²/4²  + 2⁴/4⁴  + ______  ]

Net expected value of the gambler’s prize

= ( -3/2 )[ 1  +  2²/4²  + 2⁴/4⁴  + ______  ] + (3/4) [ 1 +   2²/4²  + 2⁴/4⁴  + _____]

= (-3/4) [ 1  +  2²/4²  + 2⁴/4⁴  + ______  ]

= (-3/4)  [ 1  + 1/2²  + 1/2⁴  + ______  ]  

a = 1    r  = 1/4

(-3/4) ( 1/( 1 - 1/4))

= (-3/4) ( 1/(3/4))

= - 1

expected value of the gambler’s prize    =  -  1

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