in Probability edited by
320 views
1 vote
1 vote
A coin is such that after every toss the probability of same side coming again increases by $50\%$ from the initial value. If initially the probability of head and tail are the same, what is the expected number of tosses until we get a head?
in Probability edited by
by
320 views

1 Answer

1 vote
1 vote
$E_{\text{tosses}} = 0.5 \times 1 + 0.5 \times(1+ E_{HaT})$

$\implies E_{\text{tosses}} = 1 + 0.5 E_{HaT}$

$E_{HaT} =  0.25 \times 1 + 0.75 \times (1 + E_{HaT})$

$\implies E_{HaT} = 1 + 0.75 E_{HaT}$

$\implies E_{HaT} = \frac{1}{0.25} = 4$

$\therefore E_{\text{tosses}} = 1 + 0.5 \times 4 = 3$

So, we must toss the coin at least $3$ times to expect a Head.
by
Answer:

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true