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You have a biased coin, where the probability of flipping a heads is $70%$. You flip once, and the coin comes up tails. What is the expected number of flips from that point (so counting that as flip $\#0$) until the number of heads flipped in total equals the number of tails?

I think the answer should be $0.3x+1 = 0.7x \implies x=2.5$ but I am not sure.

For the love of maths
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rohan bagri
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1 Answers1

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I think you're right. Another way to see it is as follows: consider a biased random walk on the integers. The walk starts at 1. The question is what is the expected time until it hits 0. Let $x $ be that expectaion. Then from any position $n>0$ the expected time to hit $0$ is $nx $; the walk must make $n $ steps to the left. Therefore, $x $ must satisfy the equation $$ x=1+0.3*2x .$$ When starting at 1, after 1 step the walk either hits 0 with probability $.7$ or moves to 2 with probability $.3$. In the former event the (expected) hitting time is 1; in the latter event it is $1+2x $.

Ron
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  • To get the % sign in $70%$ to appear, you can put a \ immediately before it (so you've got 70% between the dollar signs). – timtfj Jan 25 '19 at 19:30
  • @Ron. Wait. Using recursion, shouldn't it be $X = 0.7(1) + 0.3(2+x) = 13/7$? That is, with 0.7 you are going to get the head the next flip and equalize H vs T in just 1 coin flip more. On the other hand, with 0.3 you get another T and thus you will need to flip at least 2 times more to equalize, so (2+X). – Mining Sep 08 '21 at 15:41