Suppose I need to send a message to someone. So that I will know the message was received, they send one back. Suppose there is a 10% chance that the message sent will get lost. Similarly, there is a 10% chance that the response will get lost. I know how long it takes for the round trip, so I know how long to wait before sending again (should no response come). Suppose I'm willing to try four attempts. What are the odds that all four attempts at sending the message will lose either the message sent or the response?
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2 Answers
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Consider the first round. Denote with $M$ the event that message reaches the other person and with $R$ the event that message reaches back the sender. We want to calculate the probability $$P(M'\cup R')$$ Note that we have that $$P(M)=0.90$$ and by the Total Probability Law $$P(R)=P(R|M)P(M)+P(R|M')P(M')=0.90\cdot0.90+0\cdot0.10=0.81$$ (which is not necessary to calculate). Returning to the wanted probability: $$\begin{align*}P(M'\cup R')&=P\left((M\cap R)'\right)=1-P(M\cap R)=1-P(R|M)P(M)=\\&=1-0.9\cdot0.9=1-0.81=0.19\end{align*}$$ So after the first round there is $0.19$ probability that there will be a failure.
No the solution depends on whether we will assume independece between rounds or not. If they are independent then the probability that all four will fail is equal to $$0.19^4=0.0013$$ which has the intuitive explanation, that as long as you keep trying the probability that all attempts will fail is become less or equivalently that your chances to successfully communicate with the other person (send message and receive answer) increase.
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What is the probability that all two messages (probe and answer) of one round go through? $(1-\frac{1}{10})\cdot (1-\frac{1}{10}) = 0.9^2$ (assuming the failure of message and answer are independent). So the probability of failure in the first round is $1-0.9^2$. Since you want the probability to fail in all four of your attempts (and assuming all these failures happen independently), you get $$ (1-0.9^2)^4 = 0.00130321 $$ More generally, if the probability of failure of one message is $p$, you get that the probability of all first $k$ rounds of exchange failing is $$ \left(1-(1-p)^2\right)^k $$ As a sanity check: this decreases (exponentially) with $k$, as one may expect.
answered Feb 28, 2014 at 20:25
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1$\begingroup$ Your statement about independence is not exactly correct. The answer/response has a 0% chance of success if the probe/request does not go through. $\endgroup$– BrannonCommented Feb 28, 2014 at 21:22
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$\begingroup$ Indeed -- as a matter of fact, the only independence needed (and existing) is between rounds, not within. $\endgroup$ Commented Feb 28, 2014 at 21:50
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1$\begingroup$ @ClementC. Actually you did not assume independence (although you say so :) and therefore your result is correct! $\endgroup$– Jimmy R.Commented Mar 1, 2014 at 0:08