# Probability of get a kidney donated

It is really a probability problem. I use the story of kidney donation because it is easier to describe.

Consider the following scenario:

Time is discrete.

At each period, the measure of patients in the hospital is $M$.

At each period, the measure of new kidney supply $I$ is i.i.d. positive, with mean $\mu_I$ and variance $\sigma_I^2$.

If the measure of kidneys $I$ is larger than measure of waiting patients $M$, then every patient will be matched. But unmatched kidneys cannot be used next time. If $I<M$, each will be matched by probability $I/M$. There is no first come first serve rule here.

A patient leaves the game if and only if he/she gets donated.

If one is not matched at the current period, he can go to next period. A patient never dies.

The number of patients next period will be $$M'=M-I+m$$ - a measure $m>0$ of new patients will come. $m$ is certain.

Now I assume $E(I)<m$ - the waiting pool has a positive probability of going to infinitely large.

Now my question is, whether a patient's probability of being matched with a kidney is strictly smaller than one or not at $t$ goes to infinity?

• If $I < M$, how can each person be matched with $\dfrac{I}{M}$? That means that there is a positive probability that each of the $M$ people are matched. But that's not possible because there are only $I$ kidneys and $I < M$. – adijo Jun 12 '15 at 4:48
• @adijo no, each will be matched with probability I/M, so finally only a fraction of I/M will be matched. – Andy Xu Jun 12 '15 at 4:50
• Oh okay, it seemed to me from the wording that you claimed that the probability of getting matched was constant but it obviously changes. – adijo Jun 12 '15 at 4:55
• @adijo Yes,since M is a random process and I is random shock – Andy Xu Jun 12 '15 at 5:10
• @AndyXu Is kidney supply normally distributed? – David Jun 12 '15 at 14:24

Can we not think like this that probability of a patient getting rejected first time is (M-I)/M which is less than 1.

Probability of a person not getting a kidney second time is (M'-I')/M' which also less than 1.

Now, the probability of a person not getting a kidney in first attempt and then in second attempt is (M-I)/M*(M'-I')/M.

As at any attempt probability of a person not getting a kidney will be always less than 1,the probability of a person not getting a kidney in n attempts will keep on decreasing and if n tends to infinity, it will tends to zero.

So, if probability of a person not getting a kidney in infinite attempts is zero than its inverse, probability of a person getting a kidney in infinite attempts should be 1.

• Thanks for your replay. But you cannot simply multiply all the probability together since they are not independent. Also, since kidney supply is smaller than demand, in the long run, it may be the case that there are infinite many patients, then each has little chance to be matched. – Andy Xu Jun 12 '15 at 14:47
• If we are calculating probability of a person not getting a kidney then we have to multiply and when n tends to infinity, it will tends to 1 but still be less than 1. – Sidharth Jun 12 '15 at 17:36

Let me de the calculations.

IF $M_t$ measures the patients at $t^{th}$ period, given that $\mu_I<m$, $\lim_{t\to \infty} M_t=\infty$

But what's the probability for a specific patient to recieve a kidney in the period $t$ (event $K_t$)?

\begin{align} P(K_t)&=P(I\geq M_t)\times 1+P(I<M)\times\frac{I}{M_t}\\ \lim_{t\to \infty}P(K_t)&= 0\times 1 + 1\times \frac{I}{\infty}\\ &=0 \end{align}

As you see, $P(K_t)>P(K_{t+1})$. As time goes by, probability gets nearer to $0$, making the expected waiting of periods $\infty$ for all the patients.

• Thank you for the reply. I have two problems. First, I think the probability of get a kidney at period t should be $$P(I>M)+\int_0^M g(I) I/M dM$$ Second, $P(K_t)$ is the probability of getting a kidney at current period, even if it converges to zero, it does not imply that he will not get one in the future, right? – Andy Xu Jun 12 '15 at 19:14