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i want to choose optimal decision from following problem Imagine having been bitten by an exotic, poisonous snake. Suppose the ER physician estimates that the probability you will die is $1/3$ unless you receive effective treatment immediately. At the moment, she can offer you a choice of experimental antivenins from two competing ‘‘snake farms.’’ Antivenin $X$ has been administered to ten previous victims of the same type of snake bite and nine of them survived. Antivenin $Y$, on the other hand, has only been administered to four previous patients, but all of them survived. Unfortunately, mixing the two drugs in your body would create a toxic substance much deadlier than the venom from the snake. Under these circumstances, which antivenin would you choose, and why?

so first off all i have concluded that, for substance $X$,i would have $90$% chance to be survivded,and for $Y$ ,i would have $100$%,so maybe it should be indicator for me to choose $Y$,on the other hand,if we consider it as a combinatoric problem,then we have $p=2/3$ if we don't die when get medical treatment, and $q=1/3$ if we do,so for substance $X$,we would have $(10!/9!)*((p)^{10})*q=0.05202459$ while for substance $Y$,it would be $4!/4!*(2/3)^4*(1/3)^0=16/84=0.19047619$

so it means that i have more chance for $Y$,so does it means that i should choose $Y$?

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A link to the original problem – joriki Jun 30 '12 at 11:02
yes exactly it is – dato datuashvili Jun 30 '12 at 11:07
then why it is that such question comes in exams?i am surprised – dato datuashvili Jun 30 '12 at 11:10
up vote 2 down vote accepted

This question cannot be answered without knowledge of your prior assessment of the effectiveness of the antivenins. If X was produced by a company that usually produces effective drugs and Y was produced by a quack, the result will be different than if it was the other way around. If you don't have any prior information on the likely effectiveness of the antivenins, you'll need to make some assumption that will be to some degree arbitrary. For instance, since you'd survive with probability $2/3$ without the antivenin, you could assume a uniform distribution between $2/3$ and $1$ for the chance of surviving after taking the antivenin, for each of the two antivenins (assuming you're confident that they're not harmful). Then with $j$ trials successful and $k$ trials unsuccessful for antivenin $i$, the posterior distribution for the survival probability $p_i$ of antivenin $i$ would be

$$\frac{p_i^j(1-p_i)^k}{\int_{2/3}^1p^j(1-p)^k\mathrm dp}\;,$$

and your survival probability if you take antivenin $i$ would be

$$\int_{2/3}^1\frac{p_i^j(1-p_i)^k}{\int_{2/3}^1p^j(1-p)^k\mathrm dp}p_i\mathrm dp_i\;,$$

which comes out as $502769/589806\approx0.852$ for X and $3325/3798\approx0.875$ for Y.

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actually it is very interesting coming of integral here,but it was asked in combinatorics book,where it is not mentioned integral at all ,by the way thank you very much for helping – dato datuashvili Jun 30 '12 at 10:57
yes i have copy this problem from book,it was asked in probability section in general,so i thought i could use some combination methods with probability to conclude something – dato datuashvili Jun 30 '12 at 11:01
@dato: The book may be making some unwarranted implicit assumptions. In the form it's posed, this isn't a combinatorial problem. To realize that, just think about the quack/reliable company scenario I described. It shows that the answer can't be independent of your assumptions about the sources of the antivenins. – joriki Jun 30 '12 at 11:03
yes maybe,thanks very much for helping – dato datuashvili Jun 30 '12 at 11:05

To follow up on joriki's answer...

Let's consider the case that you don't make the assumption that the antivenins aren't harmful, but rather just start from the assumption that any survival rate between 0 and 1 is equally likely for a given antivenin a priori, before any trials have been carried out.

Then the expected survival rate for an antivenin, after $j$ successful and $k$ unsuccessful trials have been carried out, is $$\int_0^1\frac{p^j(1-p)^k}{\int_0^1q^j(1-q)^k\mathrm dq}p\,\mathrm dp\,,$$ which works out to simply $$\frac{j+1}{j+k+2}.$$

Thus, under these assumptions, you should expect antivenin X to give you a $$\frac{9+1}{9+1+2} = \frac{10}{12} = \frac{5}{6} \approx 83\%$$ chance of survival, while the corresponding probability for antivenin Y should also work out to $$\frac{4+1}{4+0+2} = \frac{5}{6} \approx 83\%.$$

Thus, under these assumption, it doesn't matter which antivenin you take — either of them will cut your chance of dying down by half, from $\frac13$ down to $\frac16$.

This simple "rule of thumb", that after $x$ successful and $y$ unsuccessful trials you should expect the success rate to be $\frac{x+1}{x+y+2}$, is a very old and classic principle known as the rule of succession. It was introduced by Pierre-Simon Laplace, who famously used it to calculate the probability that the sun will rise tomorrow, given that it has done so every day for at least the past 5000 years. It may also well be what the author of your textbook was expecting you to apply.

(Of course, if you did assume a uniform prior between 0 and 1 for the survival rate after taking an untested antivenin, then you should never agree to be the first person to try one, since the $\frac12$ expected survival rate predicted by the rule of succession is less than the $\frac23$ postulated for taking no treatment. Thus, the first person who tried each antivenin must either have assumed a different prior, or they must've been very stupid or very altruistic.)

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yes it seems strange,because if student would be given to this task,then he/she will try method which i used and makes mistake – dato datuashvili Jun 30 '12 at 11:49
Nice -- I didn't realize that the answers would be the same if you allow for the antivenins to be harmful. – joriki Jun 30 '12 at 14:27

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