# Probability that a sum of potential numbers is greater than some value

Say I am about to receive 5 cash prizes and I have the probability of receiving each cash prize. Let's denote a set of cash prizes with $k$. So, below is the set of cash prizes and the set of corresponding probabilities: $$k=\{65,25,30,54,30\}$$ $$p(k)=\{0.8,0.5,0.25,0.2,0.4\}$$

Hence, the expected value of my cash prizes is $$E[k]=65*0.8+25*0.5+30*0.25+54*0.2+30*0.4=94.8$$

So, my question is: what is the probability that the sum of the cash prizes will be greater than 85\$? 90$?.

Attempt: I did not come up with anything meaningful. I do not know if the question is valid in the first place. Thanks for your help.

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if the probability of winning any of the cash prizes is independent, then there is no way that these probabilities can be correct. They can only add to 1, so this question is wrong. – mike1886 Jul 18 '14 at 14:34
@mike1886 Right! Thats why I asked it here. This is not an isolated problem. This is a real world problem. The probabilities are my opinion of receiving each cash prize. For example, I bought five lottery tickets. The probabilities are my opinion of winning each cash prize. – Koba Jul 18 '14 at 14:51
@Glen_b no. 0.8=80% – Koba Jul 18 '14 at 15:53

You don't state if the events that each of the prizes are won are independent of each other (and if not, what the form of dependence is), which you need if you want to work out the probability that the total value of prizes exceeds some amount. I'll assume independence for now.

There are also problems with the notation, since you define $k$ to be a set and then treat it like a sum of random variables. Let's construct a better notation first, or we'll have some problems.

Let there be 5 prizes, of amounts $a_1 = 65, a_2=25, a_3=30, a_4=54, a_5=30$. Let $a=(a_1, a_2, a_3, a_4, a_5)'$.

Let $X_1=1$ is prize 1 is won and $0$ otherwise. Define $X_2,...,X_5$ similarly, and let $X=(X_1, X_2, X_3, X_4, X_5)'$.

So if $X=(1,0,0,1,0)$ then the amount won is $65+54 = 119$.

Let $p_1, p_2,...,p_5$ be the respective probabilities that $X_1=1$, $X_2=1$ and so on and let $p=(p_1, p_2, p_3, p_4, p_5)'$. Then $p=(0.80,0.50,0.25,0.20,0.40)'$

There are $2^5=32$ possible outcomes, corresponding to the $2^5$ values $X$ can take.

Let $Y$ be the total amount of prize won; $Y=a'X$. We need to be able to work out the probability distribution of $Y$.

One way to proceed is simply to list all the outcomes, with their probabilities. The probability of a particular outcome, $P(X=x)$ is $p^x\cdot (1-p)^{(1-x)}$ :

$X=(0,0,0,0,0)'$, $Y=0$, probability = $(1-p_1)(1-p_2)...(1-p_5)=0.036$

$X=(0,0,0,0,1)'$, $Y=30$, probability = $(1-p_1)(1-p_2)...p_5=0.024$

$X=(0,0,0,1,0)'$, $Y=54$, probability = $(1-p_1)(1-p_2)...p_4(1-p_5)=0.009$

$X=(0,0,0,1,1)'$, $Y=84$, probability = $(1-p_1)(1-p_2)...p_4p_5=0.006$

$\vdots$

and so on up to

$X=(1,1,1,1,1)'$, $Y=204$, probability = $p_1 p_2...p_5=0.008$

If we collect them all up (summing the probability when you can get the same prize amount in more than one way), the probability (mass) function looks like this:

and the cdf like this:

(though strictly speaking step functions don't have vertical parts); or in table form:

     y  prob   cdf
1    0 0.036 0.036
2   25 0.036 0.072
3   30 0.036 0.108
4   54 0.009 0.117
5   55 0.036 0.153
6   60 0.008 0.161
7   65 0.144 0.305
8   79 0.009 0.314
9   84 0.009 0.323
10  85 0.008 0.331
11  90 0.144 0.475
12  95 0.144 0.619
13 109 0.009 0.628
14 114 0.002 0.630
15 119 0.036 0.666
16 120 0.144 0.810
17 125 0.032 0.842
18 139 0.002 0.844
19 144 0.036 0.880
20 149 0.036 0.916
21 150 0.032 0.948
22 174 0.036 0.984
23 179 0.008 0.992
24 204 0.008 1.000


from which we can find $P(Y\leq 85)$ and $P(Y\leq 90)$, and hence obtain $P(Y>85)$ and $P(Y>90)$.

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You are the man. Thank you a lot. Exactly, I am looking for the probability of their sum exceeding a value. – Koba Jul 18 '14 at 15:52
Now you have the entire pmf and cdf. – Glen_b Jul 18 '14 at 16:42
Awesome. You using R? I use R a lot as well. Thanks man. I will right an algorithm in R and Python and probably make a blog post about it. I will cite your answer. – Koba Jul 18 '14 at 16:53
Yes, I used R. I didn't do anything very clever, just used it like a calculator really. – Glen_b Jul 18 '14 at 16:53
I would say that carefully defining the important variables was critical to me being able to do this problem without getting badly confused. Notation matters a lot - if you don't get it right, you quickly become mired in confusion. (At least I do.) It's worth some effort (and takes practice) to make it right. – Glen_b Jul 18 '14 at 17:21

Now the problem at hand is trivial as expected cash prize is not random if prize values and probabilities are fixed. Consequently, this probability equals zero or one with respect to correspondence between your calculated expected value and fixed sum.

If you want instead find $P(k > 85)$, I think that as it is a real world problem you should use some brute-force algorithm to calculate this probability.

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Yeah they are fixed. Any idea how the algorithm would look like. The maximum amount of money I can receive is 204\$(just the sum). With the listed probabilities it is 94.8(expected value). In the worst case I dont receive anything, so I get 0. So, if I win some cash prizes (I dont know which) I want to know the probability of their sum being greater than some k. – Koba Jul 18 '14 at 15:05