# Which is the correct way to calculate the expected value of a shared lottery jackpot?

I want to calculate the expected value of a ticket in a lottery game which gives players a probability $p$ of winning a jackpot prize of $j$ dollars. The total number of tickets in play is $t$.

If every winning ticket gets the full prize amount, the expected value for a ticket is given by $jp$. However, if winners must evenly split the prize in case of multiple winners, then the expected value depends on the number of winners $W$.

The expected number of winners is $tp$. The probability that the number of winners $W$ is $w = 0, 1, 2, \dotsc$, follows a Poisson distribution with the expected number of winners as its parameter:

$$P(W=w) \sim Pois(tp) = \frac{tp^we^{-tp}}{w!}$$

I don't know how to get from there to calculating an accurate expected value for the ticket as a function of the number of tickets in play.

In reading online, I've found two different methods each used by several sources. If I'm following them correctly, then they give different results. My question is 1) which one is correct? 2) what is the error in reasoning (or in my understanding/implementation) in the incorrect method?

## Method 1: Number of Winners

The first method calculates the probability that the number of winners $W$ will be $w = 0, 1, \dotsc, t$, given that there is at least one winner:

$$P(W=w | W>0) = \frac{P(W>0|W=w)P(W=w)}{P(W>0)}$$

Where,

• $P(W>0|W=w)$ is $\left\{ \begin{array}{lr} 0 & : w = 0\\ 1 & : w > 0 \end{array} \right.$
• $P(W=w)$ is the probability of $w$ winners: $\frac{tp^we^{-tp}}{w!}$
• $P(W>0)$ is the probability of more than one winner: $1 - P(W=0)$

So the expected value of the ticket is given by:

$$p\sum_{w=1}^{t} \frac{j}{w}\frac{P(W=w)}{1-P(W=0)}$$

For a numerical example, we'll tabulate the first few values of $P(W=w)$ for a lottery with a 1/34,220 chance of winning \$100,000 jackpot, with 6,000 tickets in play, so$p = 1/34,220; j = 100,000; \text{and } t = 6,000$$$\begin{array}{c|c|c|c|c|} \text{Winners} & \text{Probability} & \text{Conditional Probability} & \text{Share} & \text{Contribution } \\ w & P(W=w) & P(W=w|W>0) & j/w & (j/w)P(W=w|W>0) \\ \hline 0 & 0.839 & 0 & \text{\0} & \text{\0} \\ \hline 1 & 0.147 & 0.913 & \text{\100,000} & \text{\91,300} \\ \hline 2 & 0.013 & 0.081 & \text{\50,000} & \text{\4,050} \\ \hline \end{array}$$ Summing the contribution column and multiplying by$p$gives an expected value of$2.79.

## Method 2: Number of Other Winners

The second method calculates the probability that the number of total winners $W$ is $w = 0, 1, \dotsc, t$, given that our ticket is a winner:

$$P(W=w|Winner) = \frac{P(Winner|W=w)P(W=w)}{P(Winner)}$$

Where,

• $P(Winner)$ is the probability that our ticket is a winner: $p$
• $P(Winner|W=w)$ is the probability that our ticket is a winner given $w$ winning tickets: $w/t$
• $P(W=w)$ is the probability of $w$ winners: $\frac{tp^we^{-tp}}{w!}$

Plugging those figures in shows that $P(W=w|Winner)$ reduces to $P(W=w-1)$:

$$\frac{w}{t}\frac{P(W=w)}{p} = \frac{tp^{w-1}e^{-tp}}{(w-1)!} = P(W=w-1)$$

So the expected value is given by:

$$p\sum_{w=1}^{t}\frac{j}{w}\frac{tp^{w-1}e^{-tp}}{(w-1)!}$$

Using the same lottery numbers as above, the first few values of $w$ are given in the following table.

$$\begin{array}{c|c|c|c|c|} \text{Winners} & \text{Probability} & \text{Conditional Probability} & \text{Share} & \text{Contribution } \\ w & P(W=w) & P(W=w|Winner) & j/w & (j/w)P(W=w|Winner) \\ \hline 0 & 0.839 & 0 & \text{n/a} & \text{\0} \\ \hline 1 & 0.147 & 0.839 & \text{\100,000} & \text{\83,900} \\ \hline 2 & 0.013 & 0.147 & \text{\50,000} & \text{\7,350} \\ \hline \end{array}$$

Summing the contribution column and multiplying by $p$ gives an expected value of $2.67. ### Online Resources Which Use Method 2 Clearly the expected payout for the example lottery above cannot be both \$2.79 and \$2.67, but I'm having a difficult time reasoning my way to the correct method. Any hints will be appreciated! ## 4 Answers Proof that Methods 2 and 3 are equivalent: Method 3 has a very intuitive appeal as the correct approach when interpreted as someone purchasing$allt$tickets. Now we'll show that Method 2 gives the same result if we use the binomial for the distribution of winning tickets instead of the Poisson. The expected value for Method 2 is: $$p\sum_{w=1}^t\frac jw {t-1 \choose w-1}p^{w-1}(1-p)^{t-w}$$ $$=j \sum_{w=1}^t \frac {(t-1)!}{w!(t-w)!}p^{w}(1-p)^{t-w}$$ $$=\frac jt \sum_{w=1}^t {t \choose w}p^{w}(1-p)^{t-w}$$ $$=\frac jt (1-(1-p)^t )$$ which is also the result for Method 3. If we had started with the Poisson distribution as an approximation to the binomial, then the expected value for Method 2 reduces to $$\frac jt(1-e^{-tp})$$ The Poisson will be a good approximation to the binomial when$t$is large,$p$is small and$tp$is moderate, which holds here. Method 3: By linearity of expectation, the expected value of one ticket is$\frac{1}{t}$times the expected total value of all the tickets. That expected value is$jP(W>0)=j(1-P(W=0))=j(1-(1-p)^t)$. Thus the expected value of one ticket is $$\frac{j}{t}(1-(1-p)^t))$$ For the numerical values you use, I get \$2.68. I'm not skilled enough at statistics to find the logical flaw in Method 1, but I suspect Method 2 is right (up to rounding).

Followup: it seems that this solution, or a variation, was listed under "resources" supporting the second solution. I didn't notice it on my first read-through due to wall of text.

• Thank you, vadim. Yeah, I had a link to this answer buried in the resources section, but your mentioning that expectation is linear helped me understand why it is correct without having to actually enumerate and sum the probability of each number of winners. That leaves me very confident that Method 2 is correct. However, I would like to find a good way of explaining why Method 1 is incorrect. – cristoper Jun 29 '14 at 5:25
• I suspect that the problem with Method 1 is that the probability my ticket is a winner is no longer $p$ when we have conditioned on there being exactly $w$ winners. – vadim123 Jun 29 '14 at 5:34

This is my own summary answer. If somebody has a better way of explaining why the first method in my question goes wrong (and what it does compute, if anything useful), please post an answer!

## Method 1 is incorrect

If $P(Winner)$ is the probability that our ticket is a winner, then as user159813 pointed out, the expected payout from the lottery can be expressed as:

$$\sum_{w=1}^{t} \frac{j}{t} P(Winner \cap W=w)$$

However, the payout calculated by Method 1 is:

$$\sum_{w=1}^t \frac{j}{t} P(W>0 \cap W=w)$$

Which is different because $P(W>0) \ne P(Winner)$. We want $P(W=w)$ given that our ticket won, which is a lesser probability than given that any ticket won.

## Method 2 is correct

As vadim123 points out in his answer, the correctness of Method 2 is confirmed by the nice, closed-form formula derived from the fact that the expected value of a single ticket is the same as the expected value of all tickets divided by $t$:

$$\frac{j}{t}(1-(1-p)^t)$$

## Simulation

I ran a simulation using the same numbers as the example in my question. The mean payout after the 10,000,000th iteration was about \$2.64. After about 5,000,000 iterations, the mean seemed to oscillate about the$2.68 line. There is a problem in Method 1. Letting $Y$ represent amount won and $W$ be number of winners we see that $$E(Y)=\sum_{w=0}^{t}\frac{j}{w}P(Winner\cap W=w)$$

The second method does this correctly by calculating $$P(Winner\cap W=w)=P(W=w|Winner)P(Winner)=P(W=w-1)P(Winner)=P(W=w-1)p$$

as you have above. Now the problem with Method 1 is that it says that $$P(Winner\cap W=w)=P(W=w|W>0)P(Winner)$$ which is not true. I would guess the method this person was going for is actually $$P(Winner\cap W=w)=P(Winner| W=w)P(W=w)$$

• I don't see where Method 1 says $P(Winner \cap W=w)=P(W=w|W>0)P(Winner)$. However, I do see where it says it equals $P(W=w|W>0)P(W>0)$. And looking at your formulation of the expectation function, it is easier for me to see that Method 1 is wrong because $P(W=w|W>0)P(W>0) \ne P(W=w|Winner)P(Winner)$ – cristoper Jun 29 '14 at 20:34
• And should the index in your sum start at $w=1$ to avoid the divide by zero? $P(Winner \cap W = 0) = 0$ anyway. – cristoper Jun 29 '14 at 20:47