# Positive expected value for lottery

As far as I know, to decide either you should enter a bet or not, you should get the expected value of that bet

I was wondering if the lottery has a very high expected value, is it wise to join?

There is a real life lottery with this inputs

Choose 5 numbers out of 50

If you get 3 right you win 350

If you get 4 right you win 1000,000

If you get the 5 numbers right you win 10,000,000

Entry fee is 35

So Expected value is

so total of 112.5- 35(entry fees) = 77.5

This is a very high expected value, i wonder if i did the math wrong?

it was like simple probability

3 winning balls should be
(combin(5,3) * combin(45,2)) /combin (50,5)

And so on did I calculate it wrong? or it's just too good ? and if the expected value is more than double the entry fees, is wise mathematically to join for a long run?

Noting that for i know that for even wining 4 numbers, probability to lose is 99.989%

on the long run for 52 times for example, the probability to lose them all is 99.989%^52 = 99.449%

I'm just confused how the expected value is too high, yet by intuition it seems the right decision is not to join even for a long run Or how should I think about it?

Edit : I guess the point is the 1 Million prize will be shared between winners, but i don't know how should i calculate the expected value in that case?

• Real-life lotteries are always designed to have negative expected value for players. That's because the goal of running a lottery is to make money for those who run it. Commented Jul 22, 2022 at 5:35
• This is a real lottery, but i guess i understood the trick the million will be split between the winners, so i shouldn't have multiplied by 1 million, yet i don't know how to expect how many people will share this to be able to calculate the expected value
– Mina Nessim
Commented Jul 22, 2022 at 5:49
• Can you give a source for the details of this lottery? Commented Jul 22, 2022 at 12:10
• I second @paw88789's request. Since the primary contributor to the expected value, by a large margin, is the $4$-out-of-$5$ event, I wonder if the payout was supposed to be $100,\!000$ instead of $1,\!000,\!000$. Commented Jul 22, 2022 at 16:31
• This was an unfortunate migration. Commented Aug 3, 2022 at 18:56

The expected value calculation when the pot gets to ~$$20$$ million or more has to take into account the returns from investing the winnings. If you win a pot of $$20$$ million and get about $$10$$ million after tax (being ver conservative about the tax rate here), you could reasonably expect to earn $$4\%$$ per year, which is $$400$$k per year. If you are financially responsible, a win $$20$$ million is enough for you and your kids to be financially independent.

– Community Bot
Commented Dec 11, 2022 at 2:33

I simplify the lottery a little bit. Let's say the players can win a fix prize $$W$$, no matter how many players participate in the lottery. We do not regard any fee. Each player can win this prize with a probability of $$p$$. If n players win, then the prize is equally distributed among the players. W.l.o.g. we calculate the expected value for 2 players. They win in total $$W$$ if at least one of the players win. Thus the expected value is

$$E^{(2)}(W)=W\cdot \left( p^2+2\cdot p\cdot (1-p)\right)=W\cdot (1-(1-p)^2)$$

We can generalize the result

$$E^{(n)}(W)=W\cdot (1-(1-p)^n)$$

Now the fix prize W has to be equally distributed amoung the $$n$$ players. Thus the expected value for one player is $$E^{}(W)=\frac{W}n\cdot (1-(1-p)^n)$$. We can look at some interesting values.

• $$\lim\limits_{n\to \infty} \frac{W}n\cdot (1-(1-p)^n)=0$$. If (theoretically) the number of players goes to infinity the expected value is $$0$$.

• If only one player participate at the lottery then the expected value is $$W\cdot (1-(1-p)^1)=p\cdot W$$.

If the prize W is increasing in some way by the number of players $$n$$, then this can be taken into account.