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

enter image description here

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?

  • 5
    $\begingroup$ 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. $\endgroup$ Commented Jul 22, 2022 at 5:35
  • $\begingroup$ 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 $\endgroup$
    – Mina Nessim
    Commented Jul 22, 2022 at 5:49
  • 1
    $\begingroup$ Can you give a source for the details of this lottery? $\endgroup$
    – paw88789
    Commented Jul 22, 2022 at 12:10
  • 3
    $\begingroup$ 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 $1,\!000,\!000$. $\endgroup$
    – Brian Tung
    Commented Jul 22, 2022 at 16:31
  • $\begingroup$ This was an unfortunate migration. $\endgroup$ Commented Aug 3, 2022 at 18:56

2 Answers 2


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.

  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – 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.


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