Since there are 6 numbers to select from 49, according to the basic probability calculations, the chance that some single ticket will win is 1 to 13,983,816.

Hence, the expected value for a single ticket would be 50 million PLN * 1/13983816 or 3.57 PLN. Seems, someone buying some 13.98 million tickets (each at 3 PLN, which requires an investment of 42 mln PLN) cannot lose, right?

Not exactly.

It may happen there will be more than one winner. Let's suppose, some 10 million tickets will be bought in total by various people. Then, the probabilities of 2, 3, 4 or 5 distinctive winners are as following:

> dbinom(2:5,10e6,1/choose(49,6))

[1] 0.1250688763 0.0298127636 0.0053298679 0.0007622907

In other words, there is a pretty large chance (16.1%), there will be 2 or more wining tickets.

OK, so what is the expected value given the number of tickets bought?

Assuming that between 1 and 50 million tickets will be bought, the expected value decreases from 3.512 to 1.319:

Chart: expected value [PLN] vs number of tickets sold [mln]

For 10 million tickets, the EV is equal to 2.9659, which is below the price of the ticket (3 PLN):

> Nwinners <- 100 # max. simultaneous winners in history - 80

>

> Nbets <- 10e6

>

> Vwin <- 50e6

>

> Wwin <- sum(dbinom(1:Nwinners,Nbets,1/choose(49,6))/sum(dbinom(1:Nwinners,Nbets,1/choose(49,6)))*(Vwin/(1:Nwinners)))

>

> Wwin

[1] 41475080

>

> Wwin/Vwin

[1] 0.8295016

>

> Wwin/choose(49,6)

[1] 2.965934

And there is also a 10% tax on winnings...

UPDATE: On March 30, 1994 there was 80 winning tickets. The probability of such an event is 4.985477*10^-86, even if ALL the Poles would have bought one ticket.

> dbinom(80,38*10^6,1/choose(49,6))

[1] 4.985477e-86

The probability drops to 1.529368*10^-131 with 10 million tickets bought.

Just as a reference, the number of the atoms in the observable universe is estimated at 10^80.

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