There is 45 numbers and you need 6 numbers to win the jackpot. In this game you've decided to pick 10 numbers for a better chance of winning then what is your chance of winning?


This post was edited by oenigmao on Nov 7 2011 04:33am

(10!(45-6)!) / (45!(10-6)!) = 0.000025782498839787553 = 1/38786

Gl, It's something like 70% winrate.

Less likely than getting struck by lightning

twice

never mind thought you were calculating the probability for winning with one ticket lol

still wondering why 10!...why not calculate the probability of winning with one ticket then multiply that by 10? So now instead of having one of the possible combinations, you're saying you have 10 of them.

seems like it'd be 10(6!/45*44*43*42*41*40)

the 6! comes from how many different orders of the numbers there are, and the denominator is how many numbers you have to choose from for each choice.

This post was edited by Derkaderk on Nov 7 2011 09:24pm

Having 10 tickets that each only have 6 numbers yields a lower probability of winning than having 1 ticket with 10 numbers. Consider having 45 tickets with 6 numbers each, and 1 ticket that has all 45 numbers. Obviously, the 1 ticket with 45 numbers has a 100% chance of winning, while the 45 regular tickets have a far lower winning rate.

The probability of guessing the 6 correct numbers with only 6 guesses is




This much you have calculated correctly. But this next step is where your mistake was.

The probability of guessing the 6 correct numbers with 10 guesses is equal to the above probability multiplied by the number of ways that the 6 winning numbers can be arranged within 10 guesses.



This post was edited by frail on Nov 7 2011 10:50pm

1 in 14 million.

better chance of dieing before you buy a lotto ticket then winning.. goodluck

Seems to me your probability is...thousands of tickets, and your final answer seems to reflect that. I can't imagine the odds of winning the lottery with 10 tickets being around 1/40,000. That's way too low.

I don't see how buying 10 tickets increases your chances from one in roughly 8,000,000 to one in roughly 40,000

Similarly, you wouldn't buy (45!-(45-6)!)! tickets for a guarantee of winning. You'd buy (45!-(45-6)!) tickets.

Out of curiosity, I plugged 20 into your formula and the probability now is roughly one out of 200. 30 tickets yields a probability of seven out of 100. 40 tickets is basically a coin flip.

Considering the payout, I don't think a 50/50 chance of winning with 40 tickets is economically plausible for the lottery.

from my understanding its 1 ticket with 10 #s. u need to match 6 for jackpot