ChatterBank4 mins ago
Mini Lottery
5 Answers
Hi AB
I'm curious.
If there is a lottery with possible numbers 1-36 and 3 balls are chosen at random.
How many different win combinations would there be?
I'm curious.
If there is a lottery with possible numbers 1-36 and 3 balls are chosen at random.
How many different win combinations would there be?
Answers
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Ok moving on to the more complicated bit. I haven't a clue how to work this one out!
1st jackpot = 3 numbers in order
2nd = 3 numbers random order
3rd = 2 numbers in order
4th = 2 numbers random order
What would be the odds of getting such wins?
How many different number combinations would you need to gurantee a "profit" on your original gamble?
I have a feeling someone is cheating and I'm curious how easy it would be for them to do that?
Ok moving on to the more complicated bit. I haven't a clue how to work this one out!
1st jackpot = 3 numbers in order
2nd = 3 numbers random order
3rd = 2 numbers in order
4th = 2 numbers random order
What would be the odds of getting such wins?
How many different number combinations would you need to gurantee a "profit" on your original gamble?
I have a feeling someone is cheating and I'm curious how easy it would be for them to do that?
There is no way you can ever guarantee a profit.
In your first example, although you can guarantee getting a winning combination by buying 7140 tickets, the size of the payout depends on how many other people also have a winning combination.
I think I've read somewhere that approx 55,000 have the numbers 1, 2, 3, 4, 5 & 6 in the National Lottery. If those numbers happened to come up and the jackpot was �5.5 million that would only be �100 each.
In your first example, although you can guarantee getting a winning combination by buying 7140 tickets, the size of the payout depends on how many other people also have a winning combination.
I think I've read somewhere that approx 55,000 have the numbers 1, 2, 3, 4, 5 & 6 in the National Lottery. If those numbers happened to come up and the jackpot was �5.5 million that would only be �100 each.
Once you wish to consider the order of the balls, there are 42840 different permutations. With only one ordered triple winning the Jackpot, the odds are 42839 to 1.
For the second prize there are 5 (3x2x1 -1) options and the odds fall to 7139 to 1. (see earlier answer)
For the third prize, I assume that you can have any pair (in the ordered triple) to match - and so there are 99 (3x33) options. Since 42840/99 = 433 (approx), this means the odds are 432 to 1 for two out of three correct and in the right positions.
When you remove the order the options grow to 495 (99x5). Since 42840/495 = 87 (approx) then the odds become 86 to 1.
So far we have accounted for 600 winning options meaning the odds of winning SOMETHING are 70 to 1. This then leaves 42240 options (out of 42840) that are losing options which of course carries odds of 1 to 70. )
I hope that this answers your question.
I also hope that I haven't made an error in the calculations!
For the second prize there are 5 (3x2x1 -1) options and the odds fall to 7139 to 1. (see earlier answer)
For the third prize, I assume that you can have any pair (in the ordered triple) to match - and so there are 99 (3x33) options. Since 42840/99 = 433 (approx), this means the odds are 432 to 1 for two out of three correct and in the right positions.
When you remove the order the options grow to 495 (99x5). Since 42840/495 = 87 (approx) then the odds become 86 to 1.
So far we have accounted for 600 winning options meaning the odds of winning SOMETHING are 70 to 1. This then leaves 42240 options (out of 42840) that are losing options which of course carries odds of 1 to 70. )
I hope that this answers your question.
I also hope that I haven't made an error in the calculations!