Originally posted by squaccermanYou would need to have every single combination of three numbers using the numbers between 1 and 49, inclusive. That is 49 choose 3, which equals:
What is the minimum number of tickets you need to buy for the UK national lottery to ensure winning ten pounds? For those not familiar with the lottery 6 numbers are picked from 49, and you need 3 to win ten pounds.
49! / ((49-3)! * 3!) = 18,424.
Note that every six-number combination already contains a number of three-number combinations, 6 choose 3:
6! / ((6-3)! * 3!) = 20.
This reduces the number of combinations:
18,424 / 20 = 921.2.
Round this value up to 922 tickets. All the tickets would be arranged so that every combination was accounted for. The decimal result and the rounding indicate that a few three-digit combinations would be repeated.
In any case, it is obviously not worth it to win 10$.
-Ray.
Originally posted by rgoudieI'm not convinced by this argument - how do you know that the tickets can be chosen with so little overlap? Also, why must every possible triple be covered? It is sufficient to cover only those triples made of numbers from 1 to 46, and it may be possible to reduce this number further.
You would need to have every single combination of three numbers using the numbers between 1 and 49, inclusive. That is 49 choose 3, which equals:
49! / ((49-3)! * 3!) = 18,424.
Note that every six-number combination already contains a number of three-number combinations, 6 choose 3:
6! / ((6-3)! * 3!) = 20.
This reduces the number of combinatio ...[text shortened]... binations would be repeated.
In any case, it is obviously not worth it to win 10$.
-Ray.
Your answer is probably roughly right, but I can't see how to get the exact answer and prove it.
Originally posted by AcolyteI see what you mean.
I'm not convinced by this argument - how do you know that the tickets can be chosen with so little overlap? Also, why must every possible triple be covered? It is sufficient to cover only those triples made of numbers from 1 to 46, and it may be possible to reduce this number further.
Your answer is probably roughly right, but I can't see how to get the exact answer and prove it.
This obviously needs more work.
-Ray.