Using the Lottery to Break a Tie

[MickeyHoldem]

Ignoring the bonus number.... there are 13983816 (49c6) unique tickets.

Order the tickets....
1) 1,2,3,4,5,6
2) 1,2,3,4,5,7
3) 1,2,3,4,5,8 etc.

if the ticket drawn falls in the first 6991908 tickets Brother A wins, else Brother B wins

OB Puzzle: What is the last ticket number that Brother A wins with?

[jakka]

Thanks for the help. I had been trying to find a solution to this question. MickeyHoldem's solution, I think was correct, and I could do it with a bit of programming. Nice extra question, what is that dividing nummber?

pzhon's solution - that's it. Well done.

[CardSharpCook]

HI/LO, one takes all numbers above 25, the other, all below. If the FIRST number selected by the 649 IS 25, go to the second number.

CSC

[pzhon]

[ QUOTE ]

Order the tickets....
1) 1,2,3,4,5,6
2) 1,2,3,4,5,7
3) 1,2,3,4,5,8 etc.

if the ticket drawn falls in the first 6991908 tickets Brother A wins, else Brother B wins

OB Puzzle: What is the last ticket number that Brother A wins with?

[/ QUOTE ]
That works. Ticket (49C6)/2 is [6,7,16,20,28,47].

(49 choose 6)/2 =
(49C6)-(44C6)+
(43C5)-(43C5)+
(42C4)-(34C4)+
(33C3)-(30C3)+
(29C2)-(22C2)+
(21C1)-(2C1)

[tubbyspencer]

Tell them to flip a coin. [img]/images/graemlins/smile.gif[/img]

[RocketManJames]

Enumerate via a computer program all the combinations of this lottery.

Give Bro #1 the first half, give Bro #2 the second half.

If there are an odd number of combinations, let the parity of the bonus ball decide, except when the bonus ball is 49. If this is the case, then use the parity of the next largest number.

I am assuming that there are also 49 possible numbers for the bonus number.

I think this works... not 100% though.

-RMJ

[jokerthief]

That's not 1-1 though.

[CountDuckula]

[ QUOTE ]
HI/LO, one takes all numbers above 25, the other, all below. If the FIRST number selected by the 649 IS 25, go to the second number.

[/ QUOTE ]

The problem is, the initial problem stated that the numbers are not published in the order drawn. They are sorted in ascending order (except that the bonus ball is always listed last), and the universe of combinations that include at least one ball <= 25 is much larger than that which include no balls <= 25. The only number that can be used for this purpose is the bonus ball, which is always drawn last and is always published independently of the others.

It would be possible to use odd/even on the lowest numbered ball (whether or not it happens to be the bonus ball). Since 7 total balls are drawn, we disregard all balls numbered 43 and higher to ensure that the lowest ball will be drawn from a pool of numbers containing an equal # of odd and even numbers (i.e., 1-42; however, it is necessary to hold out the possibility of a discard if the numbers 43-49 are drawn, though that is extremely unlikely).

-Mike

[Edit: Re-read initial problem statement and corrected my method to reflect the fact that 7 total balls are drawn, not 6.]

[CountDuckula]

[ QUOTE ]
Tell them to flip a coin. [img]/images/graemlins/smile.gif[/img]

[/ QUOTE ]

Did you miss the part of the problem statement which stated that they do not live in the same city, and therefore can't flip a coin? [img]/images/graemlins/grin.gif[/img]

-Mike

[pzhon]

[ QUOTE ]

It would be possible to use odd/even on the lowest numbered ball (whether or not it happens to be the bonus ball). Since 7 total balls are drawn, we disregard all balls numbered 43 and higher to ensure that the lowest ball will be drawn from a pool of numbers containing an equal # of odd and even numbers (i.e., 1-42; however, it is necessary to hold out the possibility of a discard if the numbers 43-49 are drawn, though that is extremely unlikely).


[/ QUOTE ]
The lowest numbers are not evenly distributed among the logical possibilities. There are more ways for the lowest number to be 1 than 2, more ways to have the lowest number be 3 than 4, etc. This adds up to a significant bias toward odd.

Mathematica code:
<ul type="square">
Sum[Binomial[49 - j, 6]If[EvenQ[j], even, odd], {j, 42}]
39680014 even + 46220569 odd[/list]So, even after throwing out a possibility, you are still letting one brother win 53.81% of the time.