Using the Lottery to Break a Tie

[jakka]

Two brothers, Mike and Kilby, living in two Canadian cities, place a wager on a hockey game. Not the NHL, they're not playing. But the game ends in a tie. This is a disappointment, because they need to break this tie.

Because they can't get together to flip a coin, Mike, the older brother, suggests that they somehow use the published results of the next Lotto 649 to settle their wager. Now the Lotto 649 is Canada's national lottery. They draw 6 numbers at random from 1 to 49. They also draw a 7th bonus number. When published in the newspapers, the six numbers are listed from low to high, then the bonus number is given.

Kilby says this is not so easy to do. So here's the challenge. How can they agree to use the next 649 results, such that they each have a 50% chance of winning?

[dtbog]

[ QUOTE ]
How can they agree to use the next 649 results, such that they each have a 50% chance of winning?

[/ QUOTE ]

If they really drew all of the numbers randomly, they could simply see who chose the lowest number (including the bonus ball).

If it's the same, then go to the second number.

This doesn't really use the lotto results, but it solves the problem!

An almost-complete solution would be to see who chose the bonus number closest to the actual bonus number (counting positive and negative distance as the same, and looping around at 49). Of course, these numbers could be the same, or they could be equidistant from the bonus number... so you could compare the highest number to the bonus number, etc etc.

You could also sum each player's lottery picks and see who has the highest sum. Again, doesn't use the results -- what about summing the each player's picks and seeing which player's sum comes closest to the actual sum?

Seems like there are a lot of solutions, since they chose the numbers randomly.

-DB

[Marm]

Add the values of the 6 balls into one Sum, counting 49 as zero, If the sum is Even brother 1 ones, if the sum is odd brother 2 wins.

[pzhon]

The questions is to describe an easily recognizable subset of the possible tickets of size 1/2 of the total. The following method works, but other, simpler methods may exist.

[img]/images/graemlins/diamond.gif[/img] If the bonus number is 1-24, M wins. 25-48, K wins. 49, proceed.

[img]/images/graemlins/diamond.gif[/img] If there are 0, 1, or 2 numbers from 1-24, M wins. 4, 5, or 6 from 1-24, K wins. 3, proceed.

[img]/images/graemlins/diamond.gif[/img] If the second number is 2-12, M wins. If the second number is 13-23, K wins.

This would not be possible if the numbers ranged from 1-47, since the number of possible tickets (41 * 47C6) would be odd.

[pzhon]

[ QUOTE ]
Add the values of the 6 balls into one Sum, counting 49 as zero, If the sum is Even brother 1 ones, if the sum is odd brother 2 wins.



[/ QUOTE ]
Sorry, that doesn't work. Brother 1 would win 25C6+(25C4 24C2)+(25C2 24C4)+24C6 = 6990896 times out of 13983816 (ignoring the seventh number). That's 49.9928%, 1012/13983816 less than 1/2.

[Dilbert]

Some of you guys are over thinking this.

Since the numbers are listed from low to high, just choose one of the numbers (say the 3rd lowest) and let odd/even determine who wins.

You should never choose the lowest or highest number, since they would have a slight bias to odd (since 1 and 49 are odd numbers). The 3rd or 4th out of a list of 6 from lowest to highest will not have this same bias effect.

[jakka]

"Since the numbers are listed from low to high, just choose one of the numbers (say the 3rd lowest) and let odd/even determine who wins."

I don't think that would work, in that there are more odd numbers than even numbers to choose from.

[tylerdurden]

Do you want an answer from exactly one day's result, or do you mind if it takes a few days?

If you can wait a few days, look at the "bonus" ball for two consecutive days. If it's odd on day one and even on day two, Brother 1 wins. If even then odd, Brother 2 wins. If (even, even) or (odd, odd), then start over.

[CountDuckula]

How about using the bonus number and going high/low? If the number is > 25, Mike wins, if it's < 25, Kilby wins (or vice-versa), and if it = 25, try again the next day.

-Mike

[pzhon]

[ QUOTE ]
Some of you guys are over thinking this.

[/ QUOTE ]
You may be right.

[ QUOTE ]
Since the numbers are listed from low to high, just choose one of the numbers (say the 3rd lowest) and let odd/even determine who wins.

[/ QUOTE ]
You seem to be under thinking this. You didn't bother to check whether your solution was valid.

This same technique would apply in variants with no solution, when the number of possible tickets is odd, such as if the numbers were 1-47 instead of 1-49. That doesn't mean it is wrong, but it should make you worry, particularly if you mean to correct my over thinking.

On 6990896 tickets (49.9928%), the third number is even.
On 6992920 tickets (50.0072%), the third number is odd.

Mathematica code:
<ul type="square"> Sum[Binomial[i, 2]If[EvenQ[i + 1], 1, x]Binomial[49 - i - 1, 3], {i, 2, 45}]

6990896 + 6992920 x[/list]