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.

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

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

/images/graemlins/diamond.gif 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]

[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. /images/graemlins/smile.gif

[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 &lt;= 25 is much larger than that which include no balls &lt;= 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. /images/graemlins/smile.gif

[/ 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? /images/graemlins/grin.gif

-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.

[BradleyT]

Take the published results of winners. Either add them all up and do odd/even or take the match 3 out of 6 winners and do odd/even.

[Cobra]

Would this work?

If the bonus ball is 1-24 A wins.

If the bonus ball is 26-49 B wins.

If the bonus ball is 25 then there are an even number of odd and even balls remaining therefore if the first ball is odd A wins if it is even B wins.

[TomCollins]

[ QUOTE ]
Would this work?

If the bonus ball is 1-24 A wins.

If the bonus ball is 26-49 B wins.

If the bonus ball is 25 then there are an even number of odd and even balls remaining therefore if the first ball is odd A wins if it is even B wins.

[/ QUOTE ]

I think you are close.

However, 1 is the lowest ball more often than 2 is. Do you see why?
3 is also the lowest ball more often than 4 is. A has a small advantage in this case.

[bobbyi]

I think this is a fascinating problem. Just to make sure I have the details right: are all seven numbers drawn necessarily distinct? Do they put the first six back in before drawing the bonus number?

[jakka]

[ QUOTE ]
I think this is a fascinating problem. Just to make sure I have the details right: are all seven numbers drawn necessarily distinct? Do they put the first six back in before drawing the bonus number?

[/ QUOTE ]

Yes all seven numbers are distinct. They do not put the first six back in before drawing the bonus number.

[TALLBrad]

flip a loon......

[housenuts]

[ QUOTE ]
Would this work?

If the bonus ball is 1-24 A wins.

If the bonus ball is 26-49 B wins.

good until here...
If the bonus ball is 25 then there are an even number of odd and even balls remaining therefore if the first ball is odd A wins if it is even B wins.

[/ QUOTE ]

you got me on the right track. however if the bonus ball is 25 then add up the remaining balls. there is an equal # of even and odd balls now. if the number is even Player 1 wins, if it's odd Player 2 wins.

[pzhon]

[ QUOTE ]
if the bonus ball is 25 then add up the remaining balls. there is an equal # of even and odd balls now. if the number is even Player 1 wins, if it's odd Player 2 wins.

[/ QUOTE ]
Sorry, that doesn't work. Among the 48C6 = 12,271,512 ways the bonus ball could be 25, only (24C6+(24C4)(24C2)+(24C2)(24C4)+24C6)= 6,134,744 = 49.9918% have an even sum, while 6,136,768 = 50.0082% have an odd sum.

You shouldn't expect a method to work if it would apply to an impossible case such as if the numbers were 1-47, so that the number of tickets would be odd.

[HesseJam]

The median is 25. The EV of all 6 numbers being added up is 150.
Bro's get to choose if the average of the actually drawn 6 No. is above or lower 25. Mmmh, a tie is possible.

If tied, the seventh No could well be the 25, so there is still a tie...

Nope, this doesn't work either.

[Pokerscott]

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

[/ QUOTE ]

They agree to call each other as soon as they find out the results. First person to call the other with the accurate results wins.

Since we know nothing about the probability of each one finding out the numbers first, by symmetry they have an equal chance and we have used the results to break the tie.

/images/graemlins/grin.gif

pokerscott