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#2
01-24-2005, 03:57 PM
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Re: Using the Lottery to Break a Tie
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. 
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#3
01-24-2005, 07:01 PM
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Re: Using the Lottery to Break a Tie
[ 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.
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#4
02-03-2005, 05:41 AM
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Re: Using the Lottery to Break a Tie
[ 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.
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#5
02-03-2005, 05:54 AM
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Re: Using the Lottery to Break a Tie
[ 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.
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#6
02-04-2005, 11:42 AM
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Re: Using the Lottery to Break a Tie
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.
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#7
01-27-2005, 07:46 PM
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Clarification request
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?
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#8
02-02-2005, 11:02 PM
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Re: Clarification request
[ 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.
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#10
02-05-2005, 04:20 AM
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Re: Using the Lottery to Break a Tie
[ 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. [img]/images/graemlins/grin.gif[/img] pokerscott
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