Not E.E., but a problem in logic. Which is topical for any engineer -
A village holds an election for mayor. There's a single voting booth, with a box which accumulates paper ballots. (those were the days) The voting slips are removed from the box, one by one, and tallied.
Al runs against Bob. Al receives A votes, Bob gets B; A + B = N Al take the prize.
What is the chance (probability) that, during the tally, at least one tie occurred?
Is there an even number of ballots or odd?
How many ballots, and who exactly is counting them?
Are they impartial?
Does Bob concede the election, or did he go in saying he can't lose unless Al cheats?
Inquiring minds wish to know!
John :-#)#
** If I read the problem correctly - the answer is not a number but a formula involving A, B,and N.
The formula has a range varying from 0 to 1.
0 if A or B is zero. 1 if A and B are the same...... Phil
The key idea is simple and elegant.
Each vote for Al is labeled 'a', each Bob vote is 'b'. If the first slip is b, there must occur a tie, as Al wins. Now assume the first slip is a. For every tie sequence beginning with b, there's a mirror sequence, via transformation: a <--> b Thus the total probability: 2*b/N QED
Now arises a theoretical question: is this insight the only path to the solution? It seems that way. I don't know if that's true, or whether it's provable.
It's not: eg you could reverse the order of the tie sequence instead of swappinng the votes. any bijective mapping between A and B starting shortest tie sequences of the same length will work,
reversing: bbabbaaa => aaabbabb vs translating: bbabbaaa => aabaabbb
The part about the intial B may well be a necessary
RichD wrote: ==========
** Badly worded - could mean anything.Drop the second comma, change "as" to "if" and it might make sense.
** Steps are missing and the formula is ambiguous." b " is a mark, not a number.
The problem stipulates that Al receives A votes and wins.
oops, caught me. Bob receives B votes. Prob. of tie: 2*B/N
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