Problem 8:
A decision making problem will be resolved by tossing $2n + 1$ coins. If Head comes in majority one option will be taken, for majority of tails it’ll be the other one. Initially all the coins were fair. A witty mathematician replaced $n$ pairs of fair coins with $n$ pairs of biased coins, but in each pair the probability of obtaining head in one is the same the probability of obtaining tail in the other. Will this cause any favor for any of the options available? Justify with logic.
BdMO National 2012: Higher Secondary 08
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
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abdullah al munim
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Re: BdMO National 2012: Higher Secondary 08
Yes by symmetry.
I also explained the problem this way during the olympiad.i am not sure whether i was given full mark or not(as the solution appears like literature than math).but i am pretty confident that this completes the solution
I also explained the problem this way during the olympiad.i am not sure whether i was given full mark or not(as the solution appears like literature than math).but i am pretty confident that this completes the solution
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abdullah al munim
- Posts:6
- Joined:Wed Jan 18, 2012 9:31 pm
Re: BdMO National 2012: Higher Secondary 08
Oh, sorry! I wanted to say NO
Re: BdMO National 2012: Higher Secondary 08
amar ans ascilo no change. sahitto likhecilam math er bodole
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nafistiham
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Re: BdMO National 2012: Higher Secondary 08
If there were only fair coins, the probability would be $\frac {1}{2}$
as $n$ pairs are changed, the last coin makes the decision. So, as it is fair, the probability will not change.
as $n$ pairs are changed, the last coin makes the decision. So, as it is fair, the probability will not change.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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