## BdMO National 2012: Higher Secondary 08

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
Moon
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### BdMO National 2012: Higher Secondary 08

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.
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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

abdullah al munim
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### Re: BdMO National 2012: Higher Secondary 08

Oh, sorry! I wanted to say NO

turash
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### Re: BdMO National 2012: Higher Secondary 08

amar ans ascilo no change. sahitto likhecilam math er bodole

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.
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
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