Heads or Tails challange - Answer at end

Have someone lay out a hundred coins on table with 90 of them with the obverse facing up.
Without looking at them and with your eyes closed, can you separate all the coins into two groups so each group has the same number of reverses facing up?
Without looking at them and with your eyes closed, can you separate all the coins into two groups so each group has the same number of reverses facing up?
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Keeper of the VAM Catalog • Professional Coin Imaging • Prime Number Set • World Coins in Early America • British Trade Dollars • Variety Attribution
Lance.
The rest of the coins become group #2.
Open your eyes and you will see both groups have the same number of reverses facing up.
Why it works:
10% of the coins have the reverses up. If you move 10% into a group and turn them over, then 10% will be up in both groups.
Open your eyes and you will see both groups have the same number of reverses facing up.
Each group does not have the same number of ob / re showing, only if the groups are treated as a singularity.
TO, the process is right but the explanation is wrong. If there are n heads and 100-n tails, take a group of n coins. Let q denote the number of heads in this group. So we have q heads and n-q tails. Now flip them all. We have q tails and n-q heads. In the other group we also have n-q heads (since we started with n and removed q).
Are you allowed to talk like that in a public forum?
TO, the process is right but the explanation is wrong. If there are n heads and 100-n tails, take a group of n coins. Let q denote the number of heads in this group. So we have q heads and n-q tails. Now flip them all. We have q tails and n-q heads. In the other group we also have n-q heads (since we started with n and removed q).
Doctorate in statistics, or math ?
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The answer surprised even me.
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