Gibson's Coin Paradox:(adsbygoogle = window.adsbygoogle || []).push({});

You have 2 coins with you. There is absolutely no way that you can tell the difference between them. The coins are fair, meaning the chances of getting heads is equal to the chance of getting tails so a 50% vs. 50% chance.

You put on a blindfold and drop the coins, one in each hand, into 2 separate buckets.

What is the chance that there will be at least 1 tail in your results?

The outcomes are the following:

Bucket 1: Bucket 2:

H H

H T *

T H *

T T *

The bottom 3 results satisfy the rule of there being at least one tail, there are 4 possibilities, therefore it is a 3/4 chance that there will be at least 1 tail.

However*:

You now conduct a similar experiment using only one bucket and the same coins. Now you are going to drop the coins into the same bucket using the same method as before.

The outcomes are the following:

Bucket 1:

H and H

Opposite *

T and T *

Now because the coins are indistinguishable order does not matter in the results. This combines 2 of the values into one making there only be 3 possibilities. Now 2 of the results satisfy the rule of there being at least 1 tail, there are 3 outcomes, therefore it is a 2/3 chance that there will be at least 1 tail.

Can we agree that the probabilities of these should be the same? And that's where the paradox lies, because the probabilities are not the same.

If there is any way to distinguish the coins your are more likely (3/4) to satisfy the "at least one tail rule" than if they coins are not distinguishable (2/3).

Thoughts, opinions, and comments ?

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# Did I just create a coin tossing paradox?

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