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bonanova
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Moe and May went to the beach the other day.

While May was setting out food for the picnic, Moe collected a bagful of pretty stones.

When he returned, he spread them out on the cloth for Mae to see.

Take as many of them as you like, he said, and put them in your bag.

While you finish preparing the food I'll fill my bag with more stones.

When they returned home, they counted their stones.

What is the probability they had an even number of stones between them? [edit]

Edited by bonanova
clarify - total number of stones is even
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the odds are even. npi.
they are not even, slightly favoring odd [May's bag].

Moe's bag is 50-50 by assumption.

you knew the odd number of stones Moe initially filled his bag with,

the 50% of the time his bag initially had an odd number of stones.

That's the only thing the odds depend on.

What if May doesn't like any of the stones? Wouldn't even numbers now be favored?

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What if May doesn't like any of the stones? Wouldn't even numbers now be favored?

Odd+Zero = Odd

Even+Zero = Even

Since Moe's bag has a 50/50 chance of being odd or even, May taking 0 would result in a 50/50 split, not favor even.

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Actually, bonanova, d3k3 is right in a sense.

If May picks at LEAST ONE stone (not zero) the odds favor odd.

If May could possibly pick zero, the odds favor even.

Reasoning:

Zero acts as an even number. When added to another number the resulting odd/even is unchanged. This is the same behavior as an even number. Therefore:

If May can pick zero:

Odd: 0, 1, 2, 3 : 50/50

Even: 0, 1, 2: 33/66 FAVORS EVEN

If May can not pick zero: (Our original answer)

Odd: 1, 2, 3: 33/66 FAVORS ODD

Even: 1, 2, 3, 4: 50/50

So, please clarify whether or not May can choose to take zero stones, as it changes the answer. If she can choose zero stones, the odds favor even. If she can not choose zero stones, the odds favor odd.

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Odd+Zero = Odd

Even+Zero = Even

Since Moe's bag has a 50/50 chance of being odd or even, May taking 0 would result in a 50/50 split, not favor even.

I mean if she has the option of not choosing any...

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