# »  Solutions to puzzles in my VDARE.com monthly Diary

## November 2020

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In the Math Corner of my November Diary I offered the following brainteaser, borrowed from the November 1st "Mind-Bender for the Quarantined" from Dr. Peter Winkler at the National Museum of Mathematics.

A box contains nine billiard balls numbered 1 through 9. You repeat the following ten times: reach blindly into the box, pick out a random ball, note its number, and throw the ball back in the box.

What is the probability that the sum of the numbers of the balls you picked is even?

Hmm.

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• My solution

We have ten numbers here, each of which is even with probability 4/9, odd with probability 5/9. The word "binomial" leaps to mind.

The question:

What is the probability of an even sum?

Is equivalent to:

What is the probability that ten, eight, six, four, two, or none of the ten numbers are odd?

To get that probability, just expand (5/9 + 4/9)^10 and add up the even terms. That will give you

(510 + 45*58*42 + 210*56*44 + 210*54*46 + 45*52*48 + 410) / 910

(The 45 and the 210 come from the tenth row of Pascal's Triangle, counting the apex as the zeroth row of course.)

The numerator there is 1,743,392,201. The denominator is 3,486,784,401. Divide former by latter, you get 0.500000000143398599539622… So the odds of the sum being even is a teeny tad more than fifty-fifty.

However, Dr. Winkler gets the same result a different way.

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• Dr. Winkler's solution

If there were no ball #9 available in the last round, an even sum would have probability exactly 1/2 since no matter what happens in your first 9 picks, the parity of the tenth ball will determine whether the sum is even or odd and that ball is equally likely to be either.

Now put ball #9 back in. No matter what order you choose the balls in, you can ask your assistant to reveal the numbers to you with all the 9's first. Then the above argument still applies: the parity of the last ball shown you, whose number is between 1 and 8, will determine the parity of the sum; so again, the sum is equally likely to be even or odd.

Oops, wait, there's just one possibility we haven't considered: that you got the 9 ball every time! That has probability (1/9)10 and results in an even sum. So altogether, the probability of an even sum is 1/2 x (1 - (1/9)10) + (1/9)10  = 1/2 + 1/2 x (1/9)10 which is about 0.500000014. It's probably not enough bias to be worth betting on!