# Probability of picking the red marble twice

Start with a bag containing one red marble and one white marble.

• Reach into the bag and pull out a marble at random.
• Put it back and do it again.
• If, on both times, you grabbed the red marble, stop.
• Otherwise add two white marbles to the bag and repeat these steps.

What is the probability you eventually grab the red marble twice during an iteration?

The probability

that you never draw the red marble twice in a row is the probability that you don't on the first iteration, times the probability that after that you don't on the second iteration, times the probability that after that you don't on the third iteration, and so forth.

That equals

$$\frac34\cdot\frac{15}{16}\cdot\frac{35}{36}\cdot\frac{63}{64}\cdots$$ or $$\frac{1\cdot3}{2\cdot2}\cdot\frac{3\cdot5}{4\cdot4}\cdot\frac{5\cdot7}{6\cdot6}\cdot\frac{7\cdot9}{8\cdot8}\cdots$$.

This is exactly

the reciprocal of the famous Wallis product for $$\pi$$, so the probability of never getting a red pair is $$\frac2\pi$$ and the probability of ever getting one is therefore $$1-\frac2\pi$$ or about 36.34%.

• Aye, there's my problem. I knew I was missing something. Commented Jun 3, 2020 at 13:38

The value is equivalent to an infinite series according to the function:

SUM(n=1,inf,(1/2n)^2)

Equivalent to:

(1/4)SUM(n=1,inf,1/n^2)

Which the latter returns:

pi^2/6, divided by 4 gives us pi^2/24, or roughly 41.1%. It definitely feels odd that in this case it is more likely that this procedure will continue infinitely than it is that it would eventually end. Very interesting problem!

• Good try, but the actual answer is even lower. Commented Jun 3, 2020 at 12:49
• What's the reasoning and the mistake in this answer? (Assuming the other answer is right, which I think it is) Commented Jun 4, 2020 at 2:44
• The answer didn’t account that by going with this approach, it’s an XOR problem not an OR problem. Commented Jun 4, 2020 at 14:29