Lewis Carroll is of course well known as the pen name for Charles Dodgson. And he is of course well known for having written two Alice books.

But he was a also a mathematician and devised many clever puzzles that involved math and deductive reasoning and probability. Here is one.

A bag contains one counter, known to be either white or black. A white counter is put in, the bag shaken, and a counter drawn out, which proves to be white. What is now the chance of drawing a white counter?

Okay let’s work on this. At the beginning of the experiment, we had a bag and so far as we knew, there was a fifty-fifty chance that when we draw the counter out, it might be white.

But now we have extracted some information from the bag, because we have added our own white counter to the bag. And then we pulled out a counter and it was white.

What can we say now about the likely result if we now draw the remaining counter out of the bag? Is it still 50-50 that what we draw out will turn out to be white? Or maybe now the odds are otherwise?

Surely you can say! Please post a comment below.

Urgh. These types of puzzles make me uncomfortable – probably because my intuition doesn’t match with reality.

My intuition would say that because we’re simply left with two counters again, the chance is 50:50.

But I think that reality must, as you suggest, take into account the additional information. Rather than viewing the situation as one choice out of two possibilities, we should view it as one choice out of the following three possibilities:

One black and one white, of which we picked the white (black remaining in bag, so next pick is black)

Two whites, of which we picked one (one white remaining in bag, so next pick is white)

Two whites, of which we picked the other one (again, one white remaining in bag, so next pick is white)

Which would mean that the probability of the next pick being white is actually 2 in 3. Horrible. Really horrible. I like my intuition. I don’t like reality challenging it 🙂

Thank you for commenting. Yes, you are right that one’s intuition goes one way, and the step-by-step analysis goes another way. And that this is unsettling for many. And yes you got the right answer.

There are the only two possible permutations. Adding the probabilities together, 25% + 50% = 75%.

More specifically:

There is a 50% chance we drew the newly-added counter. In that case, there’s still a 50% chance the original counter is actually white. So 50% x 50% = 25%.

There is a 50% chance we drew the original counter and it was already white. In this case there’s a 100% chance the remaining (newly-added) counter is white. So 50% x 100% = 50%.

Thank you for commenting. Your reasoning is sound and the number you get is correct.

I’d say the odds are the same as initially, 50:50, to pull a white counter.

Thank you for commenting. I think lots and lots of people have the same feeling about it that you have.

My answer is non-intuitive but I think that the probability is 66.66% that the remaining counter in the bag is white (i.e. the second drawn counter will be white). I drew out a probability tree that starts out with two branches for the two possibilities – the counter is Black or White. Then added a white counter to each and then created the possibilities of what is drawn:

B / W – Originally in Bag

BW / WW – Add White Counter

B / W / W / W – Drawn counter

W / B / W / W – Remaining counter in bag

But we know we drew a white counter so we can eliminate the case were a black counter was drawn, leaving

B / W / W as the remaining possibilities of the counter in the bag.

So there is 2 out of 3 chance that the remaining counter is white. (~67%)

Thank you for commenting. You are correct both in your prediction of the outcome, and in your methodology.

I think Bruce Young’s analytical solution is the correct one. I wrote a little simulation, the results of which follow (10 runs of 10 million shots).

tot: 7499258 blk: 2500983 wht: 4998275 fb: 0.3335 fw: 0.6665

tot: 7499250 blk: 2499660 wht: 4999590 fb: 0.3333 fw: 0.6667

tot: 7497526 blk: 2498891 wht: 4998635 fb: 0.3333 fw: 0.6667

tot: 7501022 blk: 2503364 wht: 4997658 fb: 0.3337 fw: 0.6663

tot: 7500357 blk: 2498968 wht: 5001389 fb: 0.3332 fw: 0.6668

tot: 7499978 blk: 2500542 wht: 4999436 fb: 0.3334 fw: 0.6666

tot: 7500661 blk: 2499532 wht: 5001129 fb: 0.3332 fw: 0.6668

tot: 7502712 blk: 2501651 wht: 5001061 fb: 0.3334 fw: 0.6666

tot: 7500325 blk: 2498917 wht: 5001408 fb: 0.3332 fw: 0.6668

tot: 7501282 blk: 2498249 wht: 5003033 fb: 0.3330 fw: 0.6670

I love it that you did a simulation. Thank you for commenting.

Love the simulation!

There are three ways to draw a white counter, two of which leave a white counter behind.

So the odds are 2/3.

I entered this problem into ChatGPT and it output the following:

“When I form a solution to a puzzle,’ Humpty Dumpty said in rather a scornful tone, ‘it means just what I choose it to mean — neither more nor less.’

’The question is,’ said Alice, ‘whether you can make solutions mean so many different things.’

’The question is,’ said Humpty Dumpty, ‘which is to be master — that’s all.”

A: second draw is white

B: first draw is white

P(B) = 0.5 * 1 + 0.5 * 0.5 = 0.75 (if the later-added count is white, 100% chance the first draw is white; if the later-added count is black, 50% chance that the first draw is white)

P(A∩B) = 0.5 (the probability that both draws are white – that is, the probability that the initial counter already in the bag is white)

P(A|B) = P(A∩B) / P(B) = 0.5 / 0.75 = 2/3