## Brainteaser 1657: Not a black and white case

**From The Sunday Times, 12th June 1994** [link]

Marvo has been up to his tricks again. He shuffled a normal pack of 52 cards and then looked at each one in turn without letting Alice see them. In each case she had to predict whether the card was red or black and Marvo gave no indication of whether Alice’s guesses were right or wrong. But after 51 guesses Alice had guessed “black” 26 times and “red” 25 times and Marvo announced that Alice had exactly 10 predictions wrong so far.

(1) Which colour (if either) is more likely for Alice’s last card?

Marvo repeated the game with Bob. After 51 guesses Bob too had guessed “black” 26 times and “red” 25 times and Marvo announced that Bob had exactly 11 predictions wrong so far.

(2) Which colour (if either) is more likely for Bob’s last card?

Marvo then repeated the game once more with Carol. After 51 guesses Carol too had guessed “black” 26 times and “red” 25 times and Marvo announced that Carol had got

at least10 predictions wrong so far.(3) Which colour (if either) is more likely for Carol’s last card?

This puzzle was included in the book *Brainteasers* (2002, edited by Victor Bryant). The puzzle text was, and the answers required were changed slightly, but the idea of the puzzle remains the same.

[teaser1657]

## Jim Randell 10:04 am

on15 August 2019 Permalink |Alice has guessed “black” 26 times and “red” 25 times.

If she has got

nguesses wrong, then we can supposebof the “black” guesses were wrong (i.e. they were actually red cards), and so(n – b)of the “red” guesses were wrong (i.e. they were actually black cards).So at this point the total number of actual black cards is:

And similarly for red, giving:

There can’t be more than 26 actual black cards, so:

And there can’t be more than 26 actual red cards:

So:

Which fixes the value of

b, depending on whethernis odd or even.In Alice’s case

n = 10, sob = 5. Hence there have actually been 26 black cards and 25 red cards, so the remaining card has to be red.In Bob’s case

n = 11, sob = 6. And there have actually been 25 black cards and 26 red cards, so the remaining card has to be black.And we see that if

nis even then the remaining card must be red, and if it is odd the remaining card must be black.Solution:(1) Alice’s final card is certainly red. (2) Bob’s final card is certainly black.Carol has made

at least10 incorrect predictions. So she has made 10, 11, 12, …, 51, and depending on the exact number the remaining card is red, black, red, … black. There are 24 cases where the card must be red and 24 cases where the card must be black.We need to count the number of scenarios for each outcome.

When

n = 10thenb = 5and there have been 26 black guesses (5 are guessed incorrectly) and 25 red guesses (5 are guessed incorrectly). So the number of ways of choosing the incorrect guesses are:each one giving rise to the remaining card being red.

When

n = 11andb = 6we get:each one giving rise to the remaining card being black.

(I think if we are counting the total number of possible scenarios each of these numbers would be multiplied by an additional factor to account for the order of the cards and the guesses, but we don’t need to do that to compare the relative likelihood of red and black outcomes for the final card).

This Python program counts up the number of possibilities. It runs in 85ms.

Run:[ @repl.it ]Solution:(3) It is slightly more likely that Carol’s final card is red.The calculated figures give that Carol’s final card is red 50.0001% of the time and black 49.9999% of the time. So it is close.

The puzzle as originally published in

The Sunday Timesasked for the outcome with the following numbers of wrong predictions: (a) exactly 10; (b) least 10; (c) at least 15.We have already covered (a) and (b), and for (c) the program can easily be modified to start counting from 15, rather than 10. The result is that it is slightly more likely that the card is black (50.02%) than red (49.98%).

LikeLike