Recollecting Forgotten Digit


Recollecting Forgotten Digit, problem

Solution 1

The problem is a variant of drawing balls of two colors, where, say $w=1,$ $b=0,$ and $w+b=10.$ $\displaystyle P(W_1)=P(W_2)=P(W_3)=\frac{1}{10}.$

The (first) event we are looking for is $W=W_1\cup W_2$ which comes with the probability $\displaystyle P(W)=\frac{1}{10}+\frac{1}{10}=\frac{1}{5}.$

For three attempts the probability is $\displaystyle \frac{1}{10}+ \frac{1}{10}+ \frac{1}{10}= \frac{3}{10}.$

Solution 2

The probability of a correct guess is $\displaystyle p=\frac{1}{10}.$ Two attempts will suffice if we guess the first time, or, if we fail, we guess correctly on a second attempt. This gives the probability as

$\displaystyle \frac{1}{10}+\left(1-\frac{1}{10}\right)\cdot\frac{1}{9}=\frac{2}{10}.$

For three guesses

$\displaystyle\begin{align} P(Success)&=\frac{1}{10}+\left(1-\frac{1}{10}\right)\cdot\frac{1}{9}+\left(1-\frac{1}{10}\right)\left(1-\frac{1}{9}\right)\cdot\frac{1}{8}\\ &=\frac{1}{10}+\frac{9}{10}\cdot\frac{1}{9}+\frac{9}{10}\cdot\frac{8}{9}\cdot\frac{1}{8}\\ &=\frac{3}{10}. \end{align}$

Solution 3

Imagine that we try two times, regardless of whether the first guess was correct. Then there are $10\times 9$ (ordered pairs of numbers that could be "recollected" with equal chances. Of these, $9$ include the right guess on first attempt and $9$ on the second (the right second guess and nine first failures). In all, there are $9+9=18$ pairs that realize a successful recollection. The probability of this event is $\displaystyle \frac{18}{90}=\frac{1}{5}.$

For the second part, we make three guesses regardless of whether either the first or the second were successful. There are $10\times 9\times 8=720$ ordered triples to choose from. If the first guess was a hit, there are $9\times 8=72$ variants for the second and the third guesses. With the second correct guess, there are $9$ first guess failures times $8$ distinct but irrelevant third guesses, $9\times 8=72$ in all. The right third guess comes after $9\times 8=72$ failures in the first and second guesses, giving the probability of the success in three guesses as

$\displaystyle \frac{72+72+72}{720}=\frac{3}{10}.$


This is a problem from CChallenging Mathematical Problems With Elementary Solutions, Vol. 1, by A. M. Yaglom and I. M. Yaglom.

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