Too high, too low

Probability Level 4

You play a game with a 4 sided die.

You roll it several times, and after each roll you try to predict whether the next roll will be higher or lower. If you predict wrong, you are out.

So, for example, if you roll a 2 and then say "Higher" and then roll another 2, you are out. (Unfortunately, there is no way to guard against getting the same number as the previous roll... You will always be out)

Assume that you play optimally, (to maximize the number of rolls) and let the expected number of times you will roll the die (including the final roll where you are out) be $\frac{a}{b}$ where $a$ and $b$ are coprime positive integers.

What is $a+b$ ?

Image credit: www.midlamminiatures.co.uk

The answer is 32.

1 solution

Geoff Pilling
Nov 29, 2016

Consider the following states:

$0$ : About to roll for the first time
$1$ : Last roll was a $1$ or a $4$
$2$ : Last roll was a $2$ or a $3$

$E_n =$ Expected number of rolls from state $n$

This gives us a set of linear equations to solve:

$E_0 = 1+ \frac{1}{2}E_1 + \frac{1}{2}E_2$
$E_1 = 1 + \frac{1}{4}E_1 + \frac{1}{2}E_2$
$E_2 = 1 + \frac{1}{4}E_1 + \frac{1}{4}E_2$

Solving, $E_0 = \frac{27}{5}$

$27+5 = \boxed{32}$

Too high, too low

Image credit: www.midlamminiatures.co.uk

The answer is 32.

1 solution

0 pending reports