Don't lose too fast

Probability Level 3

A game is played starting with 6 fair coins laid out with heads facing up.

Each round consists of flipping all of the coins showing heads.

If fewer than half of the flipped coins come up heads the player loses.

Rounds continue until the player either loses or gets to one head remaining.

The player wins by getting to one head without losing.

What is the probability of winning this game? The answer in lowest terms can be written as $\frac{a}{b}$ . Enter the value of $a+b$

Examples:

6→4→1 would be a loss. (1 is less than half of 4.)

6→5→3→2→2→2→1 would be a win.

1 solution

Jeremy Galvagni
Apr 21, 2018

This problem is begging to be worked backward. Let $P(x)=$ the probability of winning the game beginning with $x=$ heads.

Clearly, $P(1)=1$ since this is the winning position.

To win if $x=2$ , consider the outcomes of flipping two coins:

0 heads = lose, $p=\frac{1}{4}$

1 head = win, $p=\frac{2}{4}$

2 heads = inconsequential, $p=\frac{1}{4}$ , flip them again.

The relative probability is then

$P(2)=\frac{2}{3}$

Next x=3

0 heads = lose p= $\frac{1}{8}$

1 head = lose p= $\frac{3}{8}$

2 heads = chance to still win. p= $\frac{3}{8}$

3 heads, flip again. Thus

$P(3)=\frac{3}{7}*\frac{2}{3}=\frac{2}{7}$

x=4

0 heads = $\frac{1}{16}$ lose

1 heads = $\frac{4}{16}$ lose

2 heads = $\frac{6}{16}$ chance to continue with x=2

3 heads = $\frac{4}{16}$ chance to continue with x=3

4 heads = $\frac{1}{16}$ flip again

$P(4)= \frac{6}{15}*\frac{2}{3} + \frac{4}{15}*\frac{2}{7} = \frac{12}{35}$

x=5

must get 3 or 4 heads to continue

$P(5)= \frac{10}{31}*\frac{2}{7} + \frac{5}{31}*\frac{12}{35} = \frac{32}{217}$

x=6

must get 3, 4 or 5 heads to continue

$P(6) = \frac{20}{63}*\frac{2}{7} + \frac{15}{63}*\frac{12}{35} + \frac{6}{63}*\frac{32}{217} = \frac{52}{279}$

So the answer is $52+279=\boxed{331}$