Question

Fun question! :)) (Probability)

Answer 1

Two guys "A" and "B" play a game in which they toss a coin alternatively. The winner is the one who gets heads first. "A" starts first in the first game. The loser of the game tosses first in the next game. What is the probability that A will win the sixth game?

Answer 2

wat happened?

Answer 3

the answer was 1/2 lol

Answer 4

Nopez that not correct

Answer 5

the question you posted first. :P

Answer 6

Then also it wouldn't be correct

Answer 7

O.o okay

Answer 8

1/64?

Answer 9

Nope

Answer 10

I am confused

Answer 11

It could be multiple answers couldn't it since theres different chances like if A won all 5 and since B would go first, or if B won 5 and A wen't on the sixth etc.

Answer 12

"What is the probability that A will win the sixth game?" B has the same possibility since their both the same...

Answer 13

Hint: the probability of winning a game is more for the person who starts the game :))

And yes there can be different cases like A wins the first 5 or maybe B wins the first 5,etc so you must consider all the possible cases when A wins the 6th game while calculating the probability

Answer 14

looks fun :) its very similiar to question of expected number of rolls to get 2 Heads in a row

Answer 15

k so i think the solution simplifies to this

Answer 16

there has to be 5 heads, and umm some number of tails even or odd dunno yet, and then a Head

Answer 17

u can first solve the case where A has a 1/2 chance of winning game 6, by solving for the chance
4 Heads and x number of tails, this one is either odd or even dunno yet, but i do know this has to be odd if the statement above is even and then 1 Head

Answer 18

umm scratch that this doesnt seem helpful

Answer 19

I'm gonna start playing with this looks fun

Answer 20

kay

Answer 21

If the probability of getting H or T on each flip is 1/2 than lets say:

1. First guy goes: H
2. Second guy goes T
3. Second guy goes: H
4. First guy goes: T
5. First guy goes: H
6. Second guy goes: T

Answer 22

maybe work like this
There is 0% chance A wins in 6 rolls

Answer 23

In a way, I'm tempted to say, forget everything and just calculate the amount of times it takes to get heads 5 times by flipping a single coin successively.

Answer 24

6 times*

Answer 25

yeah i am doing that too, i think we shud just ditch the A and B and think about the coin itself

Answer 26

Or...
1. First guy goes: T
2. First guy goes: H
3. Second guy goes: T
4. Second guy goes: H
5. First guy goes: T
6. First guy goes: H

Answer 27

but u keep getting infinite series of probabilities showing up in there

Answer 28

so im thinking there has to be some nicer simpification

Answer 29

XD i dont see him winning
not sure what the probability is though

Answer 30

@Kainui ... but the winner is the person who gets heads... :P

Answer 31

& the loser flips next

Answer 32

Yeah but it doesn't matter, all that matters is who's holding the coin after it's flipped heads 5 times.

Answer 33

ohh how about like this, if 6 games in N number of throws then
okay
6 Throws, B wins 100%
7 Throws. A wins 100%

Answer 34

all even number ending will be B and odd number ending A

Answer 35

Yeah now you're on the same page

Answer 36

Oh okay makes sense :P

Answer 37

and then we can write combinatorically the solutions

Answer 38

but what if someone flips tails twice in a row :O? @518nad

Answer 39

then the game has been prolonged to more turns lets start writing an expression for 6 games ending in N turns and we can figure out a pattern

Answer 40

just an observation, the first guy to toss has a 1/2 chance of winning, since the the second guy will not even get a chance to toss if the first toss is heads,

the probability of the first gut loosing is 1//2 *1/2 =1/4
(A has to toss tails AND B has to toss heads)

Answer 41

yup baru

Answer 42

I feel like there is an infinite number of ways that A will win idk.
@imqwerty
What's the answer XD?

Answer 43

nooooooooo dont tell us!!! i think i got ittttttttt

Answer 44

Haha we'll figure it out

Answer 45

u will get infinite serieis of prob but they converge

Answer 46

there definitely is an infinite number of ways to win, just keep flipping tails lol.

Answer 47

it's easy...waiting for a solution from anyone

Answer 48

i dont trust u zarkon, u probably alreayd coded it -.-

Answer 49

how about this for the total possible cases,

there are 6 tosses of coin pairs,

on each pair, the possibilities are: (H,-) (T,H), (T,T)
-> so three per game,

so total possibilities: 3^6

Answer 50

you don't need to code it. I have two ways I can do it...one using markov chains and another slicker way.

Answer 51

ooo! cool

Answer 52

is the 2nd way a recursive way

Answer 53

yes

Answer 54

@Zarkon ... can you explain haha i'm a little slow XD

Answer 55

How is it that there are infinite ways for A to win?

Answer 56

here is the probability of winning the nth game

\[P(n)=\frac{2}{3}\sum_{k=0}^{n-1}\left(\frac{(-1)^k}{3^k}\right)\]

Answer 57

They've flipped only 5 heads and it's someone's turn. They flip tails.
So you pass it to the next person.
They've flipped only 5 heads and it's someone's turn. They flip tails.

Answer 58

@baru
Nah.. I just want someone to post the answer lol

Answer 59

woops i missed that, I assumed game terminates if there is a tie

Answer 60

use the relation\[ P(n)=\frac{2}{3}-\frac{1}{3}P(n-1)\]

which you get from

\[P(n)=P(n-1)\frac{1}{3}+[1-P(n-1)]\frac{2}{3}\]

Answer 61

\[P(6)=\frac{364}{729}\]

Answer 62

49.9%?

Answer 63

... may as well be 50%

Answer 64

that is the limit

\[\lim_{n\to\infty}P(n)=\frac{2}{3}\frac{1}{1+\frac{1}3}=\frac{2}{3}\frac{1}{\frac{4}{3}}=\frac{2}{4}=\frac{1}{2}\]

Answer 65

oh :O well okkay thanks for blowing my mind @Zarkon

Answer 66

So.... OP posted a hint..

The probability of winning the game will always be more for person A.
Is that true @Zarkon

Answer 67

depends on what you mean ...P(2)=4/9=.4444444444

so the probability B wins game 2 is higher than A

Answer 68

also \(P(6)\approx .499<.5\)

so B has a better chance to win

Answer 69

P(n)>.5 if n odd
P(n)<.5 if n even

Answer 70

@Zarkon oh it's because you were a probability mechanics major . XD
& Okay that it interesting

Answer 71

In real life my undergraduate degree is in applied mathematics

Answer 72

Gotcha :P

Answer 73

My graduate degrees are not in applied math