What is the probability of AA winning if they toss a coin more times than BB?

[philipSun]

Hello everybody. I have a problem, which is following.

AA tosses a coin 3 times and BB 2 times. AA will win, if he gets more heads than BB.
What is the probability that AA wins? Total probability is probably needed in this.

My solution:

first, this formula,
P(b) = P(a) P(b | a ) + P(a^c) P(b | a^c)

and events are;

a = happens, that there will be heads
b = happens, that there will be tails
a^c = a won't happen


In best case for AA;

a = 1-3 * a
b = 0-2 * a

a wins

In best case for BB;

b = 1-2 * a
a = 0-1 * a

now, a won't win.


P(b) = P(a) P(b | a ) + P(a^c) P(b | a^c)

P(2) = P(3) P(2 | 3 ) + P(0) P(2 | 0)

P(2) = P(3) P(2 | 3 ) + P(0) P(2 | 0)


This is as fas as I can go.

 

[daveb]

In this case you're interested in the number of heads AA gets compared to the number of heads BB gets. Just remember that if BB gets 0 heads, AA needs to get at least 1 head, and if BB gets 1, AA will have to get 2 or more, and if BB gets 2 heads, AA will have to get all 3 heads.

 

[philipSun]

Hi. Yes, that's what I meant with;

a = 1-3 * a
b = 0-2 * a

a wins

In best case for BB;

b = 1-2 * a
a = 0-1 * a.


But maybe you meant more than that.

And I want to do this in this method; P(b) = P(a) P(b | a ) + P(a^c) P(b | a^c) or something like that.
I believe that conditional probability is an issue here.

and my new events are;

a = happens, that AA gets heads
b = happens, that BB gets heads
a^c = a won't happen


Well, I think these are important, but these may be wrong..

P(a) = P(a) P(b | a ) + P(a^c) P(b | a^c)

P(b) = P(b) P(a | b ) + P(b) P(a^c | b )

P(a^c) = P(a^c) P(a | b ) + P(b) P(a^c | b ) ?

Wait a minute, both AA and BB have a 0.5 probability to have heads!

And after that we must do some multiplication, I guess.

I can't solve this problem.

 

[mathman]

For AA, P(3)=1/8, P(2)=P(1)=3/8, p(0)=1/8
For BB, P(2)=1/4, P(1)=1/2, P(0)=1/4

For AA to win - needs more heads than BB (I assume BB wins in case of ties)

P(AA wins-no. heads in bracket) = 1/8[3 heads]+(3/8)(3/4)[2 heads]+(3/8)(1/4)[1 head]=1/2

 

[philipSun]

So probability that AA wins is

P(AA wins-no. heads in bracket) = 1/8[3 heads]+(3/8)(3/4)[2 heads]+(3/8)(1/4)[1 head]=1/2
and solution is this. This wasn't easy for me. I want to say; thank you very much!

 

[scalpmaster]

Does it mean that AA will always have equal chance 1/2 of winning no matter how many additional throws he gets more than BB?

 

[mathman]

Does it mean that AA will always have equal chance 1/2 of winning no matter how many additional throws he gets more than BB?

No. If AA has more than 3 tosses, while BB has 2, then AA will win more often.

Each combination needs to be worked out.