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

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In summary: In this example, if AAA has 4 tosses, he has a 1/8 chance of winning, while if BB has 3 tosses, he has a 1/4 chance of winning.
  • #1
philipSun
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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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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
 
  • #5
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!
 
  • #6
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?
 
  • #7
scalpmaster said:
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.
 

What is the probability of getting heads when tossing a coin?

The probability of getting heads when tossing a coin is 50%, assuming the coin is fair and has an equal chance of landing on either side.

What is the sample space for tossing a coin?

The sample space for tossing a coin is {heads, tails}. This means that there are two possible outcomes for a single toss of a coin.

How many times should I toss a coin to get an accurate representation of the probability?

The number of times you should toss a coin to get an accurate representation of the probability depends on the level of accuracy you want. Generally, the more times you toss the coin, the closer your results will be to the expected probability of 50%. However, even with a small number of tosses, the results should still be relatively close to 50%.

What is the difference between theoretical probability and experimental probability?

Theoretical probability is the probability of an event occurring based on mathematical calculations and assumptions. Experimental probability, on the other hand, is the probability of an event occurring based on actual experiments or trials. Theoretical probability is often used to predict outcomes, while experimental probability is used to analyze and validate the results of an experiment.

Can the probability of getting heads change with each toss of a coin?

No, the probability of getting heads does not change with each toss of a coin. Each toss is an independent event and the probability remains constant at 50% for a fair coin.

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