# Probability problem from Pokemon cards

1. Apr 25, 2012

### Bipolarity

A long time ago when I played with Pokemon cards, I remember a Geodude card saying "Flip a coin until you get tails. This attack does 20 damage times the number of heads."

What would be the probability that the attack does more than 60 damage?

What would be the average damage of the attack if the attack were repeated indefinitely and its damage measured for each trial.

The problem popped into my mind as I am reading on probability, but any thoughts on how to solve this problem? I think of geometric probability distributions, but can't quite establish a solution.

BiP

2. Apr 26, 2012

### tiny-tim

Hi Bipolarity!

It's probably one of those ∑ xn = 1/(1-x) manoeuvres.

Show us how far you've got.

3. Apr 26, 2012

### Bipolarity

I understand that it is a geometric sum of some sort, but I'm not so clear about the connection between geometric progressions and probability.

Would it be correct in saying that the probability that the attack does more than 60 damage is sum of the probabilities that the attack does damages corresponding to 80,100,120,140... ad infinitum?

That I think I get, but beyond that, how would you calculate the individual probabilities?

And I have no clue how to calculate the average damage. I assume you would take each value, multiply it by its probability to give it a weight, and sum up all such weights, but I am used to doing this in continuous variables using calculus. I don't know how you find infinite sums for these types of problems.

BiP

4. Apr 26, 2012

### tiny-tim

yes, but easier is to say that it's 1 minus the probability of 20 40 or 60
P(20k) = P(k-1 tails then 1 heads)
Yes, ∑ 20kP(20k) …

try it and see

5. Apr 26, 2012

### HallsofIvy

It will do 0 damage if you flip tails on the first flip- the probability of that is 1/2.
It will do 20 damage if you flip tails and then heads- the probability of that is (1/2)(1/2)= 1/4.
It will do 40 damage if you flip tails twice and then heads- the probability of that (1/2)2(1/2)= 1/8.
It will do 60 damage if you flip tails three times and then heads- the probability of that is (1/2)3(1/2)= 1/16.

Now, what is the probability it will do damage of "60 or less"? What is the probability it will do damage of "more than 60"? Do you see what that has to do with a geometric series?

The damage done when rolling n consecutive tails and then a head is, of course, 20n. The probability of rolling n consecutive tails and then a head is, as above, (1/2)n+1. So the average damage (the expected value) is the sum $\sum_{n=1}^\infty n (1/2)^{n+ 1}$. We can factor $1/4= (1/2)^2$ out of that to get $(1/4)\sum_{n= 1}^\infty n(1/2)^{n-1}$

The reason I do that is that $n x^{n-1}$ is the derivative of $x^n$. Further, power series are "term by term" differentiable so that our sum is 1/4 times the derivative of the $\sum x^n= \frac{1}{1-x}$, evaluated at x= 1/2.

Last edited by a moderator: Apr 26, 2012
6. Apr 26, 2012

### Bipolarity

Thank you all. I understand it now.

BiP