# Game Theory Question

#### BennyH26

##### New Member
Hello gurus,

I got involved in a debate with my sister this evening, and despite (what I thought) was a relatively simple problem, I couldn't provide a rigorous answer. I was hoping that someone on here could give me some insight.

A casino she frequents will often give her "fake money" to use at a spinning wheel. For the purpose of this question, let's assume the wheel is a fair one and it has 10 spaces numbered 1-10. If she puts her "fake money" on any slot, and it comes up on the spin, she gets REAL money for it.

Now, let's assume she has 10 "fake" dollar bills. If she puts a single dollar on EVERY space, she is guaranteed to win $1 in real money. However, if she puts$1 on the *SAME* space for 10 spins, she is NOT guaranteed to win even though the probability of winning on each spin is 1/10.

If we spin 10 times, and the probability of a win is 1/10, shouldn't:

1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + = 1

I'm confused as to why mathematically these "independent" events don't add up to certainty the way that placing bets on EVERY space does.

I know this is a trivial question to many of you, but it has been driving me nuts.

Thanks so much,
Ben

#### Dason

Because probability doesn't work like that.

We first need to consider the actual event we care about. In your situation it is the probability of winning at least 1 time. It's actually easier to figure out the probability of not winning at least 1 time (because this is the probability of losing every time) and since {winning at least 1 time} and {not winning at all} are a partition of the sample space we know their probabilities sum to 1 so that P(winning at least 1 time) = 1 - P(not winning at all).

Now P(not winning at all) = P(lost first game AND lost second game AND ... AND lost tenth game). These are independent events so this is the same as P(lost first game)*P(lost second game)* ... *P(lost tenth game). Since the probability is the same for all of these (9/10) this is just (9/10)*(9/10)*...*(9/10) = (9/10)^10

So the probability of losing all 10 games is (9/10)^10 = 0.3486784.

So the probability of winning at least one of the games = 1 - 0.3486784 = 0.6513216

#### BennyH26

##### New Member
You rock. That is crystal clear, thanks so much for explaining it to me.

All the best,
Ben

Because probability doesn't work like that.

We first need to consider the actual event we care about. In your situation it is the probability of winning at least 1 time. It's actually easier to figure out the probability of not winning at least 1 time (because this is the probability of losing every time) and since {winning at least 1 time} and {not winning at all} are a partition of the sample space we know their probabilities sum to 1 so that P(winning at least 1 time) = 1 - P(not winning at all).

Now P(not winning at all) = P(lost first game AND lost second game AND ... AND lost tenth game). These are independent events so this is the same as P(lost first game)*P(lost second game)* ... *P(lost tenth game). Since the probability is the same for all of these (9/10) this is just (9/10)*(9/10)*...*(9/10) = (9/10)^10

So the probability of losing all 10 games is (9/10)^10 = 0.3486784.

So the probability of winning at least one of the games = 1 - 0.3486784 = 0.6513216