Calculating the Probability of Winning $25 in a Bag Game

[Yankel]

Hello all,

In a bag there are 18 paper notes. On five of them there is the digit 2, on seven the digit 3, and on six the digit 5. A man takes 3 notes by random. If the multiplicity of the notes is even, he wins 25 dollars. If for each game he pays 6 dollars, what is the average of profit he has after 90 games?

In order to solve, I think first I need the probability of getting an even multiplicity, which is tricky. Can you assist please ?

 

[Jameson]

Hello all,

In a bag there are 18 paper notes. On five of them there is the digit 2, on seven the digit 3, and on six the digit 5. A man takes 3 notes by random. If the multiplicity of the notes is even, he wins 25 dollars. If for each game he pays 6 dollars, what is the average of profit he has after 90 games?

In order to solve, I think first I need the probability of getting an even multiplicity, which is tricky. Can you assist please ?

There are $3^3=27$ possible outcomes, so you can do this by brute force.

Notice that when multiplying three values together, the result is even only if you have EEE or two odds and an even. So you need to find all of the combinations that match the pattern EEE, EOO, OEO, and OOE.

Another way to do this is to find all of the odd results after three draws, which is the complement event. Then we can use the fact that $P(E)=1-P(O)$. Either way will work. :)

 

[Yankel]

Thanks ! I see what you mean, but not sure how to count the options. Should I take any care of the order ?

 

[Jameson]

Thanks ! I see what you mean, but not sure how to count the options. Should I take any care of the order ?

Yes, we need to account for that.

The only EEE option is 222.

For EOO there could be 233,235,255, 253.

Can you write out the other lists for the remaining patterns?

 

[Yankel]

I made a tree diagram and counted. I found 20 instances, that means that P(Even) = 0.74.

Then I said that X is the number of wins in 90 games: X~Bin(90,0.74), and so E(X)=66.6

And Profit = 25X-6*90 -> E(Profit) = 25*66.6-6*90=1125

Am I even near the correct answer ?

 

[Jameson]

I made a tree diagram and counted. I found 20 instances, that means that P(Even) = 0.74.

Then I said that X is the number of wins in 90 games: X~Bin(90,0.74), and so E(X)=66.6

And Profit = 25X-6*90 -> E(Profit) = 25*66.6-6*90=1125

Am I even near the correct answer ?

I haven't worked out the problem so can't confirm without seeing your work.

Once you find $E[X]$ for 1 game, I would use the fact that $E[90X]=90E[X]$ to find the final answer. I wouldn't treat it as a binomial because we don't know the number of successes or failures. That formula is for finding probabilities when you have a specific success/failure count.

I found 222 for EEE , (233, 235, 255, 253) for EOO. Adding in the other orders you also have (323,325,523,525) for OEO, (332,352,532,552) for OOE.

Is there one I'm missing that you found before we move on to the probabilities?

 

[Yankel]

You are correct, I checked again and it's 19 options:

I found:

222, 223, 225, 232, 233, 235, 252, 253, 255, 322, 323, 325, 332, 352, 522, 523, 525, 532, 552

So the probability of an even multiplicity is 19/27=0.704.

Now let's move to the answer, I am not sure I agree with your approach, since you ignore the fact that in each game you pay 6$ to participate. If he plays 90 games, independent games, than the number of wins should be Binomial. And then we know E[X].

The profit is then 25X-540, and E[Profit] should be 25*63.36-540=1044.

What am I doing wrong here ?

 

[MarkFL]

I get a different probability for an even multiplicity. Let $E$ denote an even multiplicity and $O$ an odd multiplicity. Hence:

\(\displaystyle P(E)=1-P(O)=1-\frac{13\cdot12\cdot11}{18\cdot17\cdot16}=1-\frac{13\cdot11}{3\cdot17\cdot8}=1-\frac{143}{408}=\frac{265}{408}\)

Thus, his expected profit after 90 games would be:

\(\displaystyle 90\left(25\cdot\frac{265}{408}-6\right)=\frac{62655}{68}\)

 

[Jameson]

You are correct, I checked again and it's 19 options:

I found:

222, 223, 225, 232, 233, 235, 252, 253, 255, 322, 323, 325, 332, 352, 522, 523, 525, 532, 552

So the probability of an even multiplicity is 19/27=0.704.

Now let's move to the answer, I am not sure I agree with your approach, since you ignore the fact that in each game you pay 6$ to participate. If he plays 90 games, independent games, than the number of wins should be Binomial. And then we know E[X].

The profit is then 25X-540, and E[Profit] should be 25*63.36-540=1044.

What am I doing wrong here ?

Yes these are all the possible outcomes, but they are not equally likely. You have to find the probability of each one and add. A bit tedious I know.

For example \(\displaystyle P[222]=\frac{5}{18} \frac{4}{17} \frac{3}{16}\)

We do the same for the rest. To speed this up, we could look at all of the odd outcomes and then subtract the probability from 1, since the list of shorter. 19 even outcomes means 8 odd outcomes. (333,555,335,353,533,553,535,355).

Either way, once we find this probability we move onto $E[X]$. I didn't say we won't use the \$6 we pay to play of the \$25 prize. We just weren't at that step.

$E[X]=25\cdot P[\text{win}]-6\cdot P[\text{loss}]$, as Mark noted.