# Thread: Paper Notes in a Bag

1. ## Paper Notes in a Bag

Hello all,

I have this problem, which I am not fully sure how to solve:

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?

I have calculated the probability of getting an even multiplicity, and I got that it equals to 0.704 (I used a tree diagram for it and found 19 instances that gives this event out of 27 possible outcomes).
I am not sure how to proceed. I thought to define X as the number of wins out of 90 games, and then X is Binomial(90,0.704), which gives me E[X] = 63.36. Then I thought to define the Profit as 25X-540, and to dig E[Profit] from there. It gives me a final answer of 1044, which I am very unsure about. Am I even close to the solution?

Thank you !

2. ## Re: Paper Notes in a Bag

The expected winnings after 90 games will be 90 times the expected winnings per game because the games are independent of one another. The expectation per game is E = \$25×P(Win) – \$6.

I’m not sure what the bit about the “multiplicity of the notes” is supposed to mean. If it means that the product of the digit values of the three notes must be even then it follows that at least one of the three notes must have the digit 2 on it because the other possible digits are both odd. In this case, P(Win) = 1 – P(Lose) = 1 – (13/18)×(12/17)×(11/16) = 265/408 and E = \$4177/408 ≈ \$10.24 per game.

3. ## Re: Paper Notes in a Bag

Yes, the meaning is that the products should be even.

If you create a tree diagram, you get a different result for the probability to win. You have 27 options in this game, and in 19 of them you get an even product.

4. ## Re: Paper Notes in a Bag

It seems to me that you have a tendency to make problems much more complicated than they need to be. Rather than thinking about notes being marked “2”, “3” or “5”, think about them being marked with an even digit “e” (i.e., notes marked “2”) or an odd digit “o” (i.e., notes marked “3” or “5”). In this scenario, there are five “e” and 13 “o” notes, and there are eight possible outcomes from three draws: eee, eeo, eoe, eoo, oee, oeo, ooe, and ooo.

It should not be difficult to understand that for a product of three terms to be even, at least one of the terms must be even. Another way of saying this is that the product is odd only if all three terms are odd. The only draw scenario that yields an odd product is ooo, while the other seven yield an even product.

What is the probability of ooo occurring, P(ooo)? Again, it should not be difficult to see that P(ooo) = P(Lose) = (13/18)×(12/17)×(11/16), as stated. Why? Because you start with 13 out of 18 “o” notes. If the first note drawn is an “o”, you’re left with 12 “o” notes out of 17. If the first and second notes are both “o”, you’re left with 11 “o” notes out of 16.

Therefore, your tree diagram is either incomplete or wrong.

Now go away and think about it.

Go away ???

6. ## Re: Paper Notes in a Bag

Yes, because it’s entirely clear to me that you’re an ingrate who refuses to pay attention. I give of my time and expertise freely but you would rather be offended than study, think about, understand and be thankful for what I wrote.

So for the second time: Now go away and think about it.

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