1. ## distribution

Cards are dealt, one by one, from an ordinary deck, until the first diamond appears or 3 cards are dealt in total (whichever comes first). Find the distribution of the total number of clubs dealt.

What ive done so far

I am not exactly sure how to do this, but i started by breaking down into 4 scenarios.

1. First card is a diamond
2. Second card is a diamond
3. Third card is a diamond
4. No diamond

For the following, D = Diamond, C = Club, and X = Spade or Heart

1. D
Pr(C) = 0

2. CD, XD
Pr(C) = 1/3 (since x can be two different suites)

3. CCD, CXD, XCD, XXD
Pr(C) = 5/9

4. CCC, CCX, CXX, XXX, CXC, XCC, XXC, XCX
Pr(C) = 19/27

Would I now find the number of clubs possible in each scenario? And that would be the distribution?
for example, in scenario 3...

Pr(C = 1) = 4/9
Pr(C = 2) = 1/9

Let me know. Thanks for any help

2. Originally Posted by chrisc
Cards are dealt, one by one, from an ordinary deck, until the first diamond appears or 3 cards are dealt in total (whichever comes first). Find the distribution of the total number of clubs dealt.
Does it mean that the cards are put back in the deck before the next one is chosen? This is what you did, so I'll assume it is correct.

Would I now find the number of clubs possible in each scenario? And that would be the distribution?
for example, in scenario 3...

Pr(C = 1) = 4/9
Pr(C = 2) = 1/9
Yes, you can procede this way. What you wrote is in fact conditional probability: 4/9 is the probability to have picked one club given that scenario 3 happened. Let us write this $\displaystyle P(C=1|{\rm sc. }3)=\frac{4}{9}$. Then, to find the distribution of $\displaystyle C$ (the number of clubs), we have to multiply these numbers by the probability of the scenarios. For instance,
$\displaystyle P(C=0)=P(C=0|{\rm sc. }1)P({\rm sc. }1)+P(C=0|{\rm sc. }2)P({\rm sc. }2)$ $\displaystyle +P(C=0|{\rm sc. }3)P({\rm sc. }3)+P(C=0|{\rm sc. }4)P({\rm sc. }4)$.
And $\displaystyle P({\rm sc. }1)=\frac{1}{4}$ (choosing a D), $\displaystyle P({\rm sc. }2)=\frac{3}{4}\times\frac{1}{4}$ (anything but a D, then a D), etc.

This is a bit tedious. You could instead consider the scenarios that give a certain number of clubs:
Scenario C=0: D, XD, XXD, XXX
Scenario C=1: CD, XCD, CXD, XXC, XCX, CXX
Scenario C=2: CCD, XCC, CXC, CCX
Scenario C=3: CCC
There are $\displaystyle 4\times 4\times 4=64$ ways to choose three cards (provided they are put back in the deck one by one), and for instance the number of choices corresponding to XD is $\displaystyle 2\times 1\times 4$ (X, D, and anything), the number of choices corresponding to XCC is $\displaystyle 2\times 1\times 1$ and so on.
Finally, the distribution of C is: $\displaystyle P(C=0)=\frac{1}{64}(16+8+4+8)=\frac{36}{64}=\frac{ 9}{16}$ (you can also get this another way if you prefer: $\displaystyle P(C=0)=\frac{1}{4}+\frac{2}{4}\times\frac{1}{4}+\c dots$) and you can do the same for the other values. Notice that $\displaystyle P(C=1)=1-P(C=0)-P(C=2)-P(C=3)$ so you can deduce $\displaystyle P(C=1)$ from the others (or check that your computation is correct).