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Missing cards from the deck: $\heartsuit3$, $\spadesuit$King, $\heartsuit$King, $\clubsuit$king and $\spadesuit2$

a) If I draw $5$ cards randomly what are the chances of getting a full house ?

b) Which other card can you remove to maximize your chances of getting full house?

I don't really understand how to solve it, for my understanding there are $13\diamondsuit$, $11\spadesuit$, $11\heartsuit$ and $12\spadesuit$ left in the deck. If randomly you pick a card out you have $\frac{1}{4}$ to draw $\frac{1}{11}, \frac{1}{11}, \frac{1}{13} \text{ or }\frac{1}{12}$. So depending on what suit you draw it goes from $\frac{1}{11}$ to $\frac{1}{13}$.

RobPratt
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zellez11
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    The first part should be easy enough, no? Just count the possible full houses and the total number of possible hands. For the second part, it sure seems like taking out the $\diamondsuit K$ can't hurt. – lulu Feb 19 '20 at 00:20
  • The probability of getting a full house with a standard deck is solved here. You can modify the approach for this deck. – N. F. Taussig Feb 19 '20 at 00:26

2 Answers2

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NOTE: This is just the answer for a), not for b).

a) If I draw 5 cards randomly what are the chances of getting a full house?

You took out $5$ cards, so there would be $47$ cards leftover in the deck.

There are $\binom{47}{5}=1533939$ ways to choose any $5$ cards out of a $47$-card deck.

Then I used casework to find the ways to get a full house:

Case 1: 2 is chosen three times

There are $3$ $2$'s in the modified deck, so there is only $1$ way to choose $3$ out of $3$.

Subcase 1.a: 3 is chosen 2 times

There are $3$ $3$'s in the modified deck, so there would be $\binom{3}{2}=3$ ways to get a full house this way.

Subcase 1.b: Another rank(minus the king) is chosen 2 times

There would be $4$ cards to choose from, and there are $13-3=10$ other ranks, so that means that there would be $\binom{4}{2}*10=60$ ways to get a full house this way.

In total that would be $60+3=63$ ways for Case 1.

Case 2: 3 is chosen three times

This is essentially this same thing as Case 1 so the number of ways would stay the same. $63$ ways.

Case 3: Another rank(other than 2, 3, and king) is chosen 3 times

There are $4$ cards to choose from, so that would make it $\binom{4}{3}=4$ ways to choose $3$ cards out of $4$.

Subcase 3.a: 2 is chosen 2 times

There are $3$ cards with rank $2$, so that means the number of ways to choose $2$ cards out of $3$ is $\binom{3}{2}=3$ ways for this subcase.

Subcase 3.b: 3 is chosen 2 times

Essentially the same as subcase 3.a so $3$ ways.

Subcase 3.c: Another rank(not 2, 3, king, or the rank chosen for this case) is chosen 2 times

There are $9$ ranks leftover. Each of these ranks have $4$ choices to choose from, and you are choosing $2$ out of the $4$, so the number of ways for this subcase is $\binom{4}{2}*9=54$.

In total, that would be $10*4*(3+3+54)=2400$ ways for this case.

So, the final answer is: $$\frac{63+63+2400}{1533939}=\frac{2526}{1533939}\approx 0.0016467$$

Aiden Chow
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    Unfortunately, I think you are confused as to what a "full house" is. Nothing to do with suits....a full house consists of a three-of-a-kind and a pair. Thus $\spadesuit Q, \diamondsuit Q, \clubsuit Q, \heartsuit 10, \diamondsuit 10$ is a full house. – lulu Feb 19 '20 at 01:17
  • here is an article which discusses the various poker hands and computes the probabilities of each (for a standard deck of course). – lulu Feb 19 '20 at 01:19
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    OK thanks for replying I will edit my answer – Aiden Chow Feb 19 '20 at 01:21
  • I changed my answer, is it correct now? – Aiden Chow Feb 19 '20 at 01:41
  • Good approach, but I think you have some arithmetic errors. For instance, in case $1.b$ There are only $10$ available suits (since you have excluded $2,3,K$). And once you choose a suit you need to choose $2$ out of $4$, so I only see $10\times \binom 42$ here. But the most serious error seems to be in $3.c$. I think the count should be $10\times \binom 43\times 9\times \binom 42=2160$. Much larger than you got. – lulu Feb 19 '20 at 01:47
  • Oh I forgot to multiply by $10$ for the last case(there were $10$ ranks). I fixed it. Is it correct now? – Aiden Chow Feb 19 '20 at 01:51
  • Well, it doesn't match my number. Do you see where my calculation came from? – lulu Feb 19 '20 at 01:52
  • Yes, but by multiplying by $10$ it should account for it. $10\binom{4}{3}9\binom{4}{2}=10496=10454$ which is one of the terms in the total for case 3. – Aiden Chow Feb 19 '20 at 01:57
  • Maybe I am not clear what your case $3$ covers. And you still have not corrected your error from $1.b$. I am going to post my calculations below. You should check them! I am doing it quickly and could certainly have made some errors myself. – lulu Feb 19 '20 at 01:58
  • I've looked at some of your other answers and u seem to be way more advanced at math than me so it is likely that u r correct(maybe you could post your answer) – Aiden Chow Feb 19 '20 at 02:03
  • After all the edits, your answer matches mine exactly! As we counted by (slightly) different means, I think this counts as an arithmetic check. Good work (+1). – lulu Feb 19 '20 at 02:04
  • To your last comment: this sort of calculation is extremely prone to errors. Everybody, no matter how advanced, makes arithmetic errors in problems like this. If you look at my solution you'll see that I tried to organize the work so as to avoid errors, but it's very hard to be sure. – lulu Feb 19 '20 at 02:05
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Clearly no full house can have a King. Other than the King, the cards fall into two types:

$A$: Cards with $4$ suits: there are ten of these.

$B$: Cards with $3$ suits: there are two of these (namely $2,3$).

We consider the various types of full houses (here, for instance type $(A,B)$ means that the three of a kind is of type $A$ and the pair is of type $B$).

Type $(A,A)$. There are $10$ ways to choose the rank for the triple, then $\binom 43$ ways to choose the triple. Then there are $9$ ways to choose the rank for the pair and $\binom 42$ ways to choose the pair. Thus $$10\times \binom 43\times 9 \times \binom 42=2160$$

Type $(A,B)$. There are $10$ ways to choose the rank for the triple, then $\binom 43$ ways to choose the triple. Then there are $2$ ways to choose the rank for the pair and $\binom 32$ ways to choose the pair. Thus $$10\times \binom 43\times 2 \times \binom 32=240$$

Type $(B,B)$. There are $2$ ways to choose the rank for the triple, then $\binom 33$ ways to choose the triple. Then there is $1$ way to choose the rank for the pair and $\binom 32$ ways to choose the pair. Thus $$2\times \binom 33\times 1 \times \binom 32=6$$

Type $(B,A)$.There are $2$ ways to choose the rank for the triple, then $\binom33$ ways to choose the triple. Then there are $10$ ways to choose the rank for the pair and $\binom 42$ ways to chpose the pair. Thus $$2\times \binom 33\times 10 \times \binom 42=120$$

Finally we sum to see that there are $2526$ possible full houses. As there are $\binom {47}5$ possible hands the answer is $$\boxed {\frac {2526}{\binom {47}5}=.001647}$$

To complete the problem, note that removing any card shrinks the denominator in the the same way (doesn't matter which you remove). However, if you remove the $\diamondsuit K$ then the numerator does not change at all, since no full house can have a King using this deck. As any other removal shrinks the numerator, we see that this is the best choice.

'

lulu
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  • I feel like your answer is way better than mine. this one should get accepted(it answers both a and b). Upvoted – Aiden Chow Feb 19 '20 at 02:10
  • @AidenChow To me, these two answers are more or less the same, and yours came first. I really only posted mine so you could check your work against it. – lulu Feb 19 '20 at 02:12
  • Actually, now that I think about it, my case 1 and 2 combined is your Type $(B,A)$ and Type $(B,B)$, and my case 3 is your Type $(A,A)$ and $(A,B)$, so we essentially did it the same way. – Aiden Chow Feb 19 '20 at 02:16
  • thanks for the help so the answer for b I guest is remove any card but k diamond because it doesn't really change anything – zellez11 Feb 19 '20 at 02:25
  • Diamond can form full house, with a king @lulu – आर्यभट्ट Feb 19 '20 at 03:30
  • @N.F.Taussig Absolutely, just a typo (the division should be correct). Will edit. – lulu Feb 19 '20 at 09:58
  • @YuvrajSingh... Suits have nothing to do with a full house. A full house consists of a triple and a pair. You can't get a pair of kings, let alone a triple, so no full house can contain a king. – lulu Feb 19 '20 at 09:59
  • @zellez Not following. As I pointed out, removing $\diamondsuit K$ maximizes your chances of getting a full house. – lulu Feb 19 '20 at 10:00
  • @N.F.Taussig I am nothing if not consistent. Thanks again. – lulu Feb 19 '20 at 10:01