Thread: An interesting poker probabilty question

1. An interesting poker probabilty question

I've been wondering how to compute this one for a while, and I've done some initial work but it seems to be going nowhere.

Find the probability of holding a Royal Flush in a n hand draw, where n must be atleast 5 but no greater than 52. Also, find the value of n that gives a 100% chance to draw a royal flush.

3. Originally Posted by Dave2718
I've been wondering how to compute this one for a while, and I've done some initial work but it seems to be going nowhere.

Find the probability of holding a Royal Flush in a n hand draw, where n must be atleast 5 but no greater than 52. Also, find the value of n that gives a 100% chance to draw a royal flush.

Edit 3: You wrote "the probability of holding a Royal Flush in a n hand draw," but I suppose you meant an n-card hand? Because why else would n range from 5 to 52. Or maybe it's just because I'm unfamiliar with poker terminology. (I moved this edit to the top because my whole post is based on it.)

Well there are only 4 royal flushes, so that should help keep the numbers manageable.

I'll tackle the last question first: The probabilty of getting a royal flush is 1 if and only if $n > 48$. This is because at least one of the five cards from each royal flush must remain in the deck, so for example, if you get a 48 card hand but all 4 aces are left in the deck, then you won't have a royal flush, and this is the most cards you can have in your hand for that to happen.

I'll have to think about the other stuff. Someone will probably post an answer before I get the chance to sort it all out.

Edit: Well the probability of getting a royal flush in a 5-card hand is $\displaystyle \frac{4}{\binom{52}{5}}$.

Edit 2: For n = 6, we have 4 ways to choose a flush, and then 52-5=47 ways to choose the remaining card for each flush. So $P_6=\displaystyle\frac{4\cdot47}{\binom{52}{6}}$

And likewise $P_7=\displaystyle\frac{4\cdot\binom{47}{2}}{\binom {52}{7}}$

I think starting at n=10 we'll have to use inclusion-exclusion.

4. I'd start the problem by imagining that I am choosing n objects from 4 bins (call them A,B,C and D) each of size 13 without replacement (assume the objects in each bin are labeled 1-13).

The question "what is the probability that we have chosen at least 5 objects from at least one the bins" is a standard combinatorics question. Lets call this probability "P_1"

Now given that we have chosen at least 5 objects from one of the bins, what is probability that 5 of the objects are consecutively labeled from at least one of these bins? Says its P_2.

I believe the final answer should be P_1*P_2 using the rule P(A,B) = P(A)*(B given A).

I'll try to give this a shot. Hopefully someone soon will fill in the unfinished details.

5. Interpretting problem

Yeah at n = 10 things get tougher because we have to consider the possibility that we might have two royal flushes.

I think my above post isn't such a good strategy.

I'm wonder if the following analogy helps?

Put the 4 possible royal flushes into 4 separate piles of 5 cards, each containing one of royal flushes. Put the remaining 32 cards in a 5th pile.

If we start randomly selecting cards from one of these 5 piles without replacement, we need only know the probability that we don't use up one of the piles that initially has 5 cards in it after we have selected n cards. I'm not sure if this interpretation offers any short cuts, but I'm not really up to doing this problem the long way.

6. Originally Posted by jamix
I'd start the problem by imagining that I am choosing n objects from 4 bins (call them A,B,C and D) each of size 13 without replacement (assume the objects in each bin are labeled 1-13).

The question "what is the probability that we have chosen at least 5 objects from at least one the bins" is a standard combinatorics question. Lets call this probability "P_1"

Now given that we have chosen at least 5 objects from one of the bins, what is probability that 5 of the objects are consecutively labeled from at least one of these bins? Says its P_2.

I believe the final answer should be P_1*P_2 using the rule P(A,B) = P(A)*(B given A).

I'll try to give this a shot. Hopefully someone soon will fill in the unfinished details.
Before you do too much calculation, note that you described getting a straight flush, not a royal flush. Royal flushes must have 10-J-Q-K-A all of the same suit.

7. Originally Posted by undefined
Before you do too much calculation, note that you described getting a straight flush, not a royal flush. Royal flushes must have 10-J-Q-K-A all of the same suit.
Right, I forgot to add that in. Luckily I think that makes the question easier since we need only calculate the probability of obtaining 9,10,11,12,13.

8. Originally Posted by jamix
Yeah at n = 10 things get tougher because we have to consider the possibility that we might have two royal flushes.

I think my above post isn't such a good strategy.

I'm wonder if the following analogy helps?

Put the 4 possible royal flushes into 4 separate piles of 5 cards, each containing one of royal flushes. Put the remaining 32 cards in a 5th pile.

If we start randomly selecting cards from one of these 5 piles without replacement, we need only know the probability that we don't use up one of the piles that initially has 5 cards in it after we have selected n cards. I'm not sure if this interpretation offers any short cuts, but I'm not really up to doing this problem the long way.
I came to the same conclusion that this approach is probably best, starting at n=10 (could be done for n<10 too, but those were already covered, if I didn't make any mistakes). I was trying to make sure it worked out nicely before posting.. still working on it..

9. Isn't $P_{10}=\displaystyle\frac{4\cdot\left(\binom{47}{5 }-3\right)+\binom{4}{2}}{\binom{52}{10}}$? The first term in the numerator corresponds with getting exactly one royal flush, and the second term is getting exactly two.

I admit I'm getting bogged down by the complexity and probably am not looking at it quite right.

I think I'll set this aside for a while and hope someone comes along and answers it so that I don't have to hurt my head.

10. Well decided to plough along using my current method.

$P_{11}=\displaystyle\frac{4\cdot\left(\binom{47}{6 }-3\cdot\binom{42}{1}\right)+\binom{4}{2}\cdot\binom {42}{1}}{\binom{52}{11}}$

$P_{12}=\displaystyle\frac{4\cdot\left(\binom{47}{7 }-3\cdot\binom{42}{2}\right)+\binom{4}{2}\cdot\binom {42}{2}}{\binom{52}{12}}$

... hopefully there's a better way because n>14 will get pretty ugly.

11. Originally Posted by undefined
Well decided to plough along using my current method.

$P_{11}=\displaystyle\frac{4\cdot\left(\binom{47}{6 }-3\cdot\binom{42}{1}\right)+\binom{4}{2}\cdot\binom {42}{1}}{\binom{52}{11}}$

$P_{12}=\displaystyle\frac{4\cdot\left(\binom{47}{7 }-3\cdot\binom{42}{2}\right)+\binom{4}{2}\cdot\binom {42}{2}}{\binom{52}{12}}$

... hopefully there's a better way because n>14 will get pretty ugly.
This looks pretty close to the general solution.

For any ONE particular royal flush (RF) (ie clubs, spades, hearts, diamonds), we have that the number of ways to choose n integers from a set of 52 integers which contains THIS RF is just $\frac{\binom{47}{n-5}}{\binom{52}{n}}$.

Since there are 4 different RFs to choose from, we need to multiply the above by 4. For n < 10, this is a complete solution.

Unfortunately though for n > 9, we have added some subsets twice or more.

Clearly we need to employ the Inclusion-Exlusion principle; hence it is worthwhile to determine the number of ways to have a set containing two RFs, three RFs, and four.

For 9 < n < 15, we need only subtract the number of ways to have two RFs from our original $4\cdot \frac{\binom{47}{n-5}}{\binom{52}{n}}$

There are $\binom{4}{2} = 6$ RF pairs. For each RF pair, we have $\binom{42}{n-10}$ ways to create a set of n integers from 52 that contains this particular pair. Hence the overall probability of having a set n containing two RFs is just

$6\cdot \frac{\binom{42}{n-10}}{\binom{52}{n}}$

Hence the solution for 9 < n < 15 is the following;

$\frac{4\cdot\binom{47}{n-5} - 6\cdot\binom{42}{n-10}}{\binom{52}{n}}$

Again it is clear that this solution is complete for the above range of n since we cannot have a three or four RFs.

On the other hand, for 14 < n < 20, it is possible to have 3 RFs. Hence we need to compute this possibility as well.

Following the same reasoning we used to determine the number of subsets containing two RFs, we find that the probability of having a subset having three RFs is just $4\cdot \frac{\binom{37}{n-15}}{\binom{52}{n}}$.

Hence for 14 < n < 20, the probability of having a set with a RF is

$\frac{4\cdot\binom{47}{n-5} - 6\cdot\binom{42}{n-10} + 4\cdot\binom{37}{n-15}}{\binom{52}{n}}$

Finally for 19 < n < 53, we have the probability that a subset contains a RF is

$\frac{4\cdot\binom{47}{n-5} - 6\cdot\binom{42}{n-10} + 4\cdot\binom{37}{n-15} - \binom{32}{n-20}}{\binom{52}{n}}$.

I hope this helps.