[Math] How to get the number of ways of getting a five card hand that is a straight flush from a standard deck of cards

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I do not get the result at this page, ex. 13-7:

Suppose that Aces can be either high or low; that is, that {Ace, 2, 3, 4, 5} is a straight, and so is {10, Jack, Queen, King, Ace}. The number of ways of getting a five card hand that is a straight flush from a standard deck of cards is:

The result is 36. However I got 40 following this approach:

I have 10 starting cards, from Ace to 10, and 4 suits, so I thought to have 40 subsets:

Ace, 2, 3, 4, 5
...
10, Jack, Queen, King, Ace

for each one of the 4 suits.
Where am I doing wrong?

Thanks

Best Answer

If OR is exclusive in the question, meaning that Ace can either be high or low but not both would give you 36 possible combinations. However the question isn't exactly stated in a way that makes it easy to interpret this. Your logic is right though, and it could be an error on your sources part. The possible combos are

A 2 3 4 5

2 3 4 5 6

3 4 5 6 7

4 5 6 7 8

5 6 7 8 9

6 7 8 9 10

7 8 9 10 J

8 9 10 J Q

9 10 J Q K

10 J Q K A

Related Solutions

[Math] the probability that 5 randomly chosen cards in a deck add up to 40 or greater

Normal approximation:

Consider a single card draw from the deck:

E(X)=7

Var(X)=14

With the sum of 5 cards (drawn without replacement):

$E(X)=7 \times 5$

$\text{Var}(X)=14\times 5\times \frac{52-5}{52-1}$

(the last term being the 'finite population correction', which applies as much to the variance of the sum as it does to the variance of the mean).

The approach to normality in this situation is reasonably rapid.

This suggests the sum on 5 cards might be be roughly approximated by a normal distribution with mean $35$ and variance $70\times\frac{47}{51}\approx 64.5098$. Using a continuity correction this gives an approximate probability of totalling at least 40 on 5 cards of:

$$1-\Phi(\frac{39.5-35}{\sqrt{70\times\frac{_{47}}{^{51}}}})\approx 0.288$$

Simulation (in R) of ten million five-card draws indicates the probability is
about $0.293$ (with s.e. $\approx 1.4\times 10^{-4}$):

 res=replicate(10000000,sum(sample(rep(1:13,4),5,replace=FALSE)));mean(res>=40)  
 [1] 0.2927447

As a check on the earlier calculation of the variance, the standard deviation of of those simulated sums was 8.0316; the previous calculation gives 8.0318.

Edit: Here's a comparison of the empirical cdf of the simulated data with the above normal approximation; they're pretty close:

enter image description here

More extensive simulations are consistent with the other two answers based on complete enumeration:

enter image description here

[Math] Select cards from the Card Deck

Yes, your answer is correct: $12 \times 47 + 48 = \boxed{612}$.

Another way of getting the same answer is that there are $13$ spades and $48$ non-queens, $13 \times 48 = 624$. Then you subtract out the times that we picked the same card twice, which is once for every non-queen spade, of which there are $12$. So $624 - 12 = \boxed{612}$.

Best Answer

Related Solutions

[Math] the probability that 5 randomly chosen cards in a deck add up to 40 or greater

[Math] Select cards from the Card Deck

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