SOLUTION: (a) Find the probability of being dealt a "sevens over eights" full house (three sevens and two eights). (Round your answer to six decimal places.) (b) Find the number of diff

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Question 1201710: (a) Find the probability of being dealt a "sevens over eights" full house (three sevens and two eights). (Round your answer to six decimal places.)

(b) Find the number of different types of full houses. (Ignoring suit.)
ways
(c) Find the probability of being dealt a full house. (Round your answer to six decimal places.)
Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Answers:
  1. probability = 0.000009
  2. There are 156 types of full house when ignoring suit.
  3. probability = 0.001441
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Explanation for Part (a)

With poker hands, the order does not matter.
We aren't putting the cards back aka "no replacement".
The nCr combination formula will be used.

Determine how many ways there are to pick the three "7"s.
n = 4 copies of the card labeled "7"
r = 3 slots to fill
n C r = (n!)/(r!(n-r)!)
4 C 3 = (4!)/(3!*(4-3)!)
4 C 3 = (4!)/(3!*1!)
4 C 3 = (4*3!)/(3!*1!)
4 C 3 = (4)/(1!)
4 C 3 = (4)/(1)
4 C 3 = 4
There are 4 ways to pick three "7"s where order doesn't matter.

A shortcut: there are 4 ways to leave a certain "7" out, which leads to 4 ways to pick a trio of "7"s.

Now find out how many ways there are to pick the two "8"s
n = 4 copies of "8"
r = 2 slots to fill
n C r = (n!)/(r!(n-r)!)
4 C 2 = (4!)/(2!*(4-2)!)
4 C 2 = (4!)/(2!*2!)
4 C 2 = (4*3*2!)/(2!*2!)
4 C 2 = (4*3)/(2!)
4 C 2 = (4*3)/(2*1)
4 C 2 = 12/2
4 C 2 = 6
There are 6 ways to pick two "8"s in any order.

Here is a list of all 6 duos.
  1. duo = 8C, 8D
  2. duo = 8C, 8H
  3. duo = 8C, 8S
  4. duo = 8D, 8H
  5. duo = 8D, 8S
  6. duo = 8H, 8S
where
C = clubs
D = diamonds
H = hearts
S = spades
Order doesn't matter so something like "8C,8D" is the same as "8D,8C".

To quickly recap:Therefore, we have 4*6 = 24 ways to pick a hand of "sevens over eights", aka three "7"s and two "8"s.
We'll refer to this value 24 later on.


Now let's find out how many ways there are to pick any five cards. Again order doesn't matter.
n = 52 cards total
r = 5 slots to fill
n C r = (n!)/(r!(n-r)!)
52 C 5 = (52!)/(5!*(52-5)!)
52 C 5 = (52!)/(5!*47!)
52 C 5 = (52*51*50*49*48*47!)/(5!*47!)
52 C 5 = (52*51*50*49*48)/(5!)
52 C 5 = (52*51*50*49*48)/(5*4*3*2*1)
52 C 5 = 311875200/120
52 C 5 = 2598960
There are 2,598,960 ways to select five cards where order doesn't matter.

This number comes up a lot in poker probability problems.
I'm not saying you should memorize it, but I recommend getting familiar with its rough size (about 2.6 million) so you can check your work down the line.
Perhaps having it on a reference sheet or flash card could be useful.

We found that there are...Order doesn't matter.

One final step is to divide those items
24/2598960 = 0.00000923446301
This value is approximate.

Rounding to 6 decimal places gets us 0.000009 which is the final answer to part (a).

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Explanation for Part (b)

A full house is when you have 3 cards of one label (eg: three "7"s) and 2 cards of another label (eg: two "8"s).
This is the more generalized form of the "sevens over eights" we dealt with back in part (a).

Other examples of a full house:Here is a list of common hands
https://en.wikipedia.org/wiki/List_of_poker_hands
Scroll down a bit to reach "full house".

We have 13*12 = 156 different types of a full house if we ignore suit.

13 = number of ways to pick the first label
12 = number of ways to pick the second label
We drop down by 1 because we cannot repeat the first label.

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Explanation for Part (c)

For the first label, there are 4 ways to select three of that label (ie 4 ways to leave out a certain card). This is when we start considering the suit.

For the second label, there are 6 ways to pick two of that label.
Refer to part (a) to see where these numbers (4 and 6) are coming from.

This leads to 156*4*6 = 3744 ways to get a full house when we consider the suit.
We use the 156 from part (b).

-----------------

Another approach:

As calculated in part (a), there are 4 ways to pick the three "7"s.
If we generalize to any label, then there are 13*4 = 52 ways to pick the three starting cards of the same label.

We also calculated earlier there are 6 ways to pick the two "8"s.
This generalizes to 12*6 = 72 ways to pick two cards of the same label.
These two other cards cannot match the previous label, so that's why I went with 12 instead of 13.

All together we have 52*72 = 3744 ways to get a full house when considering the suit.

Check out this page that lists all sorts of poker probabilities.
https://en.wikipedia.org/wiki/Poker_probability
Scroll down to "full house" and notice that "3,744" is in the frequency column. This confirms we have the correct number of hands for a full house.

Just to the left of 3744 is 156, which helps confirm the answer to part (b).

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Recap:Order doesn't matter.

3744/2598960 = 0.0014405762305 = 0.001441 is the approximate probability of a full house when considering suit.

0.001441 converted to a percentage is 0.1441%

The probability of getting a full house is listed on this page
https://en.wikipedia.org/wiki/Poker_probability
which helps confirm the answer.

Further Reading
https://math.stackexchange.com/questions/808314/probability-of-getting-a-full-house
https://9to5science.com/probability-of-getting-a-full-house
https://math.stackexchange.com/questions/192788/how-to-calculate-a-fullhouse-after-dealing-5-cards
Feel free to explore other alternatives.
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