Probability of a 64 cards card game

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KhanhDo

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I need help with this problem of the card game:
Suppose i have a 32 cards deck that consist of value cards from 7 to 10 (7,8,9,10) and figure cards from J to Ace (J,Q,K,A)
Now add another deck of cards similar to the one above, making a bigger deck that have 64 cards in total.
Now draw 8 cards from the 64 cards deck
What is all the possible combinations of hands that result in two 4 cards straight (including Straight Flush and Royal Flush) from the 8 cards that you drew?
Ex: 7,8,9,10, 9,10,J,Q

I knew how to find the combinations of a 4 cards straight from a 32 cards deck which is 5C1*((4C1)^4)
But i don't know how to adjust the result if it's now 8 cards from a 64 cards deck

Thank you for your time!
 
KhanhDo said:
I need help with this problem of the card game:
Suppose i have a 32 cards deck that consist of value cards from 7 to 10 (7,8,9,10) and figure cards from J to Ace (J,Q,K,A)
Now add another deck of cards similar to the one above, making a bigger deck that have 64 cards in total.
Now draw 8 cards from the 64 cards deck
What is all the possible combinations of hands that result in two 4 cards straight (including Straight Flush and Royal Flush) from the 8 cards that you drew?
Ex: 7,8,9,10, 9,10,J,Q

I knew how to find the combinations of a 4 cards straight from a 32 cards deck which is 5C1*((4C1)^4)
But i don't know how to adjust the result if it's now 8 cards from a 64 cards deck

Thank you for your time!
Click to expand...
You failed to give a complete description the the deck of cards. You say that there are eight denomination, seven, eight,[imath]\cdots[/imath] ace.
However that is only eight cards. Are there there then four suits, [imath]\clubsuit,~\diamondsuit,~\heartsuit,~\&~\spadesuit~?[/imath]
That would make thirty-two cards in a deck. Now explain your question or correct the discription.
 
pka said:
You failed to give a complete description the the deck of cards. You say that there are eight denomination, seven, eight,[imath]\cdots[/imath] ace.
However that is only eight cards. Are there there then four suits, [imath]\clubsuit,~\diamondsuit,~\heartsuit,~\&~\spadesuit~?[/imath]
That would make thirty-two cards in a deck. Now explain your question or correct the discription.
Click to expand...
The 64 cards deck is made by combining the two exactly same 32 cards deck
So your 64 cards deck would have one set (7,8,9,10,J,Q,K,A) of the suits ♣, ♢, ♡, & ♠ and another exactly same set of (7,8,9,10,J,Q,K,A) of the suits ♣, ♢, ♡, & ♠
 
KhanhDo said:
The 64 cards deck is made by combining the two exactly same 32 cards deck
So your 64 cards deck would have one set (7,8,9,10,J,Q,K,A) of the suits ♣, ♢, ♡, & ♠ and another exactly same set of (7,8,9,10,J,Q,K,A) of the suits ♣, ♢, ♡, & ♠
Click to expand...
You are just repeating what I posted. Why? Now that we are clear as to how the deck is constituted what is your question?
What makes up a dealt hand? What are your questions about dealt hands?
 
pka said:
You are just repeating what I posted. Why? Now that we are clear as to how the deck is constituted what is your question?
What makes up a dealt hand? What are your questions about dealt hands?
Click to expand...
My question was now if you were to draw 8 cards from this 64 cards deck, what is the number of combinations of those 8 cards that will result in two 4-cards straight
 
To draw 8 cards from this deck, I would imagine 5 general patterns. P1, no repeating cards; p2, 1 pair of repeating cards, p3….p5, 4 pairs of repeating cards.
Since Pattern 1 has no repeating cards, your combination function would look like 64c1*62c1*60c1*58c1….50c1 then divide by 8! to eliminate the same combination arranged differently. Pattern 2: view 1 pair as 1 instance of the 32 pairs and the rest is just choosing 6 cards out of 62 and total would 32 times be this variant 62c6 (same as how you calculate pattern 1). And Pattern 3 would be 32*31/2! times variant 60c4… Pattern 5 is just choose 4 cards out of 32 since coincidentally out of 8 cards every of 4 different cards repeated itself. Makes sense?

Add them all together you will have your sample space.

the events would change a bit as well: so before you would have two straights forming 7to10 up to JtoA, now you would have repeating cards forming these and just add these cases into your events. This you should figure out yourself since this is simpler than the sample space.
 
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My mistake. To ensure the uniqueness of each pattern there shouldn’t exist more than 32 cards to choose from in the formula.
Pattern 1 is the same as choosing 8 cards from 32 unique cards: 32c8; pattern 2: 32*31c6; pattern 3: (32*31/2!)*30c4; pattern 4: (32*31*30/3!)*29c2; pattern 5: 32*31*30*29/4!
 
i am not sure if this is correct but what i did was considering what possible outcome of this 8 cards hand is
We have 5 possible starting point of a straight which is 7,8,9,10,J
so the first outcome is if two straights in our hand are different which means i have to chose 2 different straight from 5 possible ones so i take 5C2 and then multiply by (4C1)^8 which is the 4 possible suits of each card and we have 8 cards so ^8
the second outcome is if the two straight are the same, in which case it’s just 5C1*[(4C1)^8]
so in the end, number of all possible hands is:
[4C1^8]*(5C2+5C1)
 
The straight hand events: like you said, when you have 7 to A there are basically 5 different 4-cards straights to start with. And if you want 8 cards with 2 straights, it’s basically 5c2 + 5 = 15 combinations ( in here + 5 means the same value straights, for instance: 10JQK10JQK or JQKAJQKA)

these 15 combinations can be separated into 5 different patterns, which is: Pattern 1: 78910JQKA with no overlapping value cards. Pattern 5: 7891078910,8910J8910J, 910JQ910JQ,10JQK10JQK and JQKAJQKA, with all 4 cards’ value overlapping. Pattern 2: one card value overlapping, 7891010jqk,8910jjqka. Pattern 3: two card values overlapping 78910910jq, 8910j10jqk, 910jqjqka. Pattern 4: three card values overlapping: 789108910j, 8910j910jq, 910jq10jqk, 10jqkjqka. These add up to be 15 combinations of two sets of straights.

Pattern1 is easy to calculate, which is 4^8 for each of 8 card position has 4 suits to vary.

Pattern5 is just 4c2*4c2*4c2*4c2 and times 5 different combos.

Pattern2 is 4^3*4c2*4^3 and times 2 different combos

Pattern 3 is 4^2*4c2*4c2*4^2 and times 3 different combos

Pattern 4 is 4*4c2*4c2*4c2*4 and times 4 different combos.

Add them up that is how many events that 8 cards drawn to form 2 sets of straights, out of 32 cards.

Now when you have 64 cards, the Pattern 1 doesn’t get affected. For other patterns you will encounter a situation which you may draw identical cards from the same suit, and there are total 4 suits so 4 cases are added. So instead of 4c2 you use (4c2+4).

Apparently your formula lacks the analysis of the patterns although I didn’t totally understand your logic I wouldn’t guess it to be correct.

So once you understand how to correctly calculate the events of 2 sets of 4-card-straights in 8 out of 32 cards, it’s very simple to amend the formula for 64 cards.
 
sky2rain said:
My mistake. To ensure the uniqueness of each pattern there shouldn’t exist more than 32 cards to choose from in the formula.
Pattern 1 is the same as choosing 8 cards from 32 unique cards: 32c8; pattern 2: 32*31c6; pattern 3: (32*31/2!)*30c4; pattern 4: (32*31*30/3!)*29c2; pattern 5: 32*31*30*29/4!
Click to expand...
Use the event sum divide by the sample space sum (presented in the quoted text), you would get your probability.

And you don’t have to worry about individual probability since you added equal amount of unique cards, probability of drawing 1 card out of 32 unique cards is equal to drawing 1 card out of 2 identical sets of 32 unique cards.
 
I wrote a "Monte Carlo" simulation of this. You might be able to use it as a sanity check for your calculations. I ran it several times and these are the ouput probabilities (of getting 2 straights):- 0.011042, 0.011126, 0.011235, 0.011323 therefore it seems likely that the actual probability ought to begin with the digits 0.011

Python: 
import random

numSuits=8 # number of suits in deck
numCards=8 # number of cards in each suit
numDeal=8
numInStraight=4
desiredStraights=2


cardDeck=[]
histogram=[]
for i in range(numCards):
    histogram.append(0)
    for j in range(numSuits):
        cardDeck.append(i)

max=1000000
match=0
for i in range(max):
    for j in range(numCards):
        histogram[j] = 0
        
    random.shuffle(cardDeck)
    
    for j in range(numDeal):
        histogram[ cardDeck[j] ] += 1
        
    straightCount=0
    for j in range(numCards+1-numInStraight): # lowest card in the straight
        fnd=1
        while fnd==1:
            for k in range(numInStraight):
                if histogram[j+k]==0:
                    fnd=0
                    break
            if fnd==1:
                for k in range(numInStraight):
                    histogram[j+k] -= 1
                straightCount += 1

    if straightCount >= desiredStraights:
        match += 1
        
print(match/max)
 
I can see where the mistake lies. When drawing a royal flush of diamond 78910jqka out of a deck of 32 cards, the probability is simply 1/32c8. But doing the same from two decks of same 32 cards the probability changes.

There is actually less a chance to get a diamond royal flush 7toA in 64 cards of the same two decks of 32.
 
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If you have 4 color of balls 1 of each color the chance of drawing three balls of different color is simply 1/4c3=1/4. However if you double the 4 color sample space to have 2 of the same color balls and total 8 balls, the chance of drawing three balls of different color reduces to 8/8c3=1/7.
 
If we treat these two decks as unique cards of 64 (for example, one with red back and one with blue back) the problem becomes what’s the probability of two flashes forming when 64c8, which is about 4.5 billion hands. Our goal would be just analyze what happens when forming 2 straights out of 8 out of 32 unique face value cards, and then add the scenarios when the new deck adds in how would the face value change. With the three different color balls we can see that if there are four colors RYBW, by adding 1 additional of each color there would exist R1Y1B1W1R2Y2B2W2 totally 8 balls. And a three different color instance, for example, RYB, can be formed in 8 ways such as R1Y1B1,R1Y1B2…since each color now have two choices there would be 2^3 formations, but all 8 would give you exact same appearance. So with our 2 sets of 4-cards-straight out of 32 cards, when the total number of cards becomes 64 (2 identical face value 32 cards), each unique face-value formation would now have 2^8 variants with red/blue backs.

So simply analyze how many formations there exist in 2 sets of 4-cards-straight in the 32 cards deck and multiplying by 2^8=256 we would have a ballpark estimation of how many formations exist in the 64c8 hands.

it’s ball-park since there could be cases like d7c8h9s10 c8d9d10cA (2 club of 8 in one hand), and obviously this is not possible to happen in the 32 card deck formations, nor we should multiply this by 256 (should multiply by 64), with these additions the result would push close to the Monte Carlo simulation result of 0.011.
 
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Ballpark is (165,000*256)/4.5billion = 0.00938667, not bad.
 
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