Probability

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navm

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: If you take two cards from a standard pack of 52 and add their values together, what is more likely
a total value of 6, or a total value of 7? Aces have a value of 1; King, Queen and Jack cards score 10.
 
navm said:
: If you take two cards from a standard pack of 52 and add their values together, what is more likely
a total value of 6, or a total value of 7? Aces have a value of 1; King, Queen and Jack cards score 10.
Click to expand...
How many ways can you choose 2 cards of total value 6 from a standard pack of 52?

How many ways can you choose 2 cards of total value 7 from a standard pack of 52?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 
What confuses me "King, Queen and Jack cards score 10"

I know 1+5 =6
2+4=6
3+3=6
 
navm said:
4 set deck of cards which means 4 times x 3 combination.
Click to expand...
Is this mean 1+5 is one combination and 5+1 2nd combination.

So each set of card has 6 combination x 4 = 24 total combination
 
Steven G said:
1+5 could mean that the 1st card was a 1 and the 2nd card was a 5. 5 + 1 could mean ...
Aren't these two different when it comes to counting a sum of 6?
Click to expand...
Since we're counting combinations, so I think the order does not matter. Therefore, drawing 5 then 1, or 1 then 5 is the same combination. (Correct me if I'm wrong).
 
Steven G said:
1+5 could mean that the 1st card was a 1 and the 2nd card was a 5. 5 + 1 could mean ...
Aren't these two different when it comes to counting a sum of 6?
Click to expand...

Here is my response; kindly let me know whether I did it correctly.


Method 1​

Because we need to find whether the chance that the sum of the values of two cards is 6, is more or the chance that the sum of the values of two cards is 7, is more. Therefore, we only need to consider the first 6 cards i.e. Ace, 2, 3, 4, 5, and 6. If we consider the card with rank 7 then the minimum value of the sum of two cards is 8, which we do not want.

Keeping this in mind, tabulating the values:



Card 1
Sum of ranks1234567
Card 212345678
23456789
345678910
4567891011
56789101112
678910111213
7891011121314




The number of times we can see the sum equal to 6 = 5

The number of times we can see the sum equal to 7 = 6

The total number of outcomes remains the same for both cases. However, the favorable outcomes are more when the sum is 7 as compared to the sum of 6.

Thus, we can say that getting a total value of 7 is more likely as compared to getting a total value of 6.



Method 2​

We know that there are 4 cards for each rank in a fair deck. For the given case, we have four cards for ranks from 1 to 9. While Jack, King, and Queen have a score of 10 which means there are 16 cards for rank 10. (i.e. 9(4) + 16 = 52).

The probability that the sum of the values of two cards is 6 (assuming that cards are replaced after selecting them):

P(Sum is 6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)P(i, j): Probability that values of first and second cards are i and j respectively)=452⋅452+452⋅452+452⋅452+452⋅452+452⋅452=5×452⋅451=802,704P(Sum is 6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)P(i, j): Probability that values of first and second cards are i and j respectively)=452⋅452+452⋅452+452⋅452+452⋅452+452⋅452=5×452⋅451=802,704



The probability that the sum of the values of two cards is 7 (assuming that cards are replaced after selecting them):

P(Sum is 7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)++P(6,1)P(i, j): Probability that values of first and second cards are i and j respectively)=452⋅452+452⋅452+452⋅452+452⋅452+452⋅452+452⋅452=6×452⋅452=962,704P(Sum is 7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)++P(6,1)P(i, j): Probability that values of first and second cards are i and j respectively)=452⋅452+452⋅452+452⋅452+452⋅452+452⋅452+452⋅452=6×452⋅452=962,704



Conclusion:

P(Sum is 7)>P(Sum is 6)
 
navm said:
Here is my response; kindly let me know whether I did it correctly.


Method 1​

Because we need to find whether the chance that the sum of the values of two cards is 6, is more or the chance that the sum of the values of two cards is 7, is more. Therefore, we only need to consider the first 6 cards i.e. Ace, 2, 3, 4, 5, and 6. If we consider the card with rank 7 then the minimum value of the sum of two cards is 8, which we do not want.

Keeping this in mind, tabulating the values:



Card 1
Sum of ranks1234567
Card 212345678
23456789
345678910
4567891011
56789101112
678910111213
7891011121314




The number of times we can see the sum equal to 6 = 5

The number of times we can see the sum equal to 7 = 6

The total number of outcomes remains the same for both cases. However, the favorable outcomes are more when the sum is 7 as compared to the sum of 6.

Thus, we can say that getting a total value of 7 is more likely as compared to getting a total value of 6.



Method 2​

We know that there are 4 cards for each rank in a fair deck. For the given case, we have four cards for ranks from 1 to 9. While Jack, King, and Queen have a score of 10 which means there are 16 cards for rank 10. (i.e. 9(4) + 16 = 52).

The probability that the sum of the values of two cards is 6 (assuming that cards are replaced after selecting them):

P(Sum is 6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)P(i, j): Probability that values of first and second cards are i and j respectively)=452⋅452+452⋅452+452⋅452+452⋅452+452⋅452=5×452⋅451=802,704P(Sum is 6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)P(i, j): Probability that values of first and second cards are i and j respectively)=452⋅452+452⋅452+452⋅452+452⋅452+452⋅452=5×452⋅451=802,704



The probability that the sum of the values of two cards is 7 (assuming that cards are replaced after selecting them):

P(Sum is 7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)++P(6,1)P(i, j): Probability that values of first and second cards are i and j respectively)=452⋅452+452⋅452+452⋅452+452⋅452+452⋅452+452⋅452=6×452⋅452=962,704P(Sum is 7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)++P(6,1)P(i, j): Probability that values of first and second cards are i and j respectively)=452⋅452+452⋅452+452⋅452+452⋅452+452⋅452+452⋅452=6×452⋅452=962,704



Conclusion:

P(Sum is 7)>P(Sum is 6)
Click to expand...
While you conclusion "P(Sum is 7)>P(Sum is 6)" is correct, I do not agree with your numbers.
How many ways are there to pick cards with values 1 and 5? How about 3 and 3?
 
That table confuses me. What does the heading sum of ranks mean? Is that a label for a column or row?
The top line, labeled Card 1, has numbers 1-7. While this is not incorrect, YOU said that there is no need to go past 6.
On the left hand side you have Card 2 as a heading. My concern is under Card 2, it has the numbers 2,3,4,5,6 and 7. Why don't you have a one?

You say that P(Sum is 6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)P(i, j). What does that P(i,j) at the end mean?
How can P(Sum is 6)=802,704?? You should know that the probability of any event is between 0 and 1 inclusive!

One more point. The number of ways to say get a 1(ie ace) and then a 5 is not just 1 way as there are multiple aces and multiple 5s.
 
Questions that must be answered to solve your problem:
1)How many ways can you pick two cards from a deck? (assume the answer is A)
2) How many ways can the sum be 6? (assume the answer is B)
3) How many ways can the sum be 7? (assume the answer is C)

Then P(sum = 7) > P(sum = 6) is true if and only if C>B
P(sum = 6) =B/A.
P(sum = 7) = C/A.
 
Would you get the same results if you replaced your 52 card deck with:
a) a deck that only had 1-Ace, 1-2,...1-King (totaling 13 cards)?
b) How about with multiple decks?
 
navm said:
6×452⋅452=962,704
Click to expand...
Sorry, but 6×452⋅452=962,704 is not true!
I'm curious about where the 452 is coming from?
What surprises me the most is that neither of 6 or 452 is a factor of 962,704 = 2 x 2 x 2 x 2 x 60169
 
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