- #1

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Major help needed

I have tried everything in this problem my answer was 1/8 which was wrong I hope you guys can do better.

- Thread starter stupidkid
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- #1

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Major help needed

I have tried everything in this problem my answer was 1/8 which was wrong I hope you guys can do better.

- #2

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Well how many different sets of 4 cards can you get that are in descending order? (i.e. {36, 35, 34, 33} is one, {35, 34, 33, 32} is another... how many are there?)

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- #3

matt grime

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- #4

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- #5

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but there can 1000000000000000s of possibilties for eg 7,6,5,1 or 10,6,3,2

like that

like that

- #6

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Actually, your estimate is much larger than the actual number of possibilities. Try to think of it this way:stupidkid said:but there can 1000000000000000s of possibilties for eg 7,6,5,1 or 10,6,3,2

like that

How many possibilities are there for the first card? (Ans: Obviously 36)

How many possibilities are there for the second card? (Hint: Not 36)

How many possibilities are there for the third card?

How many possibilities are there for the fourth card?

Now you can multiply them up and it gives you the total number of possibilities for a 4-card deal.

- #7

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its not as easy as that.

- #8

matt grime

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- #9

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You can add the number of ways you can get a descending order where your highest card (call it M) is 4 (since M can't be less than 4), plus the number of possibilities when M=5, M=6, all the way to M=36.

The way of counting the number of ways to get a descending order when your highest card is M is (at least I think) easiest to see in the form of code:

possible_descending_hands = 0;

for (i=3, i < M, i = i + 1){

for (j=2, j < i, j = j + 1){

for (k=1, k < j, k = k + 1){

possible_descending_hands = possible_descending_hands + 1;

}

}

}

Then of course, you loop again for M = [4,36] and that would be your number of possible descending hands. Note that I'm not suggesting you write a program for this. It's just that it's easiest to explain how to count your descending hands with the use of loops (you can yet convert that into a mathematical expression).

[Edit: Hmm.. not allowed to put spaces at the beginning of lines. I guess that means no indentation for you :P]

- #10

matt grime

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1. dealing 1 card in descending order: 1/1

2. Dealing 36 cards in descending order 1/36!

and i thought i gave sufficient hints last time!

- #11

uart

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For the case of consecetively decending it's also very easy.

The first card must be 4 or more which has a probabilty of 33/36. After that there is no choice which card must come next so the probability is just 33/36 * 1/35 * 1/34 * 1/33 = 1/(34*35*36).

The first card must be 4 or more which has a probabilty of 33/36. After that there is no choice which card must come next so the probability is just 33/36 * 1/35 * 1/34 * 1/33 = 1/(34*35*36).

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- #12

matt grime

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the question doesn't say they are in sequence.

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