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The answer i posted is correct...i verified with Teachers Assistant.
You consider the third case i posted because you can have one card which of same suit at lets say card 2 and can also be the same value as card 1. So yea three cases, add them together and divide by total possibilities.
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Hi careless25;
You might be right, I do not know yet. I am pretty sure mine is wrong.
Hi gAr;
I am getting his 10296 and 102960 but I am getting 51480 instead of his 41184 for one suit and one rank the same.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi bobbym,
I do not understand why combinations are used here.
I think permutation's appropriate.
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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Hi gAr;
I agree, but as long as you use the same for both numerator and denominator the probability remains the same.
For instance for a badugi:
Is everything okay?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi,
Yes, I understand.
But when we use the way which is natural, there will be less confusion.
For careless25's 3 cases, I get:
52*36*22*(30)
52*36*22*(3)
52*36*22*(6)
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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Hi;
So we agree on the 51480. It is a tricky problem.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi bobbym,
That's strange, your probability would go above 60%!
Is the denominator 52C4 ?
Hi careless25,
For M.I.,
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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Hi gAr;
Yes, that is what the program is getting.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi,
So we agree on the 51480
If we agree on the numerator, what's going wrong with the final answer?!
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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Hi;
What do you get for a probability on yours?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Much less!
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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I know I had that too, I think I tried every answer so far.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Okay.
You mentioned about the program, what does it do?
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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Takes all possible 4 permutations of 52. There are only 6 497 400 ways to arange them. Then it looks for 3 patterns {3,3} which means 3 different suits and 3 different ranks.{4,3} which means 4 different ranks and 3 different suits and last {3,4} which means 3 different ranks and 4 different suits.
Takes the whole sample space and checks them all for the conditions badugi demands.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi bobbym,
Is the answer in post #52 the output of your program?
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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Yes, I finally got it to run. This turned out to be a difficult program to write in the functional style. The last few have been like that.
That is the output it is getting. But programs are written by the person solving the problem, they reflect his biases so on simulations they are sometimes unreliable.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Okay.
I wrote a procedural program, since only counting is required.
Anyway, I believe it must be less than 35%.
P(picking up the 4th card, which results in badugi or not) = 52*36*24*49 / (52*51*50*49)
So our answer must be less than that.
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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Yes, but what about picking up a second card that ruins a badugi?
Does that not assume that you choose three good cards and then a bad fourth one. How about First a good one then a bad one followed by two good ones.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hmmm, I understand why we are getting different values.
I'll check.
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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Hi bobbym,
After correcting my program:
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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Hi gAr;
What did you get for each case. On another forum they agree with my answer of 51480 instead of his 41184.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi bobbym,
1. diff value, same suit as one card : 2471040
2. diff suit, same value as one card : 247104
3. same suit, same value with another card : 988416
which agrees...
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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Hi gAr;
1) 2471040 Kayrect!
2) 247104 Kayrect!
3) 988416 This is the one that is in dispute. 1 235 520 is what I am getting.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hmmm, perhaps you overcounted something?
Did you understand the OP's solution?
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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Hi gAr;
Yes, but I never agreed with that 3rd answer, the other two yes. I posted that one particular part on another forum. This was the reply.
First we calculate in how many ways can we have exactly 2 same ranks & suits .
With 4 suits we can have exactly 2 same in[math]4 \binom{3}{2} =12 ways and with the13 ranks we can have exactly 2 same in
13 \binom{12}{2}=858 ways.
Together we can have exactly 2 same ranks & suits in 12 * 858= 10296 ways.
Now for a specific set among these....for example if among the 4 cards we have ranks K,K,J & 2,and suits s,s,d &h...let's see in how many ways can we distribute these ranks among the suits...
If both s are K,then no. ways the rest can be distributed is 2!=2
If no s are K, then no. ways the rest can be distributed is 1
If exactly one s is K the no. ways the rest can be distributed is 2!=2.
These are the only cases and hence all-together there are 5 ways for a definite of choice suits & ranks.
So total no. of possible winning arrangements is 5*10296=51480
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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