#41

Draw a non-replacement tree diagram?

#42

i feel silly putting this much thought into a childrens card game but you guys have no idea how competitive yugioh gets. to make the question easier:

40 cards

8 red 8 blue 8 green 8 yellow 8 purple

i get to draw 6 cards and keep them all in my hand

what are the odds of drawing AT LEAST 1 red?

odds of drawing at least 1 red and 1 blue

odds of drawing at least 1 red 1 blue and 1 green

thats about as much as i need to know.

Or, you could just cheat. place your best hand on the bottom, then take some cards from the middle when shuffling, and your last shuffle you just pick your bottom cards (being your best hand), you instantly destroy your oponents and become master of the millenium items!

#43

i got 50% chance

(8/40)/(32/40)/3 x 6?

I don't really know how to verify that... but

How does the /3 fit in there? Which situation did you do this for; was it the first one with only the re... oh, I see what you did there. I literally just noticed that it was the one with the 3 colors with 8 each. Would I be right to assume that you're dividing by three because of the three colours?

In that case, I think it's wrong because it doesn't seem to take into account that that you will have less cards in the deck- and possibly some cards you want right in your hand- as you pick them up. It's really a shot in the dark, but something in that equation doesn't look right to me.

If you figured it out from that formula you posted, I'd say your guess is better than mine. Just tried to do that formula as I thought Avedas explained it, and I ended up with 76,904,685.

Basically, I don't have a clue how to do this kind of math.

I'm afraid I'll have to use one of my lifelines for this: I'll call a friend and get back to you if he's up to it. He takes lots of advanced math courses in university and should know this.

Care to provide a link to wherever you got that formula?

But how does it help him to know that throwing a dice the probability of throwing a 6 gets close to 1/6 in the long run? All very true, but how does it help in a card game? There are much more different groups he has to draw from, and I doubt he can create a deck that gives him a win with >50% preset probability.

To be honest, I just saw that there was a math problem in this thread and did my best to work on it. How he plays this card game isn't really my concern.

That said, it's the statistic which should get to a more stable value of 1/6, not the probability. I'm not trying to be a dick here, but I think it's important to point out that probability is more or less a theory while the statistics are the actual results.

I guess it would be a matter of figuring out how many of these games he could potentially play without "going in the hole" and balancing it out with the probability he has of getting his desired outcome.

#44

To be honest, I just saw that there was a math problem in this thread and did my best to work on it. How he plays this card game isn't really my concern.

That said, it's the statistic which should get to a more stable value of 1/6, not the probability.I'm not trying to be a dick here, but I think it's important to point out that probability is more or less a theory while the statistics are the actual results.

I guess it would be a matter of figuring out how many of these games he could potentially play without "going in the hole" and balancing it out with the probability he has of getting his desired outcome.

Y do you act like one then? No one cares about naming it an probability or a statistic.

#45

Y do you act like one then? No one cares about naming it an probability or a statistic.

I'm not actually acting like a dick, you're just doing a failing job at trolling me. If you took time to read what I said, you can probably see the importance of making that distinction.

I don't need to elaborate on that further.

#46

I'm not actually acting like a dick, you're just doing a failing job at trolling me. If you took time to read what I said, you can probably see the importance of making that distinction.

I don't need to elaborate on that further.

Dude, you used statistics to work out the probability of 1/6 which I conceive beforehand by logic. Working all this out beforehand also gives you probabilities.

edit: yes. Your 1/6 proven by your, for this thread, useless post about the dice gives you a statistic. When I say a dice will with a probability of 1/6 become 6, it's correct to use the word probability. When talking about the chance of the deck beating people, it's also correct to use the word probability. Please go spout your unrelated crap in some other thread please.

*Last edited by Neo Evil11 at Jan 21, 2013,*

#47

Do your own damn homework, TS...jeese.

#48

This thread makes me sad.

#49

1-(32C6)/(40C6) = 0.764

1-(2*32C6-24C6)/(40C6) = 0.563

1-(3*32C6-2*16C6)/(40C6) = 0.296

I'm not completely certain of the last two, though.

EDIT: A little explanation:

The first answer has already been reached in this thread, but the more simple method is above. Essentially, it's subtracting from 1 the probability of NOT getting at least one red (i.e. getting 0 reds). 32C6 is the number of ways of choosing non-red cards. 40C6 is the number of ways of choosing cards total.

For the second question, 32C6 is the number of ways of not choosing red or not choosing blue (same number of ways because same number of each colour). It is multiplied by two because either not choosing red or not choosing blue means the conditions aren't satisfied. However, 24C6 must be taken from this total because it has been taken into account twice - it is the number of ways of choosing green/yellow/purple, and it is included in 2*32C6 twice but only needed once.

The third question is essentially the same as the second, except 32C6 is multiplied by 3 (not choosing a red, not choosing a blue, not choosing green). This means the number of ways of choosing yellow/purple is included 3 times but only needed once, so 16C6 is subtracted twice from the numerator total.

1-(2*32C6-24C6)/(40C6) = 0.563

1-(3*32C6-2*16C6)/(40C6) = 0.296

I'm not completely certain of the last two, though.

EDIT: A little explanation:

The first answer has already been reached in this thread, but the more simple method is above. Essentially, it's subtracting from 1 the probability of NOT getting at least one red (i.e. getting 0 reds). 32C6 is the number of ways of choosing non-red cards. 40C6 is the number of ways of choosing cards total.

For the second question, 32C6 is the number of ways of not choosing red or not choosing blue (same number of ways because same number of each colour). It is multiplied by two because either not choosing red or not choosing blue means the conditions aren't satisfied. However, 24C6 must be taken from this total because it has been taken into account twice - it is the number of ways of choosing green/yellow/purple, and it is included in 2*32C6 twice but only needed once.

The third question is essentially the same as the second, except 32C6 is multiplied by 3 (not choosing a red, not choosing a blue, not choosing green). This means the number of ways of choosing yellow/purple is included 3 times but only needed once, so 16C6 is subtracted twice from the numerator total.

*Last edited by CrimsonBizzare at Jan 21, 2013,*

#50

Well, I did a bit of research on the subject and I can verify that Incognito's method is correct for the first problem: drawing at least one red (8 reds) from a deck of 40 with 6 draws. Rounded up, it is 76.4%. Like he said, it's easier to find the likelihood of something not happening. Then you subtract that value from 1.

However, I'm rather skeptical about the ones involving more than one colour. If we were talking about the odds for drawing a red if there were 16 or 24 of them, I could vouch for that. But, there are two and three colours in the other problems and I suspect that might change the calculation.

Still working on it.

If you want to know the reasons I know you're trolling, just send me your guesses via PM rather than spamming up this thread. However, if you actually care you're going to have to work for it; I'll be glad to tell you if you're getting hotter/colder, or if you nail it. You're going to have to wait until I'm done in this thread though.

Keep trolling here and I'll just report you and be done with it. If you get a warning here, all it's going to do is annoy you and possibly make me an enemy on a board I visit regularly. It's in both of our interests to stay civil about this.

I'm more than willing to forget about this, and look forward to this ending right here. I sincerely apologize if I have offended you; I honestly expected you to find the comment about you trolling to be rather lighthearted rather than insulting. However, I stand by what I have said in this thread.

However, I'm rather skeptical about the ones involving more than one colour. If we were talking about the odds for drawing a red if there were 16 or 24 of them, I could vouch for that. But, there are two and three colours in the other problems and I suspect that might change the calculation.

Still working on it.

Dude, you used statistics to work out the probability of 1/6 which I conceive beforehand by logic. Working all this out beforehand also gives you probabilities.

edit: yes. Your 1/6 proven by your, for this thread, useless post about the dice gives you a statistic. When I say a dice will with a probability of 1/6 become 6, it's correct to use the word probability. When talking about the chance of the deck beating people, it's also correct to use the word probability. Please go spout your unrelated crap in some other thread please.

If you want to know the reasons I know you're trolling, just send me your guesses via PM rather than spamming up this thread. However, if you actually care you're going to have to work for it; I'll be glad to tell you if you're getting hotter/colder, or if you nail it. You're going to have to wait until I'm done in this thread though.

Keep trolling here and I'll just report you and be done with it. If you get a warning here, all it's going to do is annoy you and possibly make me an enemy on a board I visit regularly. It's in both of our interests to stay civil about this.

I'm more than willing to forget about this, and look forward to this ending right here. I sincerely apologize if I have offended you; I honestly expected you to find the comment about you trolling to be rather lighthearted rather than insulting. However, I stand by what I have said in this thread.

#51

If you want to know the reasons I know you're trolling, just send me your guesses via PM rather than spamming up this thread. However, if you actually care you're going to have to work for it; I'll be glad to tell you if you're getting hotter/colder, or if you nail it. You're going to have to wait until I'm done in this thread though.

Keep trolling here and I'll just report you and be done with it. If you get a warning here, all it's going to do is annoy you and possibly make me an enemy on a board I visit regularly. It's in both of our interests to stay civil about this.

I'm more than willing to forget about this, and look forward to this ending right here. I sincerely apologize if I have offended you; I honestly expected you to find the comment about you trolling to be rather lighthearted rather than insulting. However, I stand by what I have said in this thread.

Except I am correct and you have twice now completely acted like a retard instead of give arguments. You can;'t wiggle your way out of this. Snob.

#52

Oh fuck, I was completely wrong in my calculations of the last two. I was working on the chance of not getting red1-(32C6)/(40C6) = 0.764

1-(2*32C6-24C6)/(40C6) = 0.563

1-(3*32C6-2*24C6)/(40C6) = 0.362

I'm not completely certain of the last two, though.

*or*blue

Yeah, those last two are wrong. Can't be bothered fixing them, though, I'll just assume Crimson is right.Well, I did a bit of research on the subject and I can verify that Incognito's method is correct for the first problem: drawing at least one red (8 reds) from a deck of 40 with 6 draws. Rounded up, it is 76.4%. Like he said, it's easier to find the likelihood of something not happening. Then you subtract that value from 1.

However, I'm rather skeptical about the ones involving more than one colour. If we were talking about the odds for drawing a red if there were 16 or 24 of them, I could vouch for that. But, there are two and three colours in the other problems and I suspect that might change the calculation.

*Last edited by lncognito at Jan 21, 2013,*

#53

I ****ING HATE S1

IT IS THE DEVILS DICK

AND THEY MAKE YOU SUCK IT

THAT SHIT IS SO NOT RAVEN

):<

lowercase

IT IS THE DEVILS DICK

AND THEY MAKE YOU SUCK IT

THAT SHIT IS SO NOT RAVEN

):<

lowercase

#54

Except I am correct and you have twice now completely acted like a retard instead of give arguments. You can;'t wiggle your way out of this. Snob.

I don't argue with trolls- nevermind ones who insult me. It just doesn't accomplish anything...

Reported.

#55

I don't argue with trolls- nevermind ones who insult me. It just doesn't accomplish anything...

Reported.

How is my post about probability and statistics wrong? It comes directly from the textbook. You're really trolling.

#56

In this case, a statistic is a function of the random variable used to describe the cards. Please stop making shit up.

Apparently nobody here understands the concepts of combinations and actively chooses to ignore it.

1-(32C6)/(40C6) = 0.764

1-(2*32C6-24C6)/(40C6) = 0.563

1-(3*32C6-2*16C6)/(40C6) = 0.296

I'm not completely certain of the last two, though.

EDIT: A little explanation:

The first answer has already been reached in this thread, but the more simple method is above. Essentially, it's subtracting from 1 the probability of NOT getting at least one red (i.e. getting 0 reds). 32C6 is the number of ways of choosing non-red cards. 40C6 is the number of ways of choosing cards total.

For the second question, 32C6 is the number of ways of not choosing red or not choosing blue (same number of ways because same number of each colour). It is multiplied by two because either not choosing red or not choosing blue means the conditions aren't satisfied. However, 24C6 must be taken from this total because it has been taken into account twice - it is the number of ways of choosing green/yellow/purple, and it is included in 2*32C6 twice but only needed once.

The third question is essentially the same as the second, except 32C6 is multiplied by 3 (not choosing a red, not choosing a blue, not choosing green). This means the number of ways of choosing yellow/purple is included 3 times but only needed once, so 16C6 is subtracted twice from the numerator total.

Apparently nobody here understands the concepts of combinations and actively chooses to ignore it.

*Last edited by Avedas at Jan 21, 2013,*

#57

You can always pull a purple card and cut it in half. You'll have an extra red AND an extra blue that way.

#58

= nCk = nCr

k and r are just variables, they mean the same thing but wikipedia uses k instead of r.

most calculators have an nCr button

it is the combination button, which deals with selection where order does not matter

40 cards

8 red 8 blue 8 green 8 yellow 8 purple

i get to draw 6 cards and keep them all in my hand

ok this gets a little gay which the whole "at least" bit, but whatever

have to go 100% - the probability of getting no reds

its just way easier to do 100% - whatever

what are the odds of drawing AT LEAST 1 red?

ways of getting 5 cards without any of them red = 32 cards choose 6 = 32 C 6 =906192

ways of drawing 6 cards = 40 C 6= 3838380

906192/3838380=

odds of not getting a single red = 23.6087992%

odds of drawing at least 1 red and 1 blue

24C6/40C6

odds of drawing at least 1 red 1 blue and 1 green

16C6/40C6

^This is an e-mail I got back yesterday (with a picture missing right at the top). There's some important things not explained properly, but it doesn't seem to have a very significant inaccuracy- as far as my interpretation of it goes anyway.

I believe I'm pretty close to the correct solution, I just need to figure out the amount of possible hands you can get with some of the colours and not all of them; like 1 red, 1 green and no blue. How would that be done?

#59

are you drawing with replacement or no? makes sort of a big difference. if yes, use permutations. if no, use combinations

If my understanding of combinations/permutations is right...

The possible orders you can draw the cards in a deck of 40 is figured out through permutation. Basically, you shuffle your deck into a different permutation of the same cards. You get the combination in your hand of six from a permuted deck of 40 cards.

You could say that the order you received your combination of cards is one specific permutation of them, but other than that I don't think permutation has any practical value in this equation. One way this is true is demonstrated below:

1-(nPr(32,6))/(nPr(40,6))= 0.7639129007550061 (76.4%)

1-(nCr(32,6))/(nCr(40,6))= 0.7639129007550061

Even though the amount of permutations for this deck is much higher than the combinations, the proportions of them are identical.

There is only one problem I have with my friend's solution. It doesn't take into account that getting 1 red, 1 blue, and 0 greens still isn't getting all three cards you want; you would have to figure out the specific amount of possible hands for the in-betweens as well. At the same time, a >95% chance for the questions with >1 colour leaves little point for going further.

If you check the link Avedas put on the first page, you will find a nice explanation of the difference between the two types of permutations and combinations as well as a few ways to use these mathematically; I used that link, my friend's e-mail, and this webpage: http://www.picalc.com/ (a calculator) to figure this out.

Anything wrong with any of this?

*Last edited by JimDawson at Jan 23, 2013,*