3 spoil set odds

Post by **IcePack** » Tue Oct 11, 2011 5:25 pm

Just curious, what is the mathematical odds on someone having a set with 3 cards?
I haven't seen it anywhere.

Thanks!

chapcrap · Post by **chapcrap** » Tue Oct 11, 2011 9:03 pm

IcePack wrote:Just curious, what is the mathematical odds on someone having a set with 3 cards?
I haven't seen it anywhere.

Thanks!

Well, the first card doesn't matter. And neither doesn't the second card. So, they don't factor in the odds. It's only the third card that matters. So, the percentages are 33%. No matter what the first 2 cards are, there is a third card that will allow you to cash if you get the right card.

Porteiro · Post by **Porteiro** » Wed Oct 12, 2011 3:11 pm

Yup. No matter what the first two cards are, for the third card, there's 1 color that will give you a set, and two that will not. The odds are 1 out of 3.

If no set was made, for the 4th card there are 2 colors that will give a match, and 1 that won't, so the odds for that are 2 out of 3.

Then for the fifth card, every color will give you a set, giving you odds of 3 out of 3, or 100%.

Metsfanmax · Post by **Metsfanmax** » Wed Nov 02, 2011 12:19 am

chapcrap wrote:
IcePack wrote:Just curious, what is the mathematical odds on someone having a set with 3 cards?
I haven't seen it anywhere.

Thanks!
Well, the first card doesn't matter. And neither doesn't the second card. So, they don't factor in the odds. It's only the third card that matters. So, the percentages are 33%. No matter what the first 2 cards are, there is a third card that will allow you to cash if you get the right card.

This does not mean that the odds are 33%. Once you select two given cards and ask for the probability of the third making a set, the answer is indeed 33%. But the odds of any random group of three cards making a set are actually 40%. A little bit counter-intuitive perhaps, but anyone who has studied the Monty Hall problem in any detail should know that calculating odds is rarely intuitive.

Geger · Post by **Geger** » Wed Nov 02, 2011 1:06 am

Metsfanmax wrote:
chapcrap wrote:
IcePack wrote:Just curious, what is the mathematical odds on someone having a set with 3 cards?
I haven't seen it anywhere.

Thanks!
Well, the first card doesn't matter. And neither doesn't the second card. So, they don't factor in the odds. It's only the third card that matters. So, the percentages are 33%. No matter what the first 2 cards are, there is a third card that will allow you to cash if you get the right card.
This does not mean that the odds are 33%. Once you select two given cards and ask for the probability of the third making a set, the answer is indeed 33%. But the odds of any random group of three cards making a set are actually 40%. A little bit counter-intuitive perhaps, but anyone who has studied the Monty Hall problem in any detail should know that calculating odds is rarely intuitive.

No Idea about Monty Hall Problem, but I just wrote down all 27 possibilities in a spreadsheet and found exact 9 sets. Also 9/27 = 1/3

Metsfanmax · Post by **Metsfanmax** » Wed Nov 02, 2011 1:28 am

Geger wrote: No Idea about Monty Hall Problem, but I just wrote down all 27 possibilities in a spreadsheet and found exact 9 sets. Also 9/27 = 1/3

You're calculating a slightly different number than I was referring to. You get 1/3 if you consider sets where the order matters and 2/5 if you consider sets where the order doesn't matter, assuming equal likelihoods for each combination. From a combinatorical point of view, RRB is the same as BRR in terms of (not) meeting the criterion for making a set. But maybe that level of analysis is probably more confusing than what the OP asked for, and he can be satisfied with chap's (more intuitive) response.

chapcrap · Post by **chapcrap** » Wed Nov 02, 2011 8:33 pm

Metsfanmax wrote:
Geger wrote: No Idea about Monty Hall Problem, but I just wrote down all 27 possibilities in a spreadsheet and found exact 9 sets. Also 9/27 = 1/3
You're calculating a slightly different number than I was referring to. You get 1/3 if you consider sets where the order matters and 2/5 if you consider sets where the order doesn't matter, assuming equal likelihoods for each combination. From a combinatorical point of view, RRB is the same as BRR in terms of (not) meeting the criterion for making a set. But maybe that level of analysis is probably more confusing than what the OP asked for, and he can be satisfied with chap's (more intuitive) response.

27 Possiblities and 9 chances to cash in. Where do you get 40 %

Card 1 Card 2 Card 3
Red

Red
- Red
  Blue
  Green
Blue
- Red
  Blue
  Green
Green
- Red
  Blue
  Green

Blue

Red
- Red
  Blue
  Green
Blue
- Red
  Blue
  Green
Green
- Red
  Blue
  Green

Green

Red
- Red
  Blue
  Green
Blue
- Red
  Blue
  Green
Green
- Red
  Blue
  Green

Metsfanmax · Post by **Metsfanmax** » Wed Nov 02, 2011 10:14 pm

chapcrap wrote: 27 Possiblities and 9 chances to cash in. Where do you get 40 %

Card 1 Card 2 Card 3
Red
Red
Red
Blue
Green
Blue
Red
Blue
Green
Green
Red
Blue
Green
Blue
Red
Red
Blue
Green
Blue
Red
Blue
Green
Green
Red
Blue
Green
Green
Red
Red
Blue
Green
Blue
Red
Blue
Green
Green
Red
Blue
Green

40% came from a different analysis -- that is, 40% of the (10 total) non-ordered combinations of R+B+G can make a set:

RRR
BBB
GGG
RBG
GGR
GGB
RRB
RRG
BBG
BBR

The 33% comes from the additional information that each card is chosen randomly, that is to say, you have an equal probability of getting R/B/G, which implies non-equal probabilities of the above combinations.

chapcrap · Post by **chapcrap** » Thu Nov 03, 2011 2:30 am

Metsfanmax wrote:40% came from a different analysis -- that is, 40% of the (10 total) non-ordered combinations of R+B+G can make a set:

RRR
BBB
GGG
RBG
GGR
GGB
RRB
RRG
BBG
BBR

The 33% comes from the additional information that each card is chosen randomly, that is to say, you have an equal probability of getting R/B/G, which implies non-equal probabilities of the above combinations.

I have to disagree. I understand what you are saying, but if each card comes at an equal ratio, then it stands that each of the 27 combos come equally. Basically, BBR is more common because there are 3 different ways to acheive BBR. Just because there are 10 actual card combinations does not mean they all come equally. Your 40% is only correct if they all come equally, which they do not.

Geger · Post by **Geger** » Thu Nov 03, 2011 3:39 am

chapcrap wrote:
Metsfanmax wrote:
Geger wrote: No Idea about Monty Hall Problem, but I just wrote down all 27 possibilities in a spreadsheet and found exact 9 sets. Also 9/27 = 1/3
You're calculating a slightly different number than I was referring to. You get 1/3 if you consider sets where the order matters and 2/5 if you consider sets where the order doesn't matter, assuming equal likelihoods for each combination. From a combinatorical point of view, RRB is the same as BRR in terms of (not) meeting the criterion for making a set. But maybe that level of analysis is probably more confusing than what the OP asked for, and he can be satisfied with chap's (more intuitive) response.
27 Possiblities and 9 chances to cash in. Where do you get 40 %

Card 1 Card 2 Card 3
Red
Red
Red
Blue
Green
Blue
Red
Blue
Green
Green
Red
Blue
Green
Blue
Red
Red
Blue
Green
Blue
Red
Blue
Green
Green
Red
Blue
Green
Green
Red
Red
Blue
Green
Blue
Red
Blue
Green
Green
Red
Blue
Green

Ho ho... I wrote the combos in excel, you wrote here...

After reading about Monty Problem in wiki and math-forum, the problem here doesn't belong to Monty Problem

Metsfanmax · Post by **Metsfanmax** » Thu Nov 03, 2011 7:04 am

chapcrap wrote: I have to disagree. I understand what you are saying, but if each card comes at an equal ratio, then it stands that each of the 27 combos come equally. Basically, BBR is more common because there are 3 different ways to acheive BBR. Just because there are 10 actual card combinations does not mean they all come equally. Your 40% is only correct if they all come equally, which they do not.

Yes, each one of the 27 is equally likely, which implies that each one of those 10 is not. That's what I was saying in the post.

chapcrap · Post by **chapcrap** » Thu Nov 03, 2011 4:27 pm

Metsfanmax wrote:
chapcrap wrote: I have to disagree. I understand what you are saying, but if each card comes at an equal ratio, then it stands that each of the 27 combos come equally. Basically, BBR is more common because there are 3 different ways to acheive BBR. Just because there are 10 actual card combinations does not mean they all come equally. Your 40% is only correct if they all come equally, which they do not.
Yes, each one of the 27 is equally likely, which implies that each one of those 10 is not. That's what I was saying in the post.

Which then means it is 33% and not 40%, right? Are you agreeing with me now or still saying 40%

Teflon Kris · Post by **Teflon Kris** » Thu Nov 03, 2011 5:00 pm

From how we were taught probability at A-level the order of the combinations does matter, meaning the 9 of 27 method.

Not sure why though - it would probably take a maths genius to prove one method over the other.

Metsfanmax · Post by **Metsfanmax** » Fri Nov 04, 2011 7:33 pm

DJ Teflon wrote:From how we were taught probability at A-level the order of the combinations does matter, meaning the 9 of 27 method.

Not sure why though - it would probably take a maths genius to prove one method over the other.

It's because randomness is ensured at the individual card level when the cards are dealt, not at the set level.

3 spoil set odds

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