Odds Question (1 Viewer)

[Short Corner]

Shouldn't it just be  3/4 * 2/3 * 1/2 ? That gives us the number of "winning options" over the total number of options while accounting for no repetition. 

This isn't always the case.  If a 2 is picked first, the probability for the second card fitting is 3/3.

 

[Short Corner]

I think it might be easier to count (ie produce an algorithm for) the number of ways that fail

 

[Short Corner]

I get....

(n! - 2(n-1)+1)/n!   or    1 - (2(n-1)-1)/n!

 

[Ilov80s]

9 out of 24 is 37.5%

I'm likely thinking about it wrong, but if 2 is in the first spot, it can't be in the 3rd or 4th spot so it reduces the odds of those spaces being safe. 

Redacted my very stupid comment 

 

[Ilov80s]

This isn't always the case.  If a 2 is picked first, the probability for the second card fitting is 3/3.

True but if 2 comes first, it's improving the odds of winning the 2nd spot while reducing the odds of winning the 3rd and 4th spot. I don't know, trying to work out and think about this is likely just making me sound dumb!

 

[Short Corner]

We know that n! is the number of possible permutations.

We then need to count the number of ways, p, where p is the number of failures.  This would be nC1, nC2,....nCn

Adding these terms up would yield 2n-1-1.  Therefore the number of successes is n! - (2n-1-1)

 

[MikeMan]

I get....

(n! - 2(n-1)+1)/n!   or    1 - (2(n-1)-1)/n!

We know that n! is the number of possible permutations.

We then need to count the number of ways, p, where p is the number of failures.  This would be nC1, nC2,....nCn

Adding these terms up would yield 2n-1-1.  Therefore the number of successes is n! - (2n-1-1)

Does that yield 44/120 for 5 cards?

 

[Galileo]

I'm likely thinking about it wrong, but if 2 is in the first spot, it can't be in the 3rd or 4th spot so it reduces the odds of those spaces being safe. So you are saying there are better odds (35%) of me flipping through 4 cards and not getting a 1 in the 1 spot or 2 in the spot or a 3 in the 3 spot or a 4 in the 4 spot than there is of me just flipping over the first card and having it be a 1 (25%)? Maybe I'm misunderstanding.

I think it is an entirely different question if you know the outcome of one or more cards already.  I am just thinking about it in term of how many permutations are possible and how many of those satisfy the out come of not having a numbered card in its ordered spot.  With N=4, there are 24 possible results of shuffling the cards.  9 of those outcomes satisfy the requirement that no card is in its ordered spot.  

 

[Galileo]

We know that n! is the number of possible permutations.

We then need to count the number of ways, p, where p is the number of failures.  This would be nC1, nC2,....nCn

Adding these terms up would yield 2n-1-1.  Therefore the number of successes is n! - (2n-1-1)

This does not work for n=4 as demonstrated above in abbottjamesr's post

 

[Ilov80s]

I think it is an entirely different question if you know the outcome of one or more cards already.  I am just thinking about it in term of how many permutations are possible and how many of those satisfy the out come of not having a numbered card in its ordered spot.  With N=4, there are 24 possible results of shuffling the cards.  9 of those outcomes satisfy the requirement that no card is in its ordered spot.  

Yeah I'm dumb lol, got mixed up on a basic thing 

 

[Hot Diggity Dog]

8 ball says maybe.

 

[Ignoratio Elenchi]

2 cards - 1/2

3 cards 2/6

4 cards 9/24

Does that yield 44/120 for 5 cards?

These are all correct.  So what is it for 10 cards?

 

[Ilov80s]

These are all correct.  So what is it for 10 cards?

so the odds go from 50% to 33.3% to 37.5% to 36.6% so it's going to be some kind function where x is raised to a power and being even or odd causes it change directions, right? That is where the dips are coming from? 

 

[Ignoratio Elenchi]

so the odds go from 50% to 33.3% to 37.5% to 36.6% so it's going to be some kind function where x is raised to a power and being even or odd causes it change directions, right? That is where the dips are coming from? 

Yes, you have the right idea.

 

[shuke]

466 111 55

88

2

0457 745 008197157 00047

14783505

143555

541423

333333333333333333333

59722665

01 02 03 04 05 06 789

4

 

[Buckychudd]

466 111 55

88

2

0457 745 008197157 00047

14783505

143555

541423

333333333333333333333

59722665

01 02 03 04 05 06 789

4

The Kirkland Calculation only works if you are using the Perpetuity Therom. 

 

[abbottjamesr]

so the odds go from 50% to 33.3% to 37.5% to 36.6% so it's going to be some kind function where x is raised to a power and being even or odd causes it change directions, right? That is where the dips are coming from? 

This isn't right... the odds go 50%, 33.3%, 37.5%, 40%, 41.67%, 42.8%, 43.75%, 44.4%, 45% at n=10.

As n gets large the odds will approach 50%

You have a 1/n chance of getting a 1 on the first turn.  If you survive that card you have a 50% chance of getting through the rest regardless of n.

 

[Ilov80s]

This isn't right... the odds go 50%, 33.3%, 37.5%, 40%, 41.67%, 42.8%, 43.75%, 44.4%, 45% at n=10.

As n gets large the odds will approach 50%

You have a 1/n chance of getting a 1 on the first turn.  If you survive that card you have a 50% chance of getting through the rest regardless of n.

OK, I was just looking at the work other people were posting. I don't think others here fully agree with you. Although it looks like someone edited a previous response and removed some of their work.

 

[Galileo]

This isn't right... the odds go 50%, 33.3%, 37.5%, 40%, 41.67%, 42.8%, 43.75%, 44.4%, 45% at n=10.

As n gets large the odds will approach 50%

You have a 1/n chance of getting a 1 on the first turn.  If you survive that card you have a 50% chance of getting through the rest regardless of n.

This is not right.  When N = 5, there are 44/120 possibilities which is less than 40%.  With 5 cards, there are 120 different outcomes (5!).  I am sure we agree on that.  We can eliminate every outcome that has card 1 in the first spot (24 possibilities) leaving only the outcomes where 2, 3, 4, or 5 are in the first spot.  There are 24 more outcomes for each of these beginning cards, but only 11 of the 24 are valid.  When 2 cards, 3 cards, and 4 cards are used, half of the remaining (ones without 1 card in lead) possible out comes were valid, but less than half are valid when N=5.  For example, here are the permutations with card 2 in the lead spot...

21345 x

21354 x

21435 x

21453 

21534

21543 x

23145 x

23154

23415 x

23451

23514

23541 x

24135 x

24153

24315 x

24351 x

24513 

24531

25134

25143 x

25314 x

25341 x

25413

25431

That is 11 out of 24 with the 2 card in the lead spot.  Repeat for the 3 card, 4 card, and 5 card and you total 4 x 11 = 44 possible desired outcomes out of 120 or 36.7%

 

[abbottjamesr]

This is not right.  When N = 5, there are 44/100 possibilities which is less than 40%.  With 5 cards, there are 120 different outcomes (5!).  I am sure we agree on that.  We can eliminate every outcome that has card 1 in the first spot (24 possibilities) leaving only the outcomes where 2, 3, 4, or 5 are in the first spot.  There are 24 more outcomes for each of these beginning cards, but only 11 of the 24 are valid.  With 2 cards, 3 cards, and 4 cards, half of the remaining possible out comes were valid, but less than half are valid for N=5.  For example, here are the permutations with card 2 in the lead spot...

21345 x

21354 x

21435 x

21453 

21534

21543 x

23145 x

23154

23415 x

23451

23514

23541 x

24135 x

24153

24315 x

24351 x

24513 

24531

25134

25143 x

25314 x

25341 x

25413

25431

That is 11 out of 24 with the 2 card in the lead spot.  Repeat for the 3 card, 4 card, and 5 card and you total 4 x 11 = 44 possible desired outcomes out of 120 or 36.7%

You are correct.  I miss counted.  

 

[Short Corner]

OK, I was just looking at the work other people were posting. I don't think others here fully agree with you. Although it looks like someone edited a previous response and removed some of their work.

Not sure if you are referring to me but I edited one of my responses to take advantage of the cool super/subscript formatting.  Didn't remove any work though.

Looking back at producing am algorithm, this looks like it may be similar to a Rook Polynomial.  I have to look up the mechanics but imagine an nxn chess board where you must place n rooks where none are attacking another and none are on the main diagonal.

 

[Ilov80s]

Not sure if you are referring to me but I edited one of my responses to take advantage of the cool super/subscript formatting.  Didn't remove any work though.

Looking back at producing am algorithm, this looks like it may be similar to a Rook Polynomial.  I have to look up the mechanics but imagine an nxn chess board where you must place n rooks where none are attacking another and none are on the main diagonal.

Not you, but I actually think the word that was edited was actually correct in the first place. 

 

[Ilov80s]

This is not right.  When N = 5, there are 44/120 possibilities which is less than 40%.  With 5 cards, there are 120 different outcomes (5!).  I am sure we agree on that.  We can eliminate every outcome that has card 1 in the first spot (24 possibilities) leaving only the outcomes where 2, 3, 4, or 5 are in the first spot.  There are 24 more outcomes for each of these beginning cards, but only 11 of the 24 are valid.  With 2 cards, 3 cards, and 4 cards, half of the remaining possible out comes were valid, but less than half are valid for N=5.  For example, here are the permutations with card 2 in the lead spot...

That is 11 out of 24 with the 2 card in the lead spot.  Repeat for the 3 card, 4 card, and 5 card and you total 4 x 11 = 44 possible desired outcomes out of 120 or 36.7%

I edited that so as not to confuse anyone farther 

 

[Galileo]

I edited that so as not to confuse anyone farther 

fixed in original as well.

 

[Short Corner]

I edited that so as not to confuse anyone farther 

You should have said further.

Couldn't help myself.

 

[Ilov80s]

You should have said further.

Couldn't help myself.

I speak the Queen's English mate. 

 

[Galileo]

When N = 6, I am coming up with 225/720 or 31.25%

 

[Ignoratio Elenchi]

When N = 6, I am coming up with 225/720 or 31.25%

The numerator is incorrect (too low).  

 

[abbottjamesr]

1/e.  (Or at least, pretty close to that.)

Longer answer, provided without proof: 

Given a(1) = 0 and the recurrence relation a = n*a(n-1) + (-1)^n, then if you have n cards numbered 1 to n in random order, the probability that no card's rank will match its place in the deck is a / n!  That sequence converges to 1/e pretty quickly. 

I've changed my answer to this.  

Did a quick table in Xcel and n = 7 is 265/720 = .368056

1/e ~ .367879

 

[Galileo]

Did a quick table in Xcel and n = 7 is 265/720 = .368056

I am getting this now as well, but for N = 6, not 7.

 

[Ilov80s]

I am getting this now as well, but for N = 6, not 7.

I would think just based on 6!= 720 you are correct

 

[Galileo]

I would think just based on 6!= 720 you are correct

I think it was just a typo on his part, sort of like my 100 instead of 120 earlier...

 

[Ilov80s]

I think it was just a typo on his part, sort of like my 100 instead of 120 earlier...

Funny how some mistakes never change. Whether I am helping low level math student in pre-algebra or a strong student in Algebra 2/geometry, the same 2 problems seem to make-up 70% of the mistakes: an error with a calculation on a negative number or there was a "typo" in copying the problem/transcribing their work. 

Also, don't worry, for anyone embarrassed by my use of math here. I am not a"math teacher" (actually only took 1 math class in college) 

 

[abbottjamesr]

I am getting this now as well, but for N = 6, not 7.

Sorry typing faster than I think. yes n=6

 

[abbottjamesr]

Funny how some mistakes never change. Whether I am helping low level math student in pre-algebra or a strong student in Algebra 2/geometry, the same 2 problems seem to make-up 70% of the mistakes: an error with a calculation on a negative number or there was a "typo" in copying the problem/transcribing their work. 

Also, don't worry, for anyone embarrassed by my use of math here. I am not a"math teacher" (actually only took 1 math class in college) 

I took more math than I care to admit in college.  But it was the usefull kind like differential equations not this proof stuff

 

[Ilov80s]

I took more math than I care to admit in college.  But it was the usefull kind like differential equations not this proof stuff

My class was math for morons (I am not a moron, but I chose it because of how easy it would be). It was like basic odds (chances of picking a king out of a full deck of cards) and math with fractions and converting pounds to ounces. It was so stupid, but there actually people there that thought the class was hard. I do kind of regret not taking more as I enjoy math now. 

It's funny because my students will often say, "I won't need to use this ever." That is what I thought and now I spend 50% of my work time doing math or chemistry, basically had to relearn it all. 

 

[Short Corner]

I took more math than I care to admit in college.  But it was the usefull kind like differential equations not this proof stuff

Enumeration is pretty useful, especially if you can tweak your algorithm and apply it to subtle changes.  Still working on the Rook Polynomial solution, going to dig my combinatorics book out of the basement tonight.  Nice to give the brain a workout.

His an old standard to ponder...

A hot dog stand sells 3 types of dogs, regular, chili, and super.  How many ways can you order 6 dogs?  7?  n?

 

[Ignoratio Elenchi]

His an old standard to ponder...

A hot dog stand sells 3 types of dogs, regular, chili, and super.  How many ways can you order 6 dogs?  7?  n?

Spoiler

Let k be the number of hot dog types there are to choose from (in this case, k = 3).  The number of ways you can order n dogs is [(n+k-1) choose n], or (n+k-1)! / n!(k-1)!

So there are 8!/6!2! = 28 ways to order 6 dogs from a menu with 3 choices.  

 

[Short Corner]

Spoiler

Let k be the number of hot dog types there are to choose from (in this case, k = 3).  The number of ways you can order n dogs is [(n+k-1) choose n], or (n+k-1)! / n!(k-1)!

So there are 8!/6!2! = 28 ways to order 6 dogs from a menu with 3 choices.  

I find it easier to conceptualize using n+k-1Ck-1 but six of one....

also, how do you do a spoiler box on here?

 

[Ilov80s]

729 

 

[Short Corner]

729 

729?

 

[Man of Constant Sorrow]

N=nerd/

N raised to infinity= Infinite nerd/

N to any specific numeral = specific numeral nerd/

N to my numeral = genyus! 

Thus, N you...you Nimrods!

...sorry...just being an N-Ninny.  :(

 

[Ignoratio Elenchi]

729?

3^6, which would be right if the order mattered - e.g. if "5 regular dogs and 1 chili dog" was a different order than "1 chili dog and 5 regular dogs", there would be 729 different possible ways to order hot dogs.  I assumed the problem treats those as the same order, so the actual number of ways to order hot dogs is much smaller. 

 

[Ilov80s]

3^6, which would be right if the order mattered - e.g. if "5 regular dogs and 1 chili dog" was a different order than "1 chili dog and 5 regular dogs", there would be 729 different possible ways to order hot dogs.  I assumed the problem treats those as the same order, so the actual number of ways to order hot dogs is much smaller. 

Yeah, I guess I was still thinking about the card problem. 28 then obviously.