The official Serenes Forest riddles

[noiku]

[spoiler=This was a bit too easy]

6 and 28

[MisterIceTeaPeach]

[spoiler=This was a bit too easy]

6 and 28

[spoiler=Next time I'll create a riddle by myself again.]

Correct!

Edited August 20, 2016 by Ayama Wirdo

[MisterIceTeaPeach]

Question 8: Stochastics 2

A might be the event that a professional Fire Emblem player owns FE Birthright.

B might be the event that a professional Fire Emblem player owns FE Conquest.

• 60% of the professional FE players own Conquest.
• 42% of them own Birthright.
• 20% of them own both.

How much % of the FE players do own

1. neither Birthright nor Conquest?
2. only one of the two games?
3. Conquest but not Birthright?

Edited August 21, 2016 by Ayama Wirdo

[Rezzy]

Question 8: Stochastics 2

A might be the event that a professional Fire Emblem player owns FE Birthright.

B might be the event that a professional Fire Emblem player owns FE Conquest.

• 60% of the professional FE players own Conquest.
• 42% of them own Birthright.
• 20% of them own both.

How much % of the FE players do own

1. neither Birthright nor Conquest?
2. one of the two games?
3. Conquest but not Birthright?

[spoiler=Guess]If I understand correctly 40% ONLY own Conquests.

22% ONLY own Birthright.

20% own both.

82% own at least one

So...

1. 18% own neither

2. 82% own at least one, (62% own ONLY one of the two)

3. 40% own only Conquests

[MisterIceTeaPeach]

[spoiler=Guess]If I understand correctly 40% ONLY own Conquests.

22% ONLY own Birthright.

20% own both.

82% own at least one

So...

1. 18% own neither

2. 82% own at least one, (62% own ONLY one of the two)

3. 40% own only Conquests

Everything is correct!

As for 2. 62% was the wanted answer.

[Rezzy]

Everything is correct!

As for 2. 62% was the wanted answer.

Thanks, I wasn't sure about #2, so I included both answers. It's a bit ambiguous, so you may want to add the word "Only" to the question, but it's understandable if English isn't your first language.

[Balcerzak]

Let N = the number of professional FE players.

.6N own Conquest

.2N own both Conquest and Birthright

-> .4N own only Conquest and not Birthright

.42N own Birthright

-> .22N own only Birthright and not Conquest

.22N+.4N+.2N = .82N own some combination of Birthright and Conquest.

->

18% own neither Birthright nor Conquest.

62% own only one of the Fates titles

40% own only Conquest and not Birthright

[MisterIceTeaPeach]

I guess I won't post another riddle today.
I'm still working on another practical example of Fire Emblem about the topic for binominal distribution.
However I've problems to solve one certain question.
If I could solve it, then I'll post it.

Let N = the number of professional FE players.
.6N own Conquest
.2N own both Conquest and Birthright
-> .4N own only Conquest and not Birthright
.42N own Birthright
-> .22N own only Birthright and not Conquest

.22N+.4N+.2N = .82N own some combination of Birthright and Conquest.

->
18% own neither Birthright nor Conquest.
62% own only one of the Fates titles
40% own only Conquest and not Birthright

Correct!

[Gradivus.]

1) p(not A and not B) = 1 - p(only A) - p(only B) - p(both) = 1 - (42% - 20%) - (60% - 20%) - 20% = 18%

2) p(only A or only B) = p(A) - P(both) + p(B) - p(both) = 42% - 20% + 60% - 20% = 62%

3) p(only B) = p(B) - p(both) = 60% - 20% = 40%

[MisterIceTeaPeach]

1) p(not A and not B) = 1 - p(only A) - p(only B) - p(both) = 1 - (42% - 20%) - (60% - 20%) - 20% = 18%

2) p(only A or only B) = p(A) - P(both) + p(B) - p(both) = 42% - 20% + 60% - 20% = 62%

3) p(only B) = p(B) - p(both) = 60% - 20% = 40%

Correct!

To all new people:

You can submit your solutions of the previous riddles as well.

I still would check them.

Edit: Have to set this thread on hiatus because my internet will be off today till the rest of this month. I'll use this forced break to create new and hopefully challenging mathematical questions.

Edited August 24, 2016 by Ayama Wirdo

[MisterIceTeaPeach]

Here's a new riddle (submitted by Chegg):

[umerol]

Here's how I would put it:
Peach, Daisy, Luigi, Bowser, Yoshi, Waluigi
Mario, Toadsworth, Donkey Kong, Toad, Rosalia, Wario

Since I don't know where the head is, I put Peach and Mario to the left.

Edited August 29, 2016 by umerol

[Rezzy]

Is the head to the left or right?

I've got my answer, assuming the head is in the right, I'll post when not on mobile.[spoiler=Answer]The head has to be on the right, otherwise someone will have to sit next to Wario, who hasn't sat down already.

I think there's actually a few possible answers here, but here's one that I think meets all the criteria.

Waluigi - Rosalina - Toad - Luigi - Daisy - Peach
Wario - Toadsworth - Donkey Kong - Yoshi - Bowser - Mario

Edited August 29, 2016 by Rezzy

[MisterIceTeaPeach]

Here's how I would put it:

Peach, Daisy, Luigi, Bowser, Yoshi, Waluigi

Mario, Toadsworth, Donkey Kong, Toad, Rosalia, Wario

Since I don't know where the head is, I put Peach and Mario to the left.

Is the head to the left or right?

I've got my answer, assuming the head is in the right, I'll post when not on mobile.[spoiler=Answer]The head has to be on the right, otherwise someone will have to sit next to Wario, who hasn't sat down already.

I think there's actually a few possible answers here, but here's one that I think meets all the criteria.

Waluigi - Rosalina - Toad - Luigi - Daisy - Peach

Wario - Toadsworth - Donkey Kong - Yoshi - Bowser - Mario

[spoiler=Your answers]Also both guesses are wrong.

However I came to an other answer as well than the master solution below.

[spoiler=About the head]

It's obvious it has to be on the right.

[spoiler=Correct solution]

[Balcerzak]

WLuig |Yoshi |Luigi |Daisy |DonkeyK|Mario

-----------------------------------------

Wario |Tworth|Bowser|Toad |Rosalin|Peach

Quite a very permutations are possible, especially if you assume there is no problem with Bowser and DK sitting across from each other, or a princess across form Bowser, which isn't unrealistic, given the distinction made between sitting next to/across as far as the Waluigi/Wario situation is concerned. This was the safest one I could come up with for the most strict interpretation of the conditions, though.

[MisterIceTeaPeach]

WLuig |Yoshi |Luigi |Daisy |DonkeyK|Mario

-----------------------------------------

Wario |Tworth|Bowser|Toad |Rosalin|Peach

Quite a very permutations are possible, especially if you assume there is no problem with Bowser and DK sitting across from each other, or a princess across form Bowser, which isn't unrealistic, given the distinction made between sitting next to/across as far as the Waluigi/Wario situation is concerned. This was the safest one I could come up with for the most strict interpretation of the conditions, though.

This is the sample solution.

Idk exactly how to understand the word "next" in this riddle.

The example is shown in boxes. Even if someone sits across from each other, it still sits next because the boxes merge.

So I assumed Bowser and DK / Bowser and a princess may not sit across from each other to have exact one solution.

[MisterIceTeaPeach]

Question #10: Poker

A classical Poker game consists of 52 cards (2 - Ass in four colours).

These are all winning hands in all Poker game types.

a) In a classic Five Card Draw you get exactly - like the name already says - five cards.

What's the % possibility of the given five cards to get a...

1. pair?
2. royal flush?
3. nothing (high card)? (This requires tons of calculation since you must exclude all possible hands from 100%. A task for a math genius! (I still haven't solved it for myself since I'm not a genius).

b) In a Five Card Draw you may change your cards (0-5) once.

What's the % possibility to make a given...

1. pair into a three of a kind?
2. 8, 10 and jack into a straight?

c) How many card combinations are possible in general?

I can check each question for you except for a3 yet.

Edited September 5, 2016 by Ayama Wirdo

[Balcerzak]

The chance to get a pair:

first card

second card matches first card

third card does not match first card

fourth card does not match first card or third card

fifth card does not match first card, third card or fourth card

(1) * (3/51) * (48/50) * (44/49) * (40/48)

since the order in which the cards are matched into a pair does not matter, we must multiply by the number of equivalent ways a pair could form (5 choose 2)

5 choose 2 is 10, and can be seen through exhaustive listing as

[1,2] [1,3] [1,4] [1,5] [2,3] [2,4] [2,5] [3,4] [3,5] [4,5]

The chance to get a royal flush:

first card within A,K,Q,J,10 of any suit

second card in the same suit, that is of the remaining AKQJT

third card in the same suit, that is of the remaining AKQJT

fourth card in the same suit, that is of the remaining AKQJT

fifth card in the same suit, that is of the remaining AKQJT

(20/52) * (4/51) * (3/50) * (2/49) * (1/48)

The order in which the cards are drawn does matter for this case, as all cards are a part of the hand (five choose five).

For part 2, are you assuming the player draws optimally? e.g. they have A 2 2 7 9 and are trying to convert their pair into a 3 of a kind, do we need to account for the case where a player might only exchange their seven and their nine, while holding onto their ace (either because superstition, or to tie-break some other chap with a pair of twos)? Or do we assume they'll always dump all three of the cards that do not fit their hand?

[MisterIceTeaPeach]

The chance to get a pair:

first card

second card matches first card

third card does not match first card

fourth card does not match first card or third card

fifth card does not match first card, third card or fourth card

(1) * (3/51) * (48/50) * (44/49) * (40/48)

since the order in which the cards are matched into a pair does not matter, we must multiply by the number of equivalent ways a pair could form (5 choose 2)

5 choose 2 is 10, and can be seen through exhaustive listing as

[1,2] [1,3] [1,4] [1,5] [2,3] [2,4] [2,5] [3,4] [3,5] [4,5]

The chance to get a royal flush:

first card within A,K,Q,J,10 of any suit

second card in the same suit, that is of the remaining AKQJT

third card in the same suit, that is of the remaining AKQJT

fourth card in the same suit, that is of the remaining AKQJT

fifth card in the same suit, that is of the remaining AKQJT

(20/52) * (4/51) * (3/50) * (2/49) * (1/48)

The order in which the cards are drawn does matter for this case, as all cards are a part of the hand (five choose five).

For part 2, are you assuming the player draws optimally? e.g. they have A 2 2 7 9 and are trying to convert their pair into a 3 of a kind, do we need to account for the case where a player might only exchange their seven and their nine, while holding onto their ace (either because superstition, or to tie-break some other chap with a pair of twos)? Or do we assume they'll always dump all three of the cards that do not fit their hand?

A1: Accidently I made a wrong calculation with a wrong result because I forgot to take account 2. ; 3. ; 4. card => 1 x (3/51x4). Only calculated for the first card.

Yeah, your way is correct!

A2: Correct!

As for part 2:

Yes, drawing optimally

Not paired cards are folded.

[Balcerzak]

[spoiler=still skipping a3 and skipping c for now]

possibility of converting to a 3 of a kind:

one and only one of the new cards match. Also, the two non-matching cards cannot match each other.

(2/47) is the chance to draw a match to your pair

the rest is a bit more complicated:

If the next card we drew is from one of the ranks we previously discarded (Case A) or from a rank we have yet to see (Case B)

Case A:

(9/46) * (42/45)

3 remain from each of the 3 discards for 9 "good" cards. The following card cannot be the 1 card matching our 3oaK or the 2 remaining cards matching our "good" second pick.

Case B:

(36/46) * (41/45)

9 ranks by 4 suits gives 36 possibilities. Then it cannot be the 1 card matching our 3oaK, nor one of the 3 cards matching our "good" second.

So we get:

(2/47) * [(9/46) * (42/45) + (36/46) * (41/45)]

this needs to be multiplied by 3 choose 1 to account for the fact that only one of these 3 cards makes it into the final hand (aka any of the three draws can be the one to extend).

possibility of converting an 8TJ to a straight:

two possible straights build from the starting hand: JT987, QJT98

either way we need to draw a 9

4/47

then we need to draw either a 7 or a Q

8/46

this is multiplied by 2 choose 2, because all cards are part of the final hand

[MisterIceTeaPeach]

[spoiler=still skipping a3 and skipping c for now]

possibility of converting to a 3 of a kind:

one and only one of the new cards match. Also, the two non-matching cards cannot match each other.

(2/47) is the chance to draw a match to your pair

the rest is a bit more complicated:

If the next card we drew is from one of the ranks we previously discarded (Case A) or from a rank we have yet to see (Case B)

Case A:

(9/46) * (42/45)

3 remain from each of the 3 discards for 9 "good" cards. The following card cannot be the 1 card matching our 3oaK or the 2 remaining cards matching our "good" second pick.

Case B:

(36/46) * (41/45)

9 ranks by 4 suits gives 36 possibilities. Then it cannot be the 1 card matching our 3oaK, nor one of the 3 cards matching our "good" second.

So we get:

(2/47) * [(9/46) * (42/45) + (36/46) * (41/45)]

this needs to be multiplied by 3 choose 1 to account for the fact that only one of these 3 cards makes it into the final hand (aka any of the three draws can be the one to extend).

possibility of converting an 8TJ to a straight:

two possible straights build from the starting hand: JT987, QJT98

either way we need to draw a 9

4/47

then we need to draw either a 7 or a Q

8/46

this is multiplied by 2 choose 2, because all cards are part of the final hand

Absolute correct!

[Gradivus.]

a1)

p(1st card) = 52/52 = 1

p(rank on 1st card=2nd) = 3/51

These two cards have 10 permutations in the 5 spots (i.e. they can occupy the first two spots, or the 2nd and the 3rd, or the 3rd and the 5th, and so on).

p(a pair exists) = 3/51*10 = 30/51

The remaining probabilities are (number of cards with a different rank than the rest/number of cards remaining).

p(1 Pair) = p(a pair exists)*48/50*44/49*40/48 = 42.2569%

a2)

p(1st) = 20/52 = 5/13 [something between 10 and Ace, regardless of color]

p(2nd) to p(5th) in this case are (number of cards to go/number of cards in the stack)

p(Royal Flush) = p(1st)*p(2nd)*p(3rd)*p(4th)*p(5th) = 5/13*4/51*3/50*2/49*1/48 = 1/649740 = 0.0001539%

EDIT: removed two zeroes behind the decimal point, I incorrectly copied them from my calculator.

a3)

Everything containing a pair (One or Two Pairs, Trips, Full House and Quads) can be lumped together since the final calculation is just subtracting everything from 1 (-> same calculation as Pair, except the last 3 cards all have p=1).

p(anything with a pair) = p(a pair exists [from a1])*1*1*1 = 30/51 = 10/17

Straight (including Straight Flush and Royal Flush):

Any tier can be the highest card of a straight, except 2, 3 and 4:

p(1st) = 40/52 = 10/13

There are 5 possibilities as to where this card is located. p(2nd) through p(5th) = (cards that allow for a straight below the 1st card/cards in the stack).

p(anything with a straight) = 10/13*16/51*12/50*8/49*4/48*5 = 128/32487

Flush (including SF/RF):

p(1st) = 1

p(2nd) through p(5th) = (cards of the same color/cards on the stack)

p(anything with a flush) = 12/51*11/50*10/49*9/48 = 33/16660

Since the Straight and Flush probabilities overlap, p(Straight Flush or Royal Flush) needs to be added once at the end, to not count it twice.

p(Straight Flush or Royal Flush) = 10/13*4/51*3/50*2/49*1/48*5 = 1/64974

p(High Card) = 1 - p(anything with a pair) - p(anything with a straight) - p(anything with a flush) + p(Straight Flush or Royal Flush) = 1 - 10/17 - 128/32487 - 33/16660 + 1/64974 = 87901/216580 = 40.585926%

b1)

47 cards are on the stack since 3 cards got tossed and a pair is given. The first card is taken to be of the same rank as the pair, and its chance to appear is multiplied by 3 since it can be in 3 different spots.

p(pair to trips) = p(1st new==old)*3 + p(2nd new=/=trips) + p(3rd new=/=trips) = 2/47*45/46*44/45 = 4.0703%

b2)

One of the new cards must be a 9 and can be in 2 different spots. The other must be a 7 or a Queen.

p(8, 10, J to straight) = p(9)*p(7 or Q)*2 = 4/47*8/46*2 = 2.9602%

c)

There are 4 combinations that can yield a Royal Flush. Since all combinations are equally likely, the probability of a Royal Flush is 4 times that of any single combination.

p(any combination) = p(Royal Flush) / 4 = 1/649740 * 1/4 = 1/2598960

=> n = 1/p(any combination) = 2598960

Edited September 18, 2016 by Gradivus.

[MisterIceTeaPeach]

a1)

p(1st card) = 52/52 = 1

p(rank on 1st card=2nd) = 3/51

These two cards have 10 permutations in the 5 spots (i.e. they can occupy the first two spots, or the 2nd and the 3rd, or the 3rd and the 5th, and so on).

p(a pair exists) = 3/51*10 = 30/51

The remaining probabilities are (number of cards with a different rank than the rest/number of cards remaining).

p(1 Pair) = p(a pair exists)*48/50*44/49*40/48 = 42.2569%

a2)

p(1st) = 20/52 = 5/13 [something between 10 and Ace, regardless of color]

p(2nd) to p(5th) in this case are (number of cards to go/number of cards in the stack)

p(Royal Flush) = p(1st)*p(2nd)*p(3rd)*p(4th)*p(5th) = 5/13*4/51*3/50*2/49*1/48 = 1/649740 = 0.000001539%

a3)

Everything containing a pair (One or Two Pairs, Trips, Full House and Quads) can be lumped together since the final calculation is just subtracting everything from 1 (-> same calculation as Pair, except the last 3 cards all have p=1).

p(anything with a pair) = p(a pair exists [from a1])*1*1*1 = 30/51 = 10/17

Straight (including Straight Flush and Royal Flush):

Any tier can be the highest card of a straight, except 2, 3 and 4:

p(1st) = 40/52 = 10/13

There are 5 possibilities as to where this card is located. p(2nd) through p(5th) = (cards that allow for a straight below the 1st card/cards in the stack).

p(anything with a straight) = 10/13*16/51*12/50*8/49*4/48*5 = 128/32487

Flush (including SF/RF):

p(1st) = 1

p(2nd) through p(5th) = (cards of the same color/cards on the stack)

p(anything with a flush) = 12/51*11/50*10/49*9/48 = 33/16660

Since the Straight and Flush probabilities overlap, p(Straight Flush or Royal Flush) needs to be added once at the end, to not count it twice.

p(Straight Flush or Royal Flush) = 10/13*4/51*3/50*2/49*1/48*5 = 1/64974

p(High Card) = 1 - p(anything with a pair) - p(anything with a straight) - p(anything with a flush) + p(Straight Flush or Royal Flush) = 1 - 10/17 - 128/32487 - 33/16660 + 1/64974 = 87901/216580 = 40.585926%

b1)

47 cards are on the stack since 3 cards got tossed and a pair is given. The first card is taken to be of the same rank as the pair, and its chance to appear is multiplied by 3 since it can be in 3 different spots.

p(pair to trips) = p(1st new==old)*3 + p(2nd new=/=trips) + p(3rd new=/=trips) = 2/47*45/46*44/45 = 4.0703%

b2)

One of the new cards must be a 9 and can be in 2 different spots. The other must be a 7 or a Queen.

p(8, 10, J to straight) = p(9)*p(7 or Q)*2 = 4/47*8/46*2 = 2.9602%

c)

There are 4 combinations that can yield a Royal Flush. Since all combinations are equally likely, the probability of a Royal Flush is 4 times that of any single combination.

p(any combination) = p(Royal Flush) / 4 = 1/649740 * 1/4 = 1/2598960

=> n = 1/p(any combination) = 2598960

I still haven't solved A3 and the last question for myself yet.

The way looks correct on the first view.

You have to calculate:

1- the possibilites of

• pair
• two pairs
• three of a kind
• four of a kind
• full house
• straight
• flush

+ the possibilities of

• straight flush
• royal flush

because they're already included in straight / flush.

However wikipedia has a different result for the high card question.

All the other answers are correct!

[Gradivus.]

However wikipedia has a different result for the high card question.

Well, then my solution is probably wrong. The error seems to be in the pair calculation because the Straight and Flush ones seem to be correct. It's probably safer to calculate the chance that all cards are of different ranks - since that excludes all combinations with pairs, this chance replaces 1-p(anything with a pair) in the final calculation.

p(no pairs) = p(1st)*p(2nd=/=1st)*p(3rd=/=rest)*p(4th=/=rest)*p(5th=/=rest) = 1*48/51*44/50*40/49*36/48 = 2112/4165

p(High Card) = p(no pairs) - p(anything with a straight) - p(anything with a flush) + p(Straight Flush or Royal Flush) = 2112/4165 - 128/32487 - 33/16660 + 1/64974 = 1277/2548 = 50.1177%

Also, I had copied two zeroes too much from my calculator in A2, just forgot to omit them when converting the decimal number to a percentage. Edited that away.

Edited September 18, 2016 by Gradivus.

[MisterIceTeaPeach]

Well, then my solution is probably wrong. The error seems to be in the pair calculation because the Straight and Flush ones seem to be correct. It's probably safer to calculate the chance that all cards are of different ranks - since that excludes all combinations with pairs, this chance replaces 1-p(anything with a pair) in the final calculation.

p(no pairs) = p(1st)*p(2nd=/=1st)*p(3rd=/=rest)*p(4th=/=rest)*p(5th=/=rest) = 1*48/51*44/50*40/49*36/48 = 2112/4165

p(High Card) = p(no pairs) - p(anything with a straight) - p(anything with a flush) + p(Straight Flush or Royal Flush) = 2112/4165 - 128/32487 - 33/16660 + 1/64974 = 1277/2548 = 50.1177%

Also, I had copied two zeroes too much from my calculator in A2, just forgot to omit them when converting the decimal number to a percentage. Edited that away.

´

This is the correct result!