Easy math puzzles
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Quote: acesideLet us focus on finding the solution for this question I posted earlier:
(F_x)^2 + (F_x+1)^2 + … + (F_x+k)^2 =m^2,
where x, k, and m are all positive integers and F_x is a Fibonacci number with the index of x. Find all x, k, and m.
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Does this have a solution that you know of? I am doing a brute force search, and, other than the trivial solutions where k = 0, I cannot find any for x + k < 1000.
Quote: ThatDonGuyQuote: acesideLet us focus on finding the solution for this question I posted earlier:
(F_x)^2 + (F_x+1)^2 + … + (F_x+k)^2 =m^2,
where x, k, and m are all positive integers and F_x is a Fibonacci number with the index of x. Find all x, k, and m.
link to original post
Does this have a solution that you know of? I am doing a brute force search, and, other than the trivial solutions where k = 0, I cannot find any for x + k < 1000.
link to original post
I don’t know the solution for this problem. I just made this problem believing this can be quickly solved by an Excel spreadsheet. We know this formula,
(F_0)^2 + (F_1)^2 +(F_2)^2 … + (F_n)^2 = (F_n)*(F_n-1).
This means we can easily extend the search to many millions of terms. Thank you for your hard work.
Some mathematicians also proved that the only square Fibonacci numbers are 0, 1 and 144. This is all I know.
Quote: acesideQuote: ThatDonGuyQuote: acesideLet us focus on finding the solution for this question I posted earlier:
(F_x)^2 + (F_x+1)^2 + … + (F_x+k)^2 =m^2,
where x, k, and m are all positive integers and F_x is a Fibonacci number with the index of x. Find all x, k, and m.
link to original post
Does this have a solution that you know of? I am doing a brute force search, and, other than the trivial solutions where k = 0, I cannot find any for x + k < 1000.
link to original post
I don’t know the solution for this problem. I just made this problem believing this can be quickly solved by an Excel spreadsheet. We know this formula,
(F_0)^2 + (F_1)^2 +(F_2)^2 … + (F_n)^2 = (F_n)*(F_n-1).
This means we can easily extend the search to many millions of terms. Thank you for your hard work.
Some mathematicians also proved that the only square Fibonacci numbers are 0, 1 and 144. This is all I know.
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Another formula may be useful for the question about the Sum of Consecutive Fibonacci Number Squares:
(F_n)^2+(F_n+1)^2=F_2n+1
Quote: acesideQuote: ThatDonGuyQuote: acesideLet us focus on finding the solution for this question I posted earlier:
(F_x)^2 + (F_x+1)^2 + … + (F_x+k)^2 =m^2,
where x, k, and m are all positive integers and F_x is a Fibonacci number with the index of x. Find all x, k, and m.
link to original post
Does this have a solution that you know of? I am doing a brute force search, and, other than the trivial solutions where k = 0, I cannot find any for x + k < 1000.
link to original post
I don’t know the solution for this problem. I just made this problem believing this can be quickly solved by an Excel spreadsheet. We know this formula,
(F_0)^2 + (F_1)^2 +(F_2)^2 … + (F_n)^2 = (F_n)*(F_n-1).
This means we can easily extend the search to many millions of terms. Thank you for your hard work.
Some mathematicians also proved that the only square Fibonacci numbers are 0, 1 and 144. This is all I know.
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Note that F(n) F(n+1) = (((1 + sqrt(5))/2)^(2n+1) - (-1)^n + ((1 - sqrt(5))/2)^(2n+1)) / 5
The problem is not so much generating the terms, or the differences between nonadjacent terms, as it is trying to figure out which ones are perfect squares. I use a brute force method to determine if N is a perfect square - find the smallest power of 2 whose square >= N, then keep adding/subtracting smaller powers of 2 until either the square = N or you add/subtract 1.
For example, if N = 1,000,000, start with 1024; 1024^2 > N, so subtract 512 from 2014; 512^2 < N, so add 256 to 512; 784^2 < N, so add 128 to 784; 912^2 < N, so add 64 to 912; 976^2 < N, so add 32 to 976; 1008^2 > N, so subtract 16 from 1008; 992^2 < N, so add 8 to 992; 1000^2 = N, so N is a square.
Quote: ThatDonGuy
The problem is not so much generating the terms, or the differences between nonadjacent terms, as it is trying to figure out which ones are perfect squares. I use a brute force method to determine if N is a perfect square - find the smallest power of 2 whose square >= N, then keep adding/subtracting smaller powers of 2 until either the square = N or you add/subtract 1.
For example, if N = 1,000,000, start with 1024; 1024^2 > N, so subtract 512 from 2014; 512^2 < N, so add 256 to 512; 784^2 < N, so add 128 to 784; 912^2 < N, so add 64 to 912; 976^2 < N, so add 32 to 976; 1008^2 > N, so subtract 16 from 1008; 992^2 < N, so add 8 to 992; 1000^2 = N, so N is a square.
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Is that faster than checking if sqrt(x)>trunc(sqrt(x))?
Quote: DieterQuote: ThatDonGuy
The problem is not so much generating the terms, or the differences between nonadjacent terms, as it is trying to figure out which ones are perfect squares. I use a brute force method to determine if N is a perfect square - find the smallest power of 2 whose square >= N, then keep adding/subtracting smaller powers of 2 until either the square = N or you add/subtract 1.
For example, if N = 1,000,000, start with 1024; 1024^2 > N, so subtract 512 from 2014; 512^2 < N, so add 256 to 512; 784^2 < N, so add 128 to 784; 912^2 < N, so add 64 to 912; 976^2 < N, so add 32 to 976; 1008^2 > N, so subtract 16 from 1008; 992^2 < N, so add 8 to 992; 1000^2 = N, so N is a square.
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Is that faster than checking if sqrt(x)>trunc(sqrt(x))?
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After a certain point, sqrt(x) = trunc(sqrt(x)) starts showing false results because of the restriction on a double's precision.
Even in Excel, 11,930,465^2 + 1 has sqrt(x) = trunc(sqrt(x)). It looks like any "square + 1" > 2^47 does this.
Quote: ThatDonGuyQuote: Dieter
Is that faster than checking if sqrt(x)>trunc(sqrt(x))?
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After a certain point, sqrt(x) = trunc(sqrt(x)) starts showing false results because of the restriction on a double's precision.
Even in Excel, 11,930,465^2 + 1 has sqrt(x) = trunc(sqrt(x)). It looks like any "square + 1" > 2^47 does this.
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OK, I can understand the necessity of a workaround for spreadsheets being ill-suited to numbers of magnitude.
Quote: ThatDonGuy
After a certain point, sqrt(x) = trunc(sqrt(x)) starts showing false results because of the restriction on a double's precision.
Even in Excel, 11,930,465^2 + 1 has sqrt(x) = trunc(sqrt(x)). It looks like any "square + 1" > 2^47 does this.
You should get something better than Excel for calculations with large integers. There is a lot of free software available, if you know python, you can use SymPy, GNu Octave with the symbolic package installed can also handle large integers and I have just played around with Mathics from mathics.org, which aims to be a free alternative to Wolfram Mathematica.
Quote: GMQuote: ThatDonGuy
After a certain point, sqrt(x) = trunc(sqrt(x)) starts showing false results because of the restriction on a double's precision.
Even in Excel, 11,930,465^2 + 1 has sqrt(x) = trunc(sqrt(x)). It looks like any "square + 1" > 2^47 does this.
You should get something better than Excel for calculations with large integers. There is a lot of free software available, if you know python, you can use SymPy, GNu Octave with the symbolic package installed can also handle large integers and I have just played around with Mathics from mathics.org, which aims to be a free alternative to Wolfram Mathematica.
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I usually write my own code in Visual C# with the BigInteger class when I need to deal with large integers. I was just using Excel as an example of the limitations of floating point numbers.
The problem here is not so much how long it takes to determine if a number is a perfect square, but the number of calculations that have to be done for each Fibonacci number.
Let S(n) = F(1)^2 + F(2)^2 + ... + F(n)^2 = F(n) F(n+1); the sum of F(A)^2 + F(A+1)^2 + ... + F(B)^2 = S(B) - S(A-1).
For, say, F(1000), we know that S(1000) - S(999) is a square, since is it is F(1000)^2, but we need to determine if S(1000) - S(998), S(1000) - S(997), and so on through S(1000) - S(1) are squares.
You are on The Price Is Right, and are playing a game called 3 Strikes, but it's Dream Car Week and you are playing for a car worth over $100,000. The way the game works is this:
1. The six digits in the price of the car are all different. (Actually, there was one playing where they repeated a digit, but in this case, all six are different.)
2. There are nine balls; six of them each have a different digit of the price of the car, and the other three are "strikes." The nine balls are placed into a bag.
3. One at a time, the player pulls a ball out of the bag. If it is a strike, it is removed from the bag. If it is a digit, the contestant has to guess which digit in the price (i.e. first, second, ..., sixth) it is. If correct, the digit is revealed and the ball kept out of the bag; if incorrect, the digit is not revealed, and the ball is put back into the bag.
The contestant wins if all six digits are guessed correctly before the three strikes are removed.
Here's the question: you happen to know the price of the car in advance, so whenever you pull a number out of the bag, you will always guess its location in the price correctly. What is the probability of you winning the car? Remember, there are six digits in the price (I point that out because usually there are only five).
Quote: ThatDonGuyHere's an easy one, that I thought of when working on another thread:
You are on The Price Is Right, and are playing a game called 3 Strikes, but it's Dream Car Week and you are playing for a car worth over $100,000. The way the game works is this:
1. The six digits in the price of the car are all different. (Actually, there was one playing where they repeated a digit, but in this case, all six are different.)
2. There are nine balls; six of them each have a different digit of the price of the car, and the other three are "strikes." The nine balls are placed into a bag.
3. One at a time, the player pulls a ball out of the bag. If it is a strike, it is removed from the bag. If it is a digit, the contestant has to guess which digit in the price (i.e. first, second, ..., sixth) it is. If correct, the digit is revealed and the ball kept out of the bag; if incorrect, the digit is not revealed, and the ball is put back into the bag.
The contestant wins if all six digits are guessed correctly before the three strikes are removed.
Here's the question: you happen to know the price of the car in advance, so whenever you pull a number out of the bag, you will always guess its location in the price correctly. What is the probability of you winning the car? Remember, there are six digits in the price (I point that out because usually there are only five).
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Imagine that I pick the nine balls out of the bag and line them up. If the ninth ball is a digit, I cannot win because three strikes appear before that digit. Likewise, if the ninth ball is a strike, I would win because I would pick the nine digits before I get to the third strike.
So, the probability I win equals the probability that the ninth ball is a strike which is 3/9 = 1/3.
btw I suspect it’s an interesting and easy puzzle what’s the fewest pitches needed so (1) the Home team wins (2) the Away team wins. I assume everyone here is familiar with baseball rules!
Quote: charliepatrickbtw I suspect it’s an interesting and easy puzzle what’s the fewest pitches needed so (1) the Home team wins (2) the Away team wins. I assume everyone here is familiar with baseball rules!
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The answer to both is none, if either team forfeits.
Even without forfeits, the answer to both is still none: one team scores by having four players intentionally walked at the request of the other team's manager, and all of the other players were called out for not stepping into either batter's box (rule 5.04(b)(3)),
Quote: charliepatrickbtw I suspect it’s an interesting and easy puzzle what’s the fewest pitches needed so (1) the Home team wins (2) the Away team wins. I assume everyone here is familiar with baseball rules!
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Now in the unlikely event that weather causes a game to be stopped, it is considered an "official" game if it goes 5 innings (4 1/2 if the home team is leading). So, in this case, the minimum number of pitches would be 31 for the away team to win, and 28 for the home team to win.
I once had that in a Car Treasure Hunt in that half the field went to the wrong first place, but fortunately they all got back on track by the second village. The clue (sounds like it's on the Central Line near Southampton) I'd set didn't quite fit the wrong place as "Eling" does sound like "Ealing" but "Redbridge" is identical spelling as "Redbridge", both stations on the (London) Central line.
Quote: ChesterDogQuote: ThatDonGuyHere's an easy one, that I thought of when working on another thread:
You are on The Price Is Right, and are playing a game called 3 Strikes, but it's Dream Car Week and you are playing for a car worth over $100,000. The way the game works is this:
1. The six digits in the price of the car are all different. (Actually, there was one playing where they repeated a digit, but in this case, all six are different.)
2. There are nine balls; six of them each have a different digit of the price of the car, and the other three are "strikes." The nine balls are placed into a bag.
3. One at a time, the player pulls a ball out of the bag. If it is a strike, it is removed from the bag. If it is a digit, the contestant has to guess which digit in the price (i.e. first, second, ..., sixth) it is. If correct, the digit is revealed and the ball kept out of the bag; if incorrect, the digit is not revealed, and the ball is put back into the bag.
The contestant wins if all six digits are guessed correctly before the three strikes are removed.
Here's the question: you happen to know the price of the car in advance, so whenever you pull a number out of the bag, you will always guess its location in the price correctly. What is the probability of you winning the car? Remember, there are six digits in the price (I point that out because usually there are only five).
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Imagine that I pick the nine balls out of the bag and line them up. If the ninth ball is a digit, I cannot win because three strikes appear before that digit. Likewise, if the ninth ball is a strike, I would win because I would pick the nine digits before I get to the third strike.
So, the probability I win equals the probability that the ninth ball is a strike which is 3/9 = 1/3.
Quote: Wizard1/3
Correct on both counts - and ChesterDog's answer is the method I used to solve it.
Quote: ThatDonGuyCorrect on both counts - and ChesterDog's answer is the method I used to solve it.
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Thanks. I was thinking about the answer in the case the player had no idea of the price, including the first digit.
A MARKOV CHAIN ANALYSIS OF THE THREE STRIKES GAME addresses that question, but they show the first page of the paper only.
I could solve this with a computer, but a pure mathematical approach would seem very tedious, given the multitude of possible game states.
Question -- Let's say the digits in the price of the car are 1,2,3,4 and 5. The player picks the 1 first and guesses the first position and is wrong. He picks the 2 next. Should he guess the first position again, given that he his chances are 25% of being correct, as opposed to 18.75% in any other position? It may appear the answer is an obvious yes, but I don't know of anything other than hand-waving logic this prove this.
You are playing Texas Hold 'Em with four opponents. You are dealt pocket kings. What is the probability at least one opponent has pocket aces?
Quote: WizardMy last post seems to have stopped discussion dead, so let's try changing the topic.
You are playing Texas Hold 'Em with four opponents. You are dealt pocket kings. What is the probability at least one opponent has pocket aces?
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about 2.44464% ?
Edit: Never mind--that is my answer for 5 opponents.
For 4 opponents, I get an answer of about 1.95658%.
Quote: WizardMy last post seems to have stopped discussion dead, so let's try changing the topic.
You are playing Texas Hold 'Em with four opponents. You are dealt pocket kings. What is the probability at least one opponent has pocket aces?
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There are C(50,2) x C(48,2) x C(46,2) x C(44,2) ways to deal the 50 remaining cards
If one player has a pair of aces and no other aces are dealt, there are 4 choices for the player, 6 for the pair of aces, and C(46,2) x C(44,2) x C(42,2) for the other three pairs
If one player has a pair of aces and one other ace is dealt, there are 4 choices for the player with the pair, 6 for the pair of aces, 3 for the player with the third ace, 2 for the third ace, 46 for that player's other card, and C(45,2) x C(43,2) for the other two pairs
If one player has a pair of aces and the other two aces are dealt to two different players, there are 4 choices for the player with the pair, 6 for the pair of aces, 3 for the pair of players with one ace, 2 for the first player's ace, 46 for the first player's non-ace, 45 for the second player's non-ace, and C(44,2) for the fourth player's two cards
If two players have a pair of aces, there are 6 choices for the players with pairs, 6 pairs of aces that the first player can have, and C(46,2) x C(44,2) choices for the other two players' cards
The probability is
((24 x C(46,2) x C(44,2) x C(42,2)) + (144 x 46 x C(45,2) x C(43,2)) + (144 x 46 x 45 x C(44,2)) + (36 x C(46,2) x C(44,2)))
/ (C(50,2) x C(48,2) x C(46,2) x C(44,2))
= 51,819 / 2,648,450, or about 1 / 51.10963
Quote: ThatDonGuyQuote: WizardMy last post seems to have stopped discussion dead, so let's try changing the topic.
You are playing Texas Hold 'Em with four opponents. You are dealt pocket kings. What is the probability at least one opponent has pocket aces?
link to original post
There are C(50,2) x C(48,2) x C(46,2) x C(44,2) ways to deal the 50 remaining cards
If one player has a pair of aces and no other aces are dealt, there are 4 choices for the player, 6 for the pair of aces, and C(46,2) x C(44,2) x C(42,2) for the other three pairs
If one player has a pair of aces and one other ace is dealt, there are 4 choices for the player with the pair, 6 for the pair of aces, 3 for the player with the third ace, 2 for the third ace, 46 for that player's other card, and C(45,2) x C(43,2) for the other two pairs
If one player has a pair of aces and the other two aces are dealt to two different players, there are 4 choices for the player with the pair, 6 for the pair of aces, 3 for the pair of players with one ace, 2 for the first player's ace, 46 for the first player's non-ace, 45 for the second player's non-ace, and C(44,2) for the fourth player's two cards
If two players have a pair of aces, there are 6 choices for the players with pairs, 6 pairs of aces that the first player can have, and C(46,2) x C(44,2) choices for the other two players' cards
The probability is
((24 x C(46,2) x C(44,2) x C(42,2)) + (144 x 46 x C(45,2) x C(43,2)) + (144 x 46 x 45 x C(44,2)) + (36 x C(46,2) x C(44,2)))
/ (C(50,2) x C(48,2) x C(46,2) x C(44,2))
= 51,819 / 2,648,450, or about 1 / 51.10963
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This is very interesting to us gamblers. Can you help extend this calculation to Ultimate Texas Hold’Em?
You are playing Ultimate Texas Hold 'Em with the dealer and 3 other players. You are dealt pocket kings. What is the probability the dealer has pocket aces?
Quote: Ace2Though I used a different counting method, I concur with ThatDonGuy’s answer
I found a simpler answer than my original one
There are C(50,8) ways to draw 8 cards from 50
Of these:
C(4,2) x C(46,6) have exactly 2 aces
There are C(8,2) = 28 possible ways to distribute the aces, of which 4 result in a pair,
so the probability of a pair = 4/28 = 1/7
C(4,3) x C(46,5) have exactly 3 aces
There are C(8,3) = 56 possible ways to distribute the aces, of which 6 have the first two
as a pair, 6 have the second two as a pair, 6 have the third two as a pair, and 6 have
the fourth two as a pair, so the probability of a pair = 24/56 = 3/7
C(4,4) x C(46,4) have all 4 aces
There are C(8,4) = 70 possible ways to distribute the aces, of which C(6,2) = 15 have the first
two as a pair, 15 have the second two as a pair, 15 have the third two as a pair, and 15 have
the fourth two as a pair, but this counts the 6 ways of having two pairs twice, so there are
54 ways of having at least one pair, and the probability = 54/70 = 27/35
The probability = (6/7 C(46,6) + 12/7 C(46,5) + 27/35 C(46,4)) / C(50,8) = 2253 / 115,150
...which equals my earlier fraction of 51,819 / 2,648,450, which is (51,810 * 23) / (115,150 * 23)
Quote: acesideThis is very interesting to us gamblers. Can you help extend this calculation to Ultimate Texas Hold’Em?
You are playing Ultimate Texas Hold 'Em with the dealer and 3 other players. You are dealt pocket kings. What is the probability the dealer has pocket aces?
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In this case, the number of other players is irrelevant as their cards could be anything. This is a much easier problem.
There are 50 cards that can be used for the dealer's hole cards, so there are C(50,2) = 1225 possible pairs, of which C(4,2) = 6 are pocket aces, so the probability = 6 / 1225, or about 1 / 204.
Quote: ThatDonGuyQuote: Ace2Though I used a different counting method, I concur with ThatDonGuy’s answer
I found a simpler answer than my original one
There are C(50,8) ways to draw 8 cards from 50
Of these:
C(4,2) x C(46,6) have exactly 2 aces
There are C(8,2) = 28 possible ways to distribute the aces, of which 4 result in a pair,
so the probability of a pair = 4/28 = 1/7
C(4,3) x C(46,5) have exactly 3 aces
There are C(8,3) = 56 possible ways to distribute the aces, of which 6 have the first two
as a pair, 6 have the second two as a pair, 6 have the third two as a pair, and 6 have
the fourth two as a pair, so the probability of a pair = 24/56 = 3/7
C(4,4) x C(46,4) have all 4 aces
There are C(8,4) = 70 possible ways to distribute the aces, of which C(6,2) = 15 have the first
two as a pair, 15 have the second two as a pair, 15 have the third two as a pair, and 15 have
the fourth two as a pair, but this counts the 6 ways of having two pairs twice, so there are
54 ways of having at least one pair, and the probability = 54/70 = 27/35
The probability = (6/7 C(46,6) + 12/7 C(46,5) + 27/35 C(46,4)) / C(50,8) = 2253 / 115,150
...which equals my earlier fraction of 51,819 / 2,648,450, which is (51,810 * 23) / (115,150 * 23)
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You are so cool!
Quote: ThatDonGuyQuote: acesideThis is very interesting to us gamblers. Can you help extend this calculation to Ultimate Texas Hold’Em?
You are playing Ultimate Texas Hold 'Em with the dealer and 3 other players. You are dealt pocket kings. What is the probability the dealer has pocket aces?
link to original post
In this case, the number of other players is irrelevant as their cards could be anything. This is a much easier problem.
There are 50 cards that can be used for the dealer's hole cards, so there are C(50,2) = 1225 possible pairs, of which C(4,2) = 6 are pocket aces, so the probability = 6 / 1225, or about 1 / 204.
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Can we just simply say that the probability of the original hold’em is 4 times of the ultimate hold’em? It seems to me the numbers are really close without detailed calculation.
Quote: acesideCan we just simply say that the probability of the original hold’em is 4 times of the ultimate hold’em? It seems to me the numbers are really close without detailed calculation.
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I want to say no, because it's really two different problems.
In both cases, the probability of somebody having pocket aces = 1 - the probability of nobody having pocket aces, but the calculations are different with one person than with four; among other things, you have to allow for two opponents each having pocket aces.
Quote: Wizard
You are playing Texas Hold 'Em with four opponents. You are dealt pocket kings. What is the probability at least one opponent has pocket aces?
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There are, for instance, (c(4,3))^2 * 3! * 2^3 * p(46,5) permutations of three dealt aces. 4 ways to choose the aces times 4 ways to choose the players times 6 ways to order the players times 8 ways to sort the aces within those 3 hands times p(46,5) ways to place the five other cards
Using the same method, I summed the five cases of 0-4 Aces being dealt and divided by p(50,8). Same answer as ThatDonGuy
Quote: ThatDonGuyQuote: acesideCan we just simply say that the probability of the original hold’em is 4 times of the ultimate hold’em? It seems to me the numbers are really close without detailed calculation.
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I want to say no, because it's really two different problems.
In both cases, the probability of somebody having pocket aces = 1 - the probability of nobody having pocket aces, but the calculations are different with one person than with four; among other things, you have to allow for two opponents each having pocket aces.
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What I really push at is what is the probability when there are 8 other players at a Texas Hold’Em table.
When you have pocket kings playing with one other player, the probability the other player has pocket aces is 1/204.
When you have pocket kings playing with four other players, the probability at least one player have pocket aces is 4/204.
When you have pocket kings playing with eight other players, the probability at least one player have pocket aces is 8/204.
This is the real technic in Texas Hold’Em, so I’d like to make sure the math correct. Can you help verify these numbers?
Quote: unJonNo that’s not right. Two people has more than double the odds as one person.
Actually, the probability that at least one of two opponents in a 3-player game has pocket aces is slightly less than twice the probability that one opponent in a 2-player game would have it, because you would be counting the cases where both opponents in the 3-player game have pocket aces twice.
I suppose you could use 2/204 for two opponents, 3/204 for three, and so on, depending on how accurate you wanted to be. The exact answer for 4 opponents is about 3.99142 / 204.
Quote: ThatDonGuyQuote: unJonNo that’s not right. Two people has more than double the odds as one person.
Actually, the probability that at least one of two opponents in a 3-player game has pocket aces is slightly less than twice the probability that one opponent in a 2-player game would have it, because you would be counting the cases where both opponents in the 3-player game have pocket aces twice.
I suppose you could use 2/204 for two opponents, 3/204 for three, and so on, depending on how accurate you wanted to be. The exact answer for 4 opponents is about 3.99142 / 204.
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That’s very counterintuitive to me. But not in front of a computer. If I face two players. The chance that the first has AA given that I have KK is 1/204(ish): 4*3/(50*49). And you’re saying that the chance that the 3rd guy has AA given that I have KK and the second guy doesn’t have AA is less than 1/204.
Quote: unJonQuote: ThatDonGuyQuote: unJonNo that’s not right. Two people has more than double the odds as one person.
Actually, the probability that at least one of two opponents in a 3-player game has pocket aces is slightly less than twice the probability that one opponent in a 2-player game would have it, because you would be counting the cases where both opponents in the 3-player game have pocket aces twice.
I suppose you could use 2/204 for two opponents, 3/204 for three, and so on, depending on how accurate you wanted to be. The exact answer for 4 opponents is about 3.99142 / 204.
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That’s very counterintuitive to me. But not in front of a computer. If I face two players. The chance that the first has AA given that I have KK is 1/204(ish): 4*3/(50*49). And you’re saying that the chance that the 3rd guy has AA given that I have KK and the second guy doesn’t have AA is less than 1/204.
Remember, just because the first player doesn't have two aces doesn't mean he doesn't have one ace, which makes it much harder for the second player to have AA.
Quote: ThatDonGuyQuote: unJonQuote: ThatDonGuyQuote: unJonNo that’s not right. Two people has more than double the odds as one person.
Actually, the probability that at least one of two opponents in a 3-player game has pocket aces is slightly less than twice the probability that one opponent in a 2-player game would have it, because you would be counting the cases where both opponents in the 3-player game have pocket aces twice.
I suppose you could use 2/204 for two opponents, 3/204 for three, and so on, depending on how accurate you wanted to be. The exact answer for 4 opponents is about 3.99142 / 204.
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That’s very counterintuitive to me. But not in front of a computer. If I face two players. The chance that the first has AA given that I have KK is 1/204(ish): 4*3/(50*49). And you’re saying that the chance that the 3rd guy has AA given that I have KK and the second guy doesn’t have AA is less than 1/204.
Remember, just because the first player doesn't have two aces doesn't mean he doesn't have one ace, which makes it much harder for the second player to have AA.
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No, no, no. If the only information you have on the opponents' hands is that the first player DOESN"T have two aces then that information slightly increases the probability that the the 2nd player does have two aces.
Quote: gordonm888Quote: ThatDonGuyQuote: unJonQuote: ThatDonGuyQuote: unJonNo that’s not right. Two people has more than double the odds as one person.
Actually, the probability that at least one of two opponents in a 3-player game has pocket aces is slightly less than twice the probability that one opponent in a 2-player game would have it, because you would be counting the cases where both opponents in the 3-player game have pocket aces twice.
I suppose you could use 2/204 for two opponents, 3/204 for three, and so on, depending on how accurate you wanted to be. The exact answer for 4 opponents is about 3.99142 / 204.
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That’s very counterintuitive to me. But not in front of a computer. If I face two players. The chance that the first has AA given that I have KK is 1/204(ish): 4*3/(50*49). And you’re saying that the chance that the 3rd guy has AA given that I have KK and the second guy doesn’t have AA is less than 1/204.
Remember, just because the first player doesn't have two aces doesn't mean he doesn't have one ace, which makes it much harder for the second player to have AA.
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No, no, no. If the only information you have on the opponents' hands is that the first player DOESN"T have two aces then that information slightly increases the probability that the the 2nd player does have two aces.
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That’s what I would have thought. But still don’t have excel in front of me.
Quote: ChesterDogFor 4 opponents, I get an answer of about 1.95658%.
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I agree!
Quote: unJonQuote: gordonm888No, no, no. If the only information you have on the opponents' hands is that the first player DOESN"T have two aces then that information slightly increases the probability that the the 2nd player does have two aces.
That’s what I would have thought. But still don’t have excel in front of me.
Okay, but you still have to remember that the probability that at least one of them has two Aces = the probability that the first one does + (the probability that the first one does not * the probability that the second one does).
Since the probability that the first one does is 1/204, this is 1/204 + (203/204 * the probability that the second one does). Even if the second player probability > 1/204, if it is < 1/203, then the product < 1/204, so the sum < 2/204.
Given that cuboid A has perimeters 12, 16, and 20, and cuboid B has perimeters 12, 16, and 24, which cuboid has the greatest volume?
Quote: GialmereFor any given cuboid it is possible to measure up to three different perimeters. For example, one perimeter could be measured this way.
Given that cuboid A has perimeters 12, 16, and 20, and cuboid B has perimeters 12, 16, and 24, which cuboid has the greatest volume?
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General terms answer:
Let A, B, C be the measures of the three perimeters
Let the three side lengths be X, Y, and Z, such that A = 2X + 2Y, B = 2X + 2Z, and C = 2Y + 2Z.
A + B + C = 4 (X + Y + Z)
X + Y + Z = (A + B + C) / 4
X = (A + B + C) / 4 - C / 2
Y = (A + B + C) / 4 - B / 2
Z = (A + B + C) / 4 - A / 2
For A = 12, B = 16, C = 20:
(A + B + C) / 4 = 12
X = 12 - 10 = 2
Y = 12 - 8 = 4
Z = 12 - 6 = 6
Volume = 48
For A = 12, B = 16, C = 24:
(A + B + C) / 4 = 13
X = 13 - 12 = 1
Y = 13 - 8 = 5
Z = 13 - 6 = 7
Volume = 35
Cuboid A has the greater volume
Quote: ThatDonGuyQuote: GialmereFor any given cuboid it is possible to measure up to three different perimeters. For example, one perimeter could be measured this way.
Given that cuboid A has perimeters 12, 16, and 20, and cuboid B has perimeters 12, 16, and 24, which cuboid has the greatest volume?
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General terms answer:
Let A, B, C be the measures of the three perimeters
Let the three side lengths be X, Y, and Z, such that A = 2X + 2Y, B = 2X + 2Z, and C = 2Y + 2Z.
A + B + C = 4 (X + Y + Z)
X + Y + Z = (A + B + C) / 4
X = (A + B + C) / 4 - C / 2
Y = (A + B + C) / 4 - B / 2
Z = (A + B + C) / 4 - A / 2
For A = 12, B = 16, C = 20:
(A + B + C) / 4 = 12
X = 12 - 10 = 2
Y = 12 - 8 = 4
Z = 12 - 6 = 6
Volume = 48
For A = 12, B = 16, C = 24:
(A + B + C) / 4 = 13
X = 13 - 12 = 1
Y = 13 - 8 = 5
Z = 13 - 6 = 7
Volume = 35
Cuboid A has the greater volume
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Hi, ThatDonGuy,
Can you please help calculate more on the last problem. I am a new poker player and I really want to get the math correct.
You are playing Texas Hold 'Em with 8 opponents. You are dealt pocket kings. What is the probability at least one opponent has pocket aces? Exactly.
I agree with Don.
Will I be able to get everything that I want and need in that box for moving? The cost is $2,200 to ship it to Florida and I refuse to get a second one. We are downsizing from a 3,200 sqft house to a 2,000 sqft house.
In the game of craps, what is the probability that all six points are won (at least once) over the course of one hour?
Assume thirty passline bets resolved per hour.
Quote: aceside
Can you please help calculate more on the last problem. I am a new poker player and I really want to get the math correct.
You are playing Texas Hold 'Em with 8 opponents. You are dealt pocket kings. What is the probability at least one opponent has pocket aces? Exactly.
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Here is what I get, which matches up with simulation:
The remainder of the deck has 50 cards, of which 4 are aces.
There are C(50,16) combinations of opponents' hole cards.
Of these:
C(4,2) x C(46,14) have exactly two aces
C(4,3) x C(46,13) have exactly three aces
C(4,4) x C(46,12) have exactly four aces
If there are two aces, there are C(16,2) = 120 ways to arrange them among the eight players, of which eight of the ways have the same player holding them, so the probability is 8 / 120 = 1/15.
If there are three aces, there are C(16,3) = 560 ways to arrange them among the eight players.
For each of the eight players that can have a pair, there are 14 places where the third ace could be, so there are 8 x 14 = 112 arrangements where someone has a pair; the probability is 112 / 560 = 1/5.
If there are four aces, there are C(16,4) = 1820 ways to arrange them among the eight players.
For each of the eight players that can have a pair, there are C(14,2) = 91 ways to arrange the other two aces. However, this counts the C(8,2) ways of two players each having a pair twice, so there are 8 x 91 - 28 = 700 arrangements of at least one pair; the probability is 700 / 1820 = 5/13.
The overall probability is
((6 x 239,877,544,005 x 1/15) + (4 x 101,766,230,790 x 1/5) + (38,910,617,655 x 5/13)) / C(50,16)
= 192,329,624,409 / 4,923,689,695,575
= 2249 / 57,575, or about 1 / 25.6
(Why, yes, this is very close to 8 / 204, which is 1 / 25.5, isn't it?)
Quote: ThatDonGuyQuote: unJonQuote: gordonm888No, no, no. If the only information you have on the opponents' hands is that the first player DOESN"T have two aces then that information slightly increases the probability that the the 2nd player does have two aces.
That’s what I would have thought. But still don’t have excel in front of me.
Okay, but you still have to remember that the probability that at least one of them has two Aces = the probability that the first one does + (the probability that the first one does not * the probability that the second one does).
Since the probability that the first one does is 1/204, this is 1/204 + (203/204 * the probability that the second one does). Even if the second player probability > 1/204, if it is < 1/203, then the product < 1/204, so the sum < 2/204.
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Thank you. This explanation makes sense to me.
Quote: ThatDonGuyQuote: aceside
Can you please help calculate more on the last problem. I am a new poker player and I really want to get the math correct.
You are playing Texas Hold 'Em with 8 opponents. You are dealt pocket kings. What is the probability at least one opponent has pocket aces? Exactly.
link to original post
Here is what I get, which matches up with simulation:
The remainder of the deck has 50 cards, of which 4 are aces.
There are C(50,16) combinations of opponents' hole cards.
Of these:
C(4,2) x C(46,14) have exactly two aces
C(4,3) x C(46,13) have exactly three aces
C(4,4) x C(46,12) have exactly four aces
If there are two aces, there are C(16,2) = 120 ways to arrange them among the eight players, of which eight of the ways have the same player holding them, so the probability is 8 / 120 = 1/15.
If there are three aces, there are C(16,3) = 560 ways to arrange them among the eight players.
For each of the eight players that can have a pair, there are 14 places where the third ace could be, so there are 8 x 14 = 112 arrangements where someone has a pair; the probability is 112 / 560 = 1/5.
If there are four aces, there are C(16,4) = 1820 ways to arrange them among the eight players.
For each of the eight players that can have a pair, there are C(14,2) = 91 ways to arrange the other two aces. However, this counts the C(8,2) ways of two players each having a pair twice, so there are 8 x 91 - 28 = 700 arrangements of at least one pair; the probability is 700 / 1820 = 5/13.
The overall probability is
((6 x 239,877,544,005 x 1/15) + (4 x 101,766,230,790 x 1/5) + (38,910,617,655 x 5/13)) / C(50,16)
= 192,329,624,409 / 4,923,689,695,575
= 2249 / 57,575, or about 1 / 25.6
(Why, yes, this is very close to 8 / 204, which is 1 / 25.5, isn't it?)
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This is neat. Let me put all three numbers together.
If there is one other player, the probability is 0.999/204;
If there are four other players, the probability is 3.992/204;
If there are eight other players, the probability is 7.969/204.
This means there is not very much correlation between player’s hole cards for up to eight players.
Quote: DRich
Will I be able to get everything that I want and need in that box for moving?
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Quote: DRichSpeaking of cuboids, I am moving shortly and rented a pod from U-haul.
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If you have a yard sale, please give me first dibs.
