Easy math puzzles

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March 18th, 2021 at 12:59:38 AM permalink

One easy way to look at it is assume you add f, e, ... a to each side to get
a+b+c+d+e + f = f² + f.
a + b+c+d+e+f = a² + a.
So f² + f = a² + a etc.
giving for all a,b,c,d,e,f
(a²-b²) + (a-b) = 0;
(a-b)(a+b)+(a-b)=0
(a-b)(a+b+1)=0
So a=b or (a+b+1)=0
If a=b then a+a+a+a+a=a² giving 5a=a² giving a=0 or 5.

Not sure about (a+b+1)=0 as I think this gives a=c=e and b=d=f and a=-1-b and b²+b+3=0.

ThatDonGuy

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March 18th, 2021 at 7:31:54 AM permalink

Quote: Koi
First time poster to the forum.
Here is a nice algebra problem to chew on.
Solve the following system of equations for a, b, c, d, e, and f in real numbers other than the trivial solution of a = b = c = d = e = f = 0.
a + b + c + d + e = f^2
a + b + c + d + f = e^2
a + b + c + e + f = d^2
a + b + d + e + f = c^2
a + c + d + e + f = b^2
b + c + d + e + f = a^2

The only solution I have found so far is a = b = c = d = e = f = 5.

unJon

unJon

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March 18th, 2021 at 8:07:23 AM permalink

Quote: ThatDonGuy

The only solution I have found so far is a = b = c = d = e = f = 5.

I agree. Found two complex solutions by setting a=b=c and d=e=f and deriving a^4 - 4a^3 - 2a^2 - 15a = 0.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

ThatDonGuy

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March 18th, 2021 at 8:16:12 AM permalink

a + b + c + d + e = f^2
a + b + c + d + f = e^2
a + b + c + e + f = d^2
a + b + d + e + f = c^2
a + c + d + e + f = b^2
b + c + d + e + f = a^2

Add f to the first one, e to the second one, and so on:
a (a + 1) = b (b + 1) = c (c + 1) = d (d + 1) = e (e + 1) = f (f + 1)

a (a + 1) = b (b + 1)
a^2 + a + 1/4 = b^2 + b + 1/4
(a + 1/2)^2 = (b + 1/2)^2
a + 1/2 = +/- (b + 1/2)
a = (b + 1/2) - 1/2 or a = (-b - 1/2) - 1/2
a = b or a = -(b + 1)
However, b + 1/2 = +/- (a + 1/2)
b = (a + 1/2) - 1/2 or b = (-a - 1/2) - 1/2
b = a or b = -(a + 1)
If a <> b, then a = -(b + 1) and b = -(a + 1)

Do the same with c, d, e, and f replacing a
Each term b, c, d, e, f equals a or -(a + 1)

All five equal a:
a^2 = 5a
a = 5 or 0

Four equal a:
a^2 = 4a - (a + 1) = 3a - 1
a^2 - 3a + 1 = 0
a = 3/2 +/- sqrt(5)/2
a = (3 + sqrt(5))/2: {(3 + sqrt(5))/2, (3 + sqrt(5))/2, (3 + sqrt(5))/2, (3 + sqrt(5))/2, (3 + sqrt(5))/2, -(5 + sqrt(5))/2)

a + b + c + d + e = 5 (3 + sqrt(5))/2 = (15 + 5 sqrt(5))/2
f^2 = (-(5 + sqrt(5)))^2 / 4 = (30 + 10 sqrt(5))/4 = (15 + 5 sqrt(5))/2
This is a solution

a = (3 - sqrt(5))/2: {(3 - sqrt(5))/2, (3 - sqrt(5))/2, (3 - sqrt(5))/2, (3 - sqrt(5))/2, (3 - sqrt(5))/2, -(5 - sqrt(5))/2)
a + b + c + d + e = 5 (3 - sqrt(5))/2 = (15 - 5 sqrt(5))/2
f^2 = (-(5 - sqrt(5)))^2 / 4 = (30 - 10 sqrt(5))/4 = (15 - 5 sqrt(5))/2

Three equal a:
a^2 = 3a - 2 (a + 1) = a - 2
a^2 - a + 2 = 0
a = 2 or -1
a = 2: {2, 2, 2, 2, -1, -1} does not fit into the first equation (7 = 1)
a = -1: {-1, -1, -1, -1, 2, 2} does not fit into the first equation (-2 = 4)

Two equal a:
a^2 = 2a - 3 (a + 1) = -a - 3
a^2 + a + 3 = 0 has no real solutions

One equals a:
a^2 = a - 4 (a + 1) = -3a - 4
a^2 + 3a + 4 = 0 has no real solutions

None equal a:
a^2 = -5 (a + 1) = -5a - 5
a^2 + 5a + 5 = 0 has no real solutions

Thus, the only real nonzero solutions are:
(a) All six variables = 5;
(b) Five of the variables = (3 + sqrt(5)) / 2, and the sixth = -(5 + sqrt(5)) / 2

Koi

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March 18th, 2021 at 8:14:44 PM permalink

Quote: ThatDonGuy

a + b + c + d + e = f^2
a + b + c + d + f = e^2
a + b + c + e + f = d^2
a + b + d + e + f = c^2
a + c + d + e + f = b^2
b + c + d + e + f = a^2

Add f to the first one, e to the second one, and so on:
a (a + 1) = b (b + 1) = c (c + 1) = d (d + 1) = e (e + 1) = f (f + 1)

a (a + 1) = b (b + 1)
a^2 + a + 1/4 = b^2 + b + 1/4
(a + 1/2)^2 = (b + 1/2)^2
a + 1/2 = +/- (b + 1/2)
a = (b + 1/2) - 1/2 or a = (-b - 1/2) - 1/2
a = b or a = -(b + 1)
However, b + 1/2 = +/- (a + 1/2)
b = (a + 1/2) - 1/2 or b = (-a - 1/2) - 1/2
b = a or b = -(a + 1)
If a <> b, then a = -(b + 1) and b = -(a + 1)

Do the same with c, d, e, and f replacing a
Each term b, c, d, e, f equals a or -(a + 1)

All five equal a:
a^2 = 5a
a = 5 or 0

Four equal a:
a^2 = 4a - (a + 1) = 3a - 1
a^2 - 3a + 1 = 0
a = 3/2 +/- sqrt(5)/2
a = (3 + sqrt(5))/2: {(3 + sqrt(5))/2, (3 + sqrt(5))/2, (3 + sqrt(5))/2, (3 + sqrt(5))/2, (3 + sqrt(5))/2, -(5 + sqrt(5))/2)

a + b + c + d + e = 5 (3 + sqrt(5))/2 = (15 + 5 sqrt(5))/2
f^2 = (-(5 + sqrt(5)))^2 / 4 = (30 + 10 sqrt(5))/4 = (15 + 5 sqrt(5))/2
This is a solution

a = (3 - sqrt(5))/2: {(3 - sqrt(5))/2, (3 - sqrt(5))/2, (3 - sqrt(5))/2, (3 - sqrt(5))/2, (3 - sqrt(5))/2, -(5 - sqrt(5))/2)
a + b + c + d + e = 5 (3 - sqrt(5))/2 = (15 - 5 sqrt(5))/2
f^2 = (-(5 - sqrt(5)))^2 / 4 = (30 - 10 sqrt(5))/4 = (15 - 5 sqrt(5))/2

Three equal a:
a^2 = 3a - 2 (a + 1) = a - 2
a^2 - a + 2 = 0
a = 2 or -1
a = 2: {2, 2, 2, 2, -1, -1} does not fit into the first equation (7 = 1)
a = -1: {-1, -1, -1, -1, 2, 2} does not fit into the first equation (-2 = 4)

Two equal a:
a^2 = 2a - 3 (a + 1) = -a - 3
a^2 + a + 3 = 0 has no real solutions

One equals a:
a^2 = a - 4 (a + 1) = -3a - 4
a^2 + 3a + 4 = 0 has no real solutions

None equal a:
a^2 = -5 (a + 1) = -5a - 5
a^2 + 5a + 5 = 0 has no real solutions

Thus, the only real nonzero solutions are:
(a) All six variables = 5;
(b) Five of the variables = (3 + sqrt(5)) / 2, and the sixth = -(5 + sqrt(5)) / 2

You mentioned all of them except for one. The case where a = (3 - sqrt(5))/2 is also a valid solution. You did correctly solve for that value of a, but didn't mention that as a solution at the bottom of your post.

Nice explanation on why the other values of a don't give valid solutions.

gordonm888
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March 29th, 2021 at 9:55:08 AM permalink

This is an original problem that I have created, and I hope you find it interesting.
*****************
Statement: The game is Texas Hold'em with a 55 card deck - there are 3 jokers in the deck and they all play as being completely wild. (This is an actual game that is offered on-line on a "play money" site.) Hand rankings are as customary in a wild card game: FIve of a KInd is the highest hand category, followed by Royal Flush, and so on.

You have Qs-Js (Queen and Jack of spades) as your two hole cards. The flop has come as Joker-Joker-Joker. Congratulations. you have flopped a Royal Flush! However, there are two more common cards to be dealt and . . .

a) You have one opponent with two random cards. What is your probability of winning, tieing and losing?

b) Same question, but two opponents, each with two random cards.

c) Same question, but three opponents, each with two random cards.
****************
Comments: I have started to analyze this but do not yet have the answers. This problem is easily stated but has "subtleties", so I decided to share it. Results of looping codes or simulations are welcome and would be much appreciated, but I believe a combinatronics solution for the 1 and 2 opponent cases should be obtainable. A combinatronics analysis of the 3 opponent case, however, might turn your brain to jelly and therefore should only be attempted by forum members whose brains are already jelly.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

charliepatrick

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March 29th, 2021 at 10:55:48 AM permalink

I can see a method of approaching this which look at possible board types and then you can determine what other players need to tie or beat you.
Pairs
(i) Pair on the board (AA KK 10s-2s) so you are playing the board and can only be beaten by a higher hidden pair.
(ii) QQ,JJ similar to (i) except everyone has the same as you so you can only be beaten by AA or KK (or QQ).
Suited
(iii) Qx Jx - similar to (ii) except you have the Quins and AA KK (Qx) beats you and (Qx) Jx ties with you.
(iv) A/K with Q/J - similar to (iii) except you lose to AA/KK/(QQ) Ax/Kx/(Qx), as appropriate, beats you and Qx/(Jx) ties you.
(v) AK/AT/KT - here the board has RF but you can be beaten by any hidden pair or matching the board to make Quins.
(vi) Ax/Kx/Tx - here a player only needs a matching high suited card to tie your RF as well as hidden pair or matching a board card.
(vii) 9x or worse - player needs a hidden pair or to match a board card to beat you, or two suited cards that make RF to tie you.
Un-suited
Most of these are as above except sometimes the board won't have the RF.

gordonm888
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March 29th, 2021 at 1:34:17 PM permalink

Quote: charliepatrick
I can see a method of approaching this which look at possible board types and then you can determine what other players need to tie or beat you.
Pairs
(i) Pair on the board (AA KK 10s-2s) so you are playing the board and can only be beaten by a higher hidden pair.
(ii) QQ,JJ similar to (i) except everyone has the same as you so you can only be beaten by AA or KK (or QQ).
Suited
(iii) Qx Jx - similar to (ii) except you have the Quins and AA KK (Qx) beats you and (Qx) Jx ties with you.
(iv) A/K with Q/J - similar to (iii) except you lose to AA/KK/(QQ) Ax/Kx/(Qx), as appropriate, beats you and Qx/(Jx) ties you.
(v) AK/AT/KT - here the board has RF but you can be beaten by any hidden pair or matching the board to make Quins.
(vi) Ax/Kx/Tx - here a player only needs a matching high suited card to tie your RF as well as hidden pair or matching a board card.
(vii) 9x or worse - player needs a hidden pair or to match a board card to beat you, or two suited cards that make RF to tie you.
Un-suited
Most of these are as above except sometimes the board won't have the RF.

Yes, that is exactly what I came up with as an analytical approach. However, as you implement it you need to differentiate between As,Ks,Ts on the board versus A,K,T of other suits, and you need to be careful to avoid doublecounting opponent hands that can beat you two ways, or both beat you and tie you.

Last edited by: gordonm888 on Mar 29, 2021

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

charliepatrick

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March 30th, 2021 at 3:09:26 AM permalink

Just treat the game as a two-card poker game formed from your two cards and the two community cards.
If you can form a pair then that's the highest ranking hand (AA thru 22). If you can create a "top-five" flush then that's RF (i.e. any two from AKQJT in the same suit). All other hands are essentially the same and "losers".

This probably makes the brute force method easier as you just look for a pair or a "top-five", everything else can be ignored.

Ace2

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March 30th, 2021 at 4:12:35 PM permalink

A new hand grenade has been developed. From the time the lever is released, it takes an average of 6 seconds for it to explode. This average time follows the exponential distribution.

If you drop this grenade into a bottomless pit, what is the expected distance it will fall before exploding?

Assume gravitational force of 32 feet per second per second and no air resistance

It�s all about making that GTA

ThatDonGuy

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April 8th, 2021 at 6:28:04 PM permalink

Quote: Ace2
A new hand grenade has been developed. From the time the lever is released, it takes an average of 6 seconds for it to explode. This average time follows the exponential distribution.

Somebody is going to have to explain to me just what "this average time follows the exponential distribution" means.

Meanwhile, here's one from one of this year's American Invitational Mathematics Exams:

A teacher has four consecutive integers from 10 to 99 inclusive. The teacher gives a different number to each of four students. The students cannot see or otherwise communicate with each other.
The teacher tells each of them, "Each of you has been given one of four consecutive integers from 10 to 99 inclusive. Exactly one of the four numbers is a multiple of 6; exactly one of the other three numbers is a multiple of 7. Raise your hand if you know what the largest of the four numbers is."
Nobody raises their hands.
The teacher then tells them, "Nobody raised their hand. Given that information, raise your hand if you know what the largest of the four numbers is."
All four raise their hands.
What are all of the possible values of the largest of the four numbers?

chevy

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April 8th, 2021 at 7:42:44 PM permalink

A first try of
8 (5, 6, 7, 8)
37 (34, 35, 36, 37)
50 (47, 48, 49, 50)
79 (76, 77, 78, 79)
92 (89, 90, 91, 92)

The two multiples numbers (mult of 6 and mult of 7) must be consecutive.
If there are 2 numbers between, one of the middle two knows the higher multiple is max.
If there is 1 number between,
If the other is lower, then that person knows the higher multiple is max.
If the other is higher, then that person knows they have the max.

So the two multiples are consecutive.
If two others are above, the person 2 higher than a multiple has max
If two others are below, the person 2 lower than multiple knows the higher multiple is max.

So must be one above and one below the consecutive multiples. And since nobody could deduce a max, then they know the person 1 above the higher multiple has max

or so it seems

Ace2

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April 8th, 2021 at 7:50:13 PM permalink

Quote: ThatDonGuy
Somebody is going to have to explain to me just what "this average time follows the exponential distribution" means.

You can think of it as the continuous version of the binomial (discrete) distribution. For instance, the probability of a 00 not hitting in 60 roulette spins is (37/38)^60 = 20.2%.

Using the exponential distribution it�s 1/e^(60/38) = 20.6%. That�s just for illustration purposes since spins are a discrete variable. But most things measured in time, for instance, are continuous not discrete. This example could be worded as: it takes an average of 38 minutes for a 00 to appear, following the exponential distribution. What is the probability that it hasn�t appeared after waiting 60 minutes? 20.6%

It�s all about making that GTA

charliepatrick

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April 9th, 2021 at 12:57:43 AM permalink

Firstly if there is a solution x, x+1, x+2, x+3 then x+42, x+43, x+44, x+45 has similar logic. So only consider 10...13 up to 51...54 and any solution found creates similar ones 42, 84 etc. up.

If the sequence contains 42 then the holder of 42 can never know whether the other numbers are higher or lower. It is true either 39/45 would know on the first round and 40/44 would know on the second round; but the other people wouldn't.

If the multiple of 7 is at either end, then the other end is either 7x+3 or 7x-3, and they would know the situation on the first round. So the multiple of 7 must be in the middle.

Similarly if the multiple of 6 is at either end (A), the the other end (D) is either 6x+3 or 6x-3. D knows they are at one end and can work it out: knowing there's a multiple of 7 between will enable them to decide between these two. For instance holding 51 they could be 48,49,50,51 or 51,52,53,54 however the former has a multiple of 7, the other doesn't. There can only be one multiple of 7 in seven consecutive numbers, so D will always be able to find out which end of the sequence they lie.

So the 6x and 7x are in the middle, e.g. 34 35 36 37.

On the first round (if it is say 34 35 36 37) no-one can be sure whether they're part of 33 34 35 36 or 35 36 37 38 or 34 35 36 37. On the second round they know that 33 or 38 would have spoken up and didn't, that rules out the other two options, so everyone now knows the situation. It doesn't matter which way round the 6x and 7x are, the same logic applies (e.g. 47 48 49 50 not 46/51)

Working through the lower list of numbers these are the only two solutions. So there are also ones 42 larger.

This gives 37 (34 35 36 37) 50 (47 48 49 50) 79 (76 77 78 79) 92 (89 90 91 92). There is also a solution 8 (5 6 7 8) but that lies outside the range asked in the original question.

Wizard
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April 9th, 2021 at 6:40:09 AM permalink

Quote: Ace2
If you drop this grenade into a bottomless pit, what is the expected distance it will fall before exploding?

I get 1152 feet. If I'm right, I'll provide a solution

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ThatDonGuy

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April 9th, 2021 at 8:01:47 AM permalink

Quote: chevy

A first try of
8 (5, 6, 7, 8)
37 (34, 35, 36, 37)
50 (47, 48, 49, 50)
79 (76, 77, 78, 79)
92 (89, 90, 91, 92)

The two multiples numbers (mult of 6 and mult of 7) must be consecutive.
If there are 2 numbers between, one of the middle two knows the higher multiple is max.
If there is 1 number between,
If the other is lower, then that person knows the higher multiple is max.
If the other is higher, then that person knows they have the max.

So the two multiples are consecutive.
If two others are above, the person 2 higher than a multiple has max
If two others are below, the person 2 lower than multiple knows the higher multiple is max.

So must be one above and one below the consecutive multiples. And since nobody could deduce a max, then they know the person 1 above the higher multiple has max

or so it seems

Quote: charliepatrick
Firstly if there is a solution x, x+1, x+2, x+3 then x+42, x+43, x+44, x+45 has similar logic. So only consider 10...13 up to 51...54 and any solution found creates similar ones 42, 84 etc. up.

If the sequence contains 42 then the holder of 42 can never know whether the other numbers are higher or lower. It is true either 39/45 would know on the first round and 40/44 would know on the second round; but the other people wouldn't.

If the multiple of 7 is at either end, then the other end is either 7x+3 or 7x-3, and they would know the situation on the first round. So the multiple of 7 must be in the middle.

Similarly if the multiple of 6 is at either end (A), the the other end (D) is either 6x+3 or 6x-3. D knows they are at one end and can work it out: knowing there's a multiple of 7 between will enable them to decide between these two. For instance holding 51 they could be 48,49,50,51 or 51,52,53,54 however the former has a multiple of 7, the other doesn't. There can only be one multiple of 7 in seven consecutive numbers, so D will always be able to find out which end of the sequence they lie.

So the 6x and 7x are in the middle, e.g. 34 35 36 37.

On the first round (if it is say 34 35 36 37) no-one can be sure whether they're part of 33 34 35 36 or 35 36 37 38 or 34 35 36 37. On the second round they know that 33 or 38 would have spoken up and didn't, that rules out the other two options, so everyone now knows the situation. It doesn't matter which way round the 6x and 7x are, the same logic applies (e.g. 47 48 49 50 not 46/51)

Working through the lower list of numbers these are the only two solutions. So there are also ones 42 larger.

This gives 37 (34 35 36 37) 50 (47 48 49 50) 79 (76 77 78 79) 92 (89 90 91 92). There is also a solution 8 (5 6 7 8) but that lies outside the range asked in the original question.

Both correct.
Chevy - note that the numbers range from 10 to 99, not 1 to 99.
Charlie - note that 42 and 84 were left out as the multiple of 6 and the multiple of 7 have to be different numbers.

Since the multiple of 6 and the multiple of 7 are different numbers, 42 and 84 are not among them.

The possible multiples of 7, with their possible number ranges, are:
14: 11-14, 12-15
21: 18-21, 21-24
28: 27-30, 28-31
35: 33-36, 34-37, 35-38
49: 46-49, 47-50, 48-51
56: 53-56, 54-57
63: 60-63, 63-66
70: 69-72, 70-73
77: 75-78, 76-79, 77-80
91: 88-91, 89-92, 90-93
98: 95-98, 96-99

If one of the students has, say, 14, then either another has 11 or another has 15;
11 and 15 each appear in only one set of numbers, so if someone has one of those numbers,
they would know what the set is.
Similarly, if one has 21, then either another has 18 or another has 24, and would know the set.
However, with the four numbers in three sets, the numbers in the middle set are all in at least
two sets, so none of the students could be sure what the set is - until they are told that none
of the students know what the set is.
The highest numbers of the four "middle sets" are 37, 50, 79, and 92.

Ace2

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April 9th, 2021 at 1:04:52 PM permalink

Quote: Wizard
I get 1152 feet. If I'm right, I'll provide a solution

I agree with that

It�s all about making that GTA

Wizard
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April 9th, 2021 at 1:33:21 PM permalink

Quote: Ace2
I agree with that

Yay! Here is a quick solution:

Let's start with an equation for how far the grenade has fallen after t seconds.

Given acceleration is 32 ft/sec^2, the velocity at time t is the integral of that = 32t. We know the velocity is 0 at t=0, so we don't need to fuss with the constant of integration.

The integral of that is the distance fallen, which is 16t^2. The distance fallen at t=0 is zero, so the constant of integration is 0.

We're given the expected lifetime of the grenade is six seconds and it follows the memoryless property of the exponential distribution.

The density function for the probability of exploding at time t is 1/6 * exp(-x/6).

So, to solve the problem we integrate the product of the distance fallen and the probability of explosion for t = 0 to infinity.

This product is f(t) = 16t^2 * 1/6 * exp(-t/6)

Then you run through integration by parts twice to get:

(8/3)*(-6t^2 * exp(-t/6) - 72t * exp(-t/6) - 432 exp(-t/6)) from 0 to infinity = 1152.

Good problem!

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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April 12th, 2021 at 5:55:39 PM permalink

Quote: Ace2
A new hand grenade has been developed. From the time the lever is released, it takes an average of 6 seconds for it to explode. This average time follows the exponential distribution.

If you drop this grenade into a bottomless pit, what is the expected distance it will fall before exploding?

Assume gravitational force of 32 feet per second per second and no air resistance

Here is a first draft of my solution (PDF) to this problem. I welcome all comments and corrections.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Ace2

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April 13th, 2021 at 9:22:17 AM permalink

My approach was:

The integral over all time of 1/6 * e^(-t/6) gives us the probability distribution of the grenade exploding at any point in time t. We need to know the weighted average/sum of t^2 for all of these points, so taking the integral of 1/6 * e^(-t/6) * t^2 we get 72.

Plug that into .5 * 32 * t^2 to get 1152 feet.

It�s all about making that GTA

Ace2

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April 15th, 2021 at 6:49:43 PM permalink

Two gamblers have ten black chips each. They decide to roll a pair of dice with player A winning one chip from B if the total is seven or higher, otherwise he loses one chip to B. The game ends when one player holds all twenty chips. How many rolls, on average, should the game last? Looking for an exact expression of the answer

It�s all about making that GTA

gordonm888
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April 16th, 2021 at 5:07:17 PM permalink

Here's a simple one that I hope will make you think a bit and surprise you.

Definition: a Texas hold'em hand is two hole cards and five community cards which is rated according to the best five card poker hand it can make.

Problem: What is a (7-card) Texas hold'em hand that has 0% chance of winning and 0% chance of tieing, given one or more opponents? To be clear, I want a 7-card Texas Hold-em hand with 100% chance of losing to any opponent,, which makes it the worst possible Texas hold-em hand.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

gordonm888
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April 16th, 2021 at 5:07:22 PM permalink

Here's a simple one that I hope will make you think a bit and surprise you.

Definition: a Texas hold'em hand is two hole cards and five community cards which is rated according to the best five card poker hand it can make.

Problem: What is a (7-card) Texas hold'em hand that has 0% chance of winning and 0% chance of tieing, given one or more opponents? To be clear, I want a 7-card Texas Hold-em hand with 100% chance of losing to any opponent,, which makes it the worst possible Texas hold-em hand.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

gordonm888
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April 16th, 2021 at 5:07:25 PM permalink

Here's a simple one that I hope will make you think a bit and surprise you.

Definition: a Texas hold'em hand is two hole cards and five community cards which is rated according to the best five card poker hand it can make.

Problem: What is a (7-card) Texas hold'em hand that has 0% chance of winning and 0% chance of tieing, given one or more opponents? To be clear, I want a 7-card Texas Hold-em hand with 100% chance of losing to any opponent with any two hole cards, which makes it the worst possible Texas hold-em hand.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

ChesterDog

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April 16th, 2021 at 5:18:22 PM permalink

Quote: gordonm888
Here's a simple one that I hope will make you think a bit and surprise you.

Definition: a Texas hold'em hand is two hole cards and five community cards which is rated according to the best five card poker hand it can make.

Problem: What is a (7-card) Texas hold'em hand that has 0% chance of winning and 0% chance of tieing, given one or more opponents? To be clear, I want a 7-card Texas Hold-em hand with 100% chance of losing to any opponent with any two hole cards, which makes it the worst possible Texas hold-em hand.

board = 3 3 3 3 2
my hand = 2 2

I have four 3s with a kicker of 2. Everyone else has to have a better kicker than I have, so I can't tie or win.

It looks like I have a great hand with four-of-a-kind and three-of-a-kind, but I have to lose.

gordonm888
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April 16th, 2021 at 5:28:30 PM permalink

Quote: ChesterDog

board = 3 3 3 3 2
my hand = 2 2

I have four 3s with a kicker of 2. Everyone else has to have a better kicker than I have, so I can't tie or win.

It looks like I have a great hand with four-of-a-kind and three-of-a-kind, but I have to lose.

CORRECT.

You posted the correct answer 11 minutes after the problem was posted! What took you so long?

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

ChesterDog

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April 16th, 2021 at 5:51:25 PM permalink

Quote: gordonm888
Quote: ChesterDog

board = 3 3 3 3 2
my hand = 2 2

I have four 3s with a kicker of 2. Everyone else has to have a better kicker than I have, so I can't tie or win.

It looks like I have a great hand with four-of-a-kind and three-of-a-kind, but I have to lose.

CORRECT.

You posted the correct answer 11 minutes after the problem was posted! What took you so long?

Your clues were very helpful.

Actually, a board of 2 with any four-of-a-kind would do.

It would be fun to see that hand on TV. The player would have a pair of 2s, and if the flop was 2 and any pair, the player would feel really good with a full house.

ThatDonGuy

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April 16th, 2021 at 6:46:20 PM permalink

Quote: Ace2
Two gamblers have ten black chips each. They decide to roll a pair of dice with player A winning one chip from B if the total is seven or higher, otherwise he loses one chip to B. The game ends when one player holds all twenty chips. How many rolls, on average, should the game last? Looking for an exact expression of the answer

Let E(N) = the expected number of rolls when A has N chips
E(20) = E(0) = 0
For all N from 1 to 19 inclusive, E(N) = 1 + 7/12 E(N+1) + 5/12 E(N-1)
12 E(N) - 7 E(N+1) - 5 E(N-1) = 12
Since E(20) and E(0) are known, this is 19 equations in 19 unknowns
12 E(19) - 5 E(18) = 12
12 E(18) - 7 E(19) - 5 E(17) = 12
12 E(17) - 7 E(18) - 5 E(16) = 12
...
12 E(2) - 7 E(3) - 5 E(1) = 12
12 E(1) - 7 E(2) = 12
Solve for E(10):

The solution is 8,181,288,720 / 146,120,437, or about 55.99.

Ace2

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April 16th, 2021 at 7:08:25 PM permalink

Quote: ThatDonGuy

Let E(N) = the expected number of rolls when A has N chips
E(20) = E(0) = 0
For all N from 1 to 19 inclusive, E(N) = 1 + 7/12 E(N+1) + 5/12 E(N-1)
12 E(N) - 7 E(N+1) - 5 E(N-1) = 12
Since E(20) and E(0) are known, this is 19 equations in 19 unknowns
12 E(19) - 5 E(18) = 12
12 E(18) - 7 E(19) - 5 E(17) = 12
12 E(17) - 7 E(18) - 5 E(16) = 12
...
12 E(2) - 7 E(3) - 5 E(1) = 12
12 E(1) - 7 E(2) = 12
Solve for E(10):

The solution is 8,181,288,720 / 146,120,437, or about 55.99.

That is the correct answer!

I got to it using this formula:

[2 * (1 - (q/p)^n) / (1 - (q/p)^N) - 1] * n / (p - q)

Where p=7/12, q=1-p, n=10, N=20

I still haven�t entirely figured out why it works though.

(1 - (q/p)^n) / (1 - (q/p)^N =~ 96.7% is the risk of ruin for player B.

n / (p - q) = 60 is the average number of rolls it will take for A to have 10 more wins than B

It�s all about making that GTA

ThatDonGuy

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April 18th, 2021 at 1:10:14 PM permalink

Quote: Ace2
I got to it using this formula:

[2 * (1 - (q/p)^n) / (1 - (q/p)^N) - 1] * n / (p - q)

Where p=7/12, q=1-p, n=10, N=20

I should have noticed that this was a Gamblers' Ruin problem...

I assume n = the starting bankroll and N = the target success bankroll (i.e. you start with n and make bets of 1 with winning probability p, and continue until your bankroll is either N or 0).
The formula I have, from Ethier's The Doctrine of Chances, is:

n / (q - p) - N / (q - p) * ((q/p)^n - 1) / ((q/p)^N - 1)

I'm not exactly sure how it was derived either; the proof is in the book, but you have to jump through some hoops to get it.

Note that, where N = 2n, as in your case, this reduces to:

n / (q - p) * ((q/p)^n - 1) / ((q/p)^n + 1)

Wizard
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April 24th, 2021 at 6:17:33 PM permalink

Two ferries leave at the same time from ports A and B. Each is bound for the opposite port and then immediately turns back towards its point of origin. Assume no time spent making the u-turn.

200 miles from port A, they cross the first time.

In making the return trip, they cross 100 miles from port B.

How long is the channel between A and B?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

chevy

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April 24th, 2021 at 7:34:42 PM permalink

My first attempt

Assuming they cross both times in opposite direction....i.e. the second crossing is after they both turn around....
Let locations x_A= Origin, x_B = Z
Let speeds v_A = speed of ship leaving port A, v_B = speed of ship leaving port B
Let times t_1 = first meeting, t_2 = second meeting

First meeting:
v_A * t_1 = 200
v_B * t_1 = Z-200

take ratio v_B/v_A = (Z-200)/200

Second Meeting:
v_A * t_2 = Z+100
v_B * t_2 = 2Z - 100

Take ratio v_B/v_A = (2Z-100)/(Z+100)

Set ratios equal

(Z-200)/200 = (2Z-100)/(Z+100)

Z^2-100Z-20,000 = 400Z-20,000

Z^2-500Z=0

Z=0 or 500.

Z=500miles being the physical relevant answer

But I got to thinking, you don't say the second crossing is after BOTH are returning. (though I imagine that is implied). But if only one is on return........

1) ship B is traveling fast enough to turn and catch ship A on it's first traverse.

First meeting:
v_A * t_1 = 200
v_B * t_1 = Z-200
take ratio v_B/v_A = (Z-200)/200

Second Meeting:
v_A * t_2 = Z-100
v_B * t_2 = 2Z - 100
Take ratio v_B/v_A = (2Z-100)/(Z-100)

Set ratios equal, cross multiply and rearrange, I get qudratic
Z^2-700Z+40,000=0
Complex solutions
xxxxxxx

2) ship A is traveling fast enough to turn and catch ship B on it's first traverse.

First meeting:
v_A * t_1 = 200
v_B * t_1 = Z-200
take ratio v_B/v_A = (Z-200)/200

Second Meeting:
v_A * t_2 = Z+100
v_B * t_2 = 100
Take ratio v_B/v_A = (100)/(Z+100)

Set ratios equal, cross multiply and rearrange, I get qudratic
Z^2-100Z-40,000=0

Solutions of
Z=50 +/- 50sqrt(17)

Only + gives physical distance. Z=50+50sqrt(17)

SO: either
Z=500 miles or 50+50sqrt(17) = 256.155281280883027 miles
v_B/v_A = 3/2. or. v_B/v_A = (-3/4 +(1/4)*sqrt(17)) = 0.280776406404415

Wizard
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April 24th, 2021 at 7:47:26 PM permalink

Sorry, the second meeting is when both are returning.

Maybe I'm the one in error, but I don't agree with any of your answers, chevy.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

chevy

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April 24th, 2021 at 8:03:55 PM permalink

Doing a lot of the algebra in my head, but checking with some numbers

1) Most I can find is a ratio for speeds. v_B/v_A=3/2

let v_A = 100 mph. Then v_B = 150 mph

First meeting is 200miles from A.....so t_1 = 2 hrs.

at t_1= 2hrs, B has travelled 2*150 = 300miles.

If total is 500 miles 200 + 300 works.

Second meeting is 100 miles from B......so A travels 500+100(return) = 600 miles.....so t_2 = 6hrs

at t_2 = 6 hrs, B has travelled 6*150 = 900 miles. = 500 + 400(return)

So that works, and again any speeds giving that ratio seems to work.

2) Same construction
v_B/v_A = 0.28
v_A = 100mph.......so v_B = 28mph (rounding)

first meeting 200 from A .... so t_1= 2 hr.
at t_1 = 2 hr.....B travelled 2*28 = 56 miles

If total is 256 miles 200 + 56 works.

Second meeting is 100 miles from B......so A travels 256+100(return) =356 miles.....so t_2 = 3.56hrs

at t_2 = 3.56 hrs, B has travelled 3.56*28 = 99.7 miles. = ~100 from B

So that works, and again any speeds giving that ratio seems to work.

?????????

charliepatrick

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April 25th, 2021 at 2:12:00 AM permalink

Let a=speed of ferry A, and b=speed of ferry B. Let x=distance B travels to first meeting.
To first meeting A goes 200, and B goes x.
Time to first meeting: 200/a = x/b.
Distance from A to B = 200+x.
To second meeting A goes 200+x there and 100 back = 300+x. B goes 200+x then (200+x-100) back = 300+2x.
Time to second meeting: (300+x)/a = (300+2x)/b.

Consider a/b from both equations gives 200/x = (300+x)/(300+2x).
gives x²-100x-60000=0
equals (x-300)*(x+200)=0.
So x=300. AB=500 miles.
(A goes 200 while B goes 300 to first meeting; A goes 500+100=600, B goes 500+(500-100)=900 , so ratio still 2:3.)

ThatDonGuy

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April 25th, 2021 at 8:38:11 AM permalink

Let D be the length of the channel in miles, A be A's speed, and B be B's speed

The first meeting is when A moves 200 miles in time 200 / A and B moves (D - 200) miles in time (D - 200) / B
200 / A = (D - 200) / B
200 B = (D - 200) A
B = A (D - 200) / 200

The second meeting is when A moves (D + 100) miles in time (D + 100) / A and B moves (2D - 100) miles in time (2D - 100) / B
(D + 100) / A = (2D - 100) / B
B (D + 100) = A (2D - 100)
A (D - 200) (D + 100) / 200 = A (2D - 100)
(D - 200) (D + 100) = 200 (2D - 100)
D^2 - 100 D - 20,000 = 400 D - 20,000
D^2 = 500 D
D > 0, so D = 500 miles

The generic solution is, the channel length = 3 times the distance from A to the first meeting point - the distance from B to the second meeting point

chevy

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April 25th, 2021 at 9:23:09 AM permalink

Turns out I can't do all the math in my head after all. Corrected mistake in my alternate possibilities.

1) ship B is traveling fast enough to turn and catch ship A on it's first traverse.

First meeting:
v_A * t_1 = 200
v_B * t_1 = Z-200
take ratio v_B/v_A = (Z-200)/200

Second Meeting:
v_A * t_2 = Z-100
v_B * t_2 = 2Z - 100
Take ratio v_B/v_A = (2Z-100)/(Z-100)

Set ratios equal, cross multiply and rearrange, I get qudratic
Z^2-700Z+40,000=0

********EDITED because I can't square 700 in my head******

Solutions of
Z=350 +/- 50sqrt(33)

Only + gives physical distance. Z=350+50sqrt(33)

the "-" gives Z =62.771867673098567 which is not possible given the setup (first meet 200 from A, etc)

*********END EDIT***********************************************

2) ship A is traveling fast enough to turn and catch ship B on it's first traverse.

First meeting:
v_A * t_1 = 200
v_B * t_1 = Z-200
take ratio v_B/v_A = (Z-200)/200

Second Meeting:
v_A * t_2 = Z+100
v_B * t_2 = 100
Take ratio v_B/v_A = (100)/(Z+100)

Set ratios equal, cross multiply and rearrange, I get qudratic
Z^2-100Z-40,000=0

Solutions of
Z=50 +/- 50sqrt(17)

Only + gives physical distance. Z=50+50sqrt(17)

SO: If question is just "Ships meet first at 200 miles from A, and meet second at 100 miles from B". (independent of direction). I get 3 possibilities

0) Both on return trip
Z=500 miles. with v_B/v_A = 3/2.

1) ship B is traveling fast enough to turn and catch ship A on it's first traverse.
Z=350 + 50sqrt(33) = 637.228132326901433 miles with v_B/v_A = (+3/4 +(1/4)*sqrt(17)) =2.186140661634507

2) ship A is traveling fast enough to turn and catch ship B on it's first traverse.
Z=50+50sqrt(17) = 256.155281280883027 miles with v_B/v_A = (-3/4 +(1/4)*sqrt(17)) = 0.280776406404415

charliepatrick

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April 25th, 2021 at 11:22:01 AM permalink

Quote: chevy
...**EDITED** Other Possibilities...

If ferry a sets out from Port A, however slowly, and ferry b sets out from Port B, then ferry b must meet ferry a before reaching Port A.

Similar to other solution work out a/b for both meetings. A to B is 200+x.
First one is 200/x.
Second one: a has just 100 miles to go to reach Port B, so has travelled 200+x-100 = 100+x; b has almost gone from B to A back to B, so has gone (200+x)*2-100 = 300+2x.
200/x=100+x/(300+2x) gives x²-300x-60000=0.
x=(300 +/- SQRT(90k+240k))/2 = 150 +/- SQRT(82500) = about 437.228
So the distance A to B is 637.228.

Check ratio of speeds 437.228/200 or 1174.456/537.228 is the same.

chevy

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April 25th, 2021 at 2:08:57 PM permalink

Quote: charliepatrick
If ferry a sets out from Port A, however slowly, and ferry b sets out from Port B, then ferry b must meet ferry a before reaching Port A.
Similar to other solution work out a/b for both meetings. A to B is 200+x.
First one is 200/x.
Second one: a has just 100 miles to go to reach Port B, so has travelled 200+x-100 = 100+x; b has almost gone from B to A back to B, so has gone (200+x)*2-100 = 300+2x.
200/x=100+x/(300+2x) gives x²-300x-60000=0.
x=(300 +/- SQRT(90k+240k))/2 = 150 +/- SQRT(82500) = about 437.228
So the distance A to B is 637.228.

Check ratio of speeds 437.228/200 or 1174.456/537.228 is the same.

CharliePatrick: I am not sure if you were saying my other options were wrong? Or if you were confirming one of them....we agree on one of them, but I think I also found another.

Though you use different variables, we Got the same answer for my options listed as 0) and 1) at end of my earlier solution. I think there is still the case where B is traveling so slow that A passes it and catches it on the return (B still on initial transverse)

Since you comment about the first meeting, maybe I was unclear. I previously commented to Wiz about if both were on return for second passing. THEN my additional options were meant to apply to the SECOND passing. I may have not made that clear.

------------Repeating my solutions without the work, and with Clarifying descriptions--------------

SO: If question is just "Ships meet first at 200 miles from A, and meet second at 100 miles from B". (independent of direction). I get 3 possibilities....in all of them the first meeting was passing each other in initial opposite directions

0) *****Second Meeting is when ******** Both on return trip
Z=500 miles.
with v_B/v_A = 3/2.

1) *****Second Meeting is when ******** ship B is traveling fast enough to turn and catch ship A ***FROM BEHIND*** on A's first traverse.
Z=350 + 50sqrt(33) = 637.228132326901433 miles
with v_B/v_A = (+3/4 +(1/4)*sqrt(17)) =2.186140661634507

2) *****Second Meeting is when ******** ship A is traveling fast enough to turn and catch ship B ***FROM BEHIND*** on B's first traverse.
Z=50+50sqrt(17) = 256.155281280883027 miles
with v_B/v_A = (-3/4 +(1/4)*sqrt(17)) = 0.280776406404415

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

charliepatrick

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April 25th, 2021 at 2:46:28 PM permalink

^ Yes I'm agree with 0) and 1) and, because I misread it, can see that 2) is correct using a similar method.

Wizard
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April 25th, 2021 at 3:50:12 PM permalink

Congratulations to all who got 500. I made an error in my original answer, so I apologize to chevy and everyone else for saying otherwise.

I'll reword the question this way the next time I ask it, unless anyone thinks the wording can be improved.

Quote:

Ferry A leaves Avonlea at the same time as ferry B leaves Bree. Both head directly across the same channel towards the opposite village, make an immediate u-turn and head back to their port of original port. Each ferry travels at a constant speed.

The two ferries cross the first time 200 miles from Avonlea.

On their return trip, both ferries cross 100 miles from Bree.

What is the distance between A and B?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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April 25th, 2021 at 5:56:53 PM permalink

Five red dice and five blue dice are rolled. What is the probability the two rolls are the same, disregarding the order the dice are thrown?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

chevy

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April 25th, 2021 at 6:39:45 PM permalink

Quote: Wizard

I'll reword the question this way the next time I ask it, unless anyone thinks the wording can be improved.

I think the wording is fine if you just want the one answer. If you want to allow for the three cases I would change "cross" to "meet" and remove the part about return trips. e.g.

Ferry A leaves Avonlea at the same time as ferry B leaves Bree. Both head directly across the same channel towards the opposite village, make an immediate u-turn and head back to their port of original port. Each ferry travels at a constant speed.

The two ferries meet the first time 200 miles from Avonlea.

The two ferries meet the second time 100 miles from Bree.

What are the possible distances between A and B?

Wizard
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April 25th, 2021 at 8:02:36 PM permalink

Quote: chevy
Ferry A leaves Avonlea at the same time as ferry B leaves Bree. Both head directly across the same channel towards the opposite village, make an immediate u-turn and head back to their port of original port. Each ferry travels at a constant speed.

The two ferries meet the first time 200 miles from Avonlea.

The two ferries meet the second time 100 miles from Bree.

What are the possible distances between A and B?

Thank you!

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Ace2

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April 25th, 2021 at 8:09:51 PM permalink

Quote: Wizard
Five red dice and five blue dice are rolled. What is the probability the two rolls are the same, disregarding the order the dice are thrown?

Assuming I understand the question, about 1 in 5,666

Last edited by: Ace2 on Apr 25, 2021

It�s all about making that GTA

chevy

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April 25th, 2021 at 8:17:14 PM permalink

Just noticed that by naming the ports, the last line should be

What are the possible distances between Avonlea and Bree?

Wizard
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April 25th, 2021 at 9:18:58 PM permalink

Quote: chevy
Just noticed that by naming the ports, the last line should be

What are the possible distances between Avonlea and Bree?

I want to phrase it in a way that there is only one reasonable answer. That is why I added wordage about it being a round trip.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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