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charliepatrick
charliepatrick
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Thanks for this post from:
rsactuary
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
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
Itís all about making that GTA
ThatDonGuy
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
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
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 itís 1/e^(60/38) = 20.6%. Thatís 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 hasnít appeared after waiting 60 minutes? 20.6%
Itís all about making that GTA
charliepatrick
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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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
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
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
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
Itís all about making that GTA
Wizard
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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!

It's not whether you win or lose; it's whether or not you had a good bet.

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