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Wizard
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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.
It's not whether you win or lose; it's whether or not you had a good bet.
Ace2
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.
Itís 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
Itís all about making that GTA
gordonm888
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
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
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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Thanks for this post from:
gordonm888
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
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
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.


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