Dice problem
February 20th, 2012 at 9:51:19 PM
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Start with 40. Roll a dice, and subtract that number from the count until you reach 0. But there is a twist, if you roll a 1, you start from 40 again.
Actually, this is the math homework for my grade 3 son to work on his subtraction. But as a father who is stuck with it, you want to get it over with as quick as possible. So what is the expected number of times you have to roll the dice to reach 0?
I know the chance of getting through in one pass t = (5/6)^x.
My guess of x is : 40/mean of (2 to 6) = 40/4 = 10 . This way of thinking is probably correct if not for the "bankrupt" feature, but I am not sure if it is still good with it.
Actually, this is the math homework for my grade 3 son to work on his subtraction. But as a father who is stuck with it, you want to get it over with as quick as possible. So what is the expected number of times you have to roll the dice to reach 0?
I know the chance of getting through in one pass t = (5/6)^x.
My guess of x is : 40/mean of (2 to 6) = 40/4 = 10 . This way of thinking is probably correct if not for the "bankrupt" feature, but I am not sure if it is still good with it.
February 20th, 2012 at 9:54:20 PM
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Let me get this straight.
You are asking for help with your eight year old son's math homework?
You are asking for help with your eight year old son's math homework?
In a bet, there is a fool and a thief.
- Proverb.
February 20th, 2012 at 10:11:28 PM
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You know why I love this post? It allows me to say the following:
So I ran a million trial simulation on your son's third grade homework and got the following results:
Rolls: 33,360,950
Average Rolls Until Total <= 0: 33.36
So I ran a million trial simulation on your son's third grade homework and got the following results:
Rolls: 33,360,950
Average Rolls Until Total <= 0: 33.36
February 20th, 2012 at 10:18:32 PM
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I don't think he should be able to turn that in.
Question: does a roll that puts you below zero not count, equivalent to Chutes and Ladders? Or is the ending criteria zero or lower, equivalent to Candy Land?
(It's a safe bet I have kids, huh?)
Question: does a roll that puts you below zero not count, equivalent to Chutes and Ladders? Or is the ending criteria zero or lower, equivalent to Candy Land?
(It's a safe bet I have kids, huh?)
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
February 20th, 2012 at 10:18:49 PM
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Quote: PopCanYou know why I love this post? It allows me to say the following:
So I ran a million trial simulation on your son's third grade homework and got the following results:
Rolls: 33,360,950
Average Rolls Until Total <= 0: 33.36
But it is homework. Must show math!!! ;)
February 20th, 2012 at 10:21:29 PM
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Quote: WongBoLet me get this straight.
You are asking for help with your eight year old son's math homework?
He's taking his son's homework, which is simply to help with subtraction, and asking a new question of it. What is the expected number of rolls before you would be complete.
The average value of each roll that would have subtraction value would be (2+3+4+5+6)/5, which is 4. This means that 5/6 of the time, you can expect a subtraction of 4. The other 1/6 of the time, you would reset to 40.
February 20th, 2012 at 10:28:29 PM
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Ummm ok. My post was a joke. Sorry you didn't get it.
In a bet, there is a fool and a thief.
- Proverb.
February 20th, 2012 at 10:30:13 PM
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Quote: WongBoUmmm ok. My post was a joke. Sorry you didn't get it.
Don't be sorry. Sarcasm is often lost over the internet. I was simply pointing out that the problem he asked is not typical of 8 year old math curriculum.
February 20th, 2012 at 10:32:52 PM
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Quote: TriplellDon't be sorry. Sarcasm is often lost over the internet. I was simply pointing out that the problem he asked is not typical of 8 year old math curriculum.
If that was typical 3rd grade math homework casinos would be out of business.
February 20th, 2012 at 10:37:26 PM
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It doesn't really have an answer does it?
It just has a range of probabilities?
It just has a range of probabilities?
In a bet, there is a fool and a thief.
- Proverb.
February 20th, 2012 at 10:46:00 PM
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Quote: WongBoIt doesn't really have an answer does it?
It just has a range of probabilities?
I'm sure the Wizard could use on of his iterative techniques to figure out the exact number of average rolls. That's well beyond my highschool math education, however.
February 21st, 2012 at 1:24:02 AM
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Quote: MathExtremistI don't think he should be able to turn that in.
Question: does a roll that puts you below zero not count, equivalent to Chutes and Ladders? Or is the ending criteria zero or lower, equivalent to Candy Land?
(It's a safe bet I have kids, huh?)
lets say is zero or below
February 21st, 2012 at 1:37:07 AM
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Quote: WongBoIt doesn't really have an answer does it?
It just has a range of probabilities?
actually, getting an "expected number of roll" is the easier part of the question. "a range of probabilities" would be more difficult as you would have to figure out the standard deviation (in order to find a 95% confidence interval etc).
February 21st, 2012 at 7:41:02 AM
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I calculate 33.37217893, which comes very close to another poster's simulation.
So, how many times did it take your son?
So, how many times did it take your son?
I heart Crystal Math.
February 21st, 2012 at 9:22:13 AM
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something tells me what the teacher thinks the answer is, is actually incorrect
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
February 21st, 2012 at 5:47:33 PM
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Quote: CrystalMathI calculate 33.37217893, which comes very close to another poster's simulation.
So, how many times did it take your son?
before i started i think it is going to take FOREVER, but luckily i reached 0 in about 20 rolls
February 21st, 2012 at 6:18:04 PM
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i tried calculating using excel assuming it takes 10 rolls to reach 0 (without hitting the 1), and in average you hit the 1 in every 6 rolls.
it shows you need about 40.2 rolls.
it shows you need about 40.2 rolls.
February 21st, 2012 at 8:14:24 PM
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Yeah, I'm having a hard time getting this calculated.
I know that if you just discarded 1s, the probabilities are:
The problem is what happens when a one is rolled?
I know that if you just discarded 1s, the probabilities are:
| # of Rolls | Probability |
|---|---|
| 7 | 0.0004608 |
| 8 | 0.0291072 |
| 9 | 0.1779712 |
| 10 | 0.336336384 |
| 11 | 0.28534636544 |
| 12 | 0.129685766144 |
| 13 | 0.034703417344 |
| 14 | 0.00574724915200001 |
| 15 | 0.000600678760448001 |
| 16 | 0.0000393485418496 |
| 17 | 0.00000155612348416 |
| 18 | 0.00000003414556672 |
| 19 | 0.000000000347602944 |
| 20 | 0.000000000001048576 |
The problem is what happens when a one is rolled?
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You want the truth! You can't handle the truth!
February 21st, 2012 at 8:14:25 PM
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Doubled
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You want the truth! You can't handle the truth!
February 22nd, 2012 at 11:21:04 AM
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This, I calculated through the use of a Markov chain. Yesterday, I used a stochastic matrix with an absorbing state to calculate only the average number of rolls to end. There is almost exactly a 40% chance of finishing in 20 rolls or fewer. About 1 in 20 kids would have to go to 85 or more rolls.
| # of Rolls | Probability |
|---|---|
| 7 | 0.000128601 |
| 8 | 0.006790838 |
| 9 | 0.035641678 |
| 10 | 0.06121854 |
| 11 | 0.054355896 |
| 12 | 0.036897524 |
| 13 | 0.028020105 |
| 14 | 0.025764809 |
| 15 | 0.025430765 |
| 16 | 0.025400113 |
| 17 | 0.025389245 |
| 18 | 0.025307461 |
| 19 | 0.024970031 |
| 20 | 0.024189534 |
| 21 | 0.023035933 |
| 22 | 0.021817628 |
| 23 | 0.020776793 |
| 24 | 0.019942031 |
| 25 | 0.019230398 |
| 26 | 0.018567709 |
| 27 | 0.017919202 |
| 28 | 0.017274663 |
| 29 | 0.016634267 |
| 30 | 0.016003349 |
| 31 | 0.015389895 |
| 32 | 0.01480094 |
| 33 | 0.01423942 |
| 34 | 0.013703981 |
| 35 | 0.013191251 |
| 36 | 0.012698135 |
| 37 | 0.012222728 |
| 38 | 0.011764137 |
| 39 | 0.011322 |
| 40 | 0.010896106 |
I heart Crystal Math.
February 22nd, 2012 at 11:32:31 AM
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As a kid, I would have cheated. My chart would have likely looked like this:
| 40 |
| 34 |
| 28 |
| 22 |
| 16 |
| 10 |
| 4 |
| 0 |
February 22nd, 2012 at 11:35:08 AM
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It's actually quite easy to do using the blanket roll. Just put the 1/6 axis on the side and let the die roll forward off your open palm. You'll never roll a 1 and you'll be done in roughly 11.4 rolls.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
