My Math Forum Tossing dice

 August 6th, 2011, 01:09 PM #1 Senior Member   Joined: Apr 2011 From: Recife, BR Posts: 352 Thanks: 0 Tossing dice We have three dice in a box. A round consists in shaking the box and then removing all the dice which faced up 1. The game ends once all dice have been removed. Find the probability $P(n)$ that the game ends after exactly $n$ rounds. After cumbersome calculations and casework, I reached a recursion which seems hard to be converted into closed form. Obviously $f(1)= \frac{1}{216}.$ $P(n+1)= \frac{11}{10} (\frac{1}{6} (\frac{5}{6})^{n+1} - \frac{1}{5} (\frac{5}{6})^{2n+2}) + \frac{125}{216} P(n)$ My question is, can this be generalized for $k$ dice? Or even for different probabilities, i.e. dice numbered from one to ten? And what if I want to know the expected value of the number of rounds of a game?
 August 6th, 2011, 01:34 PM #2 Newbie   Joined: Aug 2011 Posts: 5 Thanks: 0 Re: Tossing dice To do expected value, wouldn't you just roll the dice and then divide the totals by the amount of times? Because if you had .30 as an expected value and you wanted to know that it's .30 it would mean that you gain .30 profit for every time. If you had like 3.00 profit and you got 10 the answer would be .30. For the question, since they're dice, you would probably have to do an actual experiment.
 August 6th, 2011, 01:35 PM #3 Newbie   Joined: Aug 2011 Posts: 5 Thanks: 0 Re: Tossing dice And that's because you made an even value for every time. If you roll the dice it will never be even every time so you'd probably have to have numbers for every try right?
August 6th, 2011, 02:16 PM   #4
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Re: Tossing dice

Quote:
 Originally Posted by XeLious23 For the question, since they're dice, you would probably have to do an actual experiment.
This is mathematics, not physics or chemistry.

August 6th, 2011, 04:14 PM   #5
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Re: Tossing dice

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Originally Posted by proglote
Quote:
 Originally Posted by XeLious23 For the question, since they're dice, you would probably have to do an actual experiment.
This is mathematics, not physics or chemistry.
It doesn't have to be. You can still experiment. If what I said is wrong then you don't need to bash it. You can say that he's talking about something else or help me understand better.

What an attitude

August 6th, 2011, 04:44 PM   #6
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Re: Tossing dice

Quote:
 Originally Posted by XeLious23 It doesn't have to be. You can still experiment. If what I said is wrong then you don't need to bash it. You can say that he's talking about something else or help me understand better. What an attitude
What I meant is that we may construct a mathematical argument to find the expected value, we don't need to make experiments. And even though we do make experiments, it's hard to get 100% honest dice, and the amount of trials we would need to make a solid conclusion would be very large.

Sorry if I sounded a bit too rough.

August 6th, 2011, 08:19 PM   #7
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Re: Tossing dice

Quote:
 Originally Posted by proglote My question is, can this be generalized for $k$ dice? Or even for different probabilities, i.e. dice numbered from one to ten?
Generalizing to different probabilities is easy. In this game, you care only about the two probabilities, rolling a one (p = 1/6) and rolling a non-one (q = 5/6). You can safely replace these with other values, even values other than inverse natural numbers.

The probability that a given die will be removed on round n is $X=q^{n-1}\cdot p$. The probability that it's removed before round n is$Y=\sum_{a=0}^{n-2}q^a\cdot p=p\frac{1-q^{n-1}}{1-q}$. So knowing those two probabilities you can find the probability of the game ending in n rounds as
$\sum_{a=1}^k{k\choose a}X^aY^{k-a}=p^k\sum_{a=1}^k{k\choose a}q^{an-a}\left(\frac{1-q^{n-1}}{1-q}\right)^{k-a}$.

August 7th, 2011, 05:54 AM   #8
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Re: Tossing dice

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 Originally Posted by CRGreathouse So knowing those two probabilities you can find the probability of the game ending in n rounds as $\sum_{a=1}^k{k\choose a}X^aY^{k-a}=p^k\sum_{a=1}^k{k\choose a}q^{an-a}\left(\frac{1-q^{n-1}}{1-q}\right)^{k-a}$.
Thanks! But why is the binomial ${k\choose a}$ there? I thought only $X^aY^{k-a}$ would suffice.. maybe it's because we have to consider the different combinations for the $a$ dice at the end?

EDIT: Amazingly, W|A simplifies your summation to $\left(1-q^{n-1}\right)^k \left(\left(\frac{1 - q^{n}}{1 - q^{n-1}}\right)^{k} -1\right)= (1-q^n)^k - (1-q^{n-1})^k.$

And this seems correct; surprisingly, the solution to the recursive relation in the original post is indeed $\left(1-\left(\frac{5}{6}\right)^n\right)^3 - \left(1-\left(\frac{5}{6}\right)^{n-1}\right)^3$.

With respect to the expected value; W|A gives $\frac{10566}{1001}$ for the case k = 3. How can it be generalized for any k? (Wolfram Alpha couldn't)

http://www.wolframalpha.com/input/?i...n-1%29%29^3%29

How can those infinite sums be computed by hand?

August 7th, 2011, 10:30 AM   #9
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Re: Tossing dice

Quote:
 Originally Posted by proglote Thanks! But why is the binomial ${k\choose a}$ there? I thought only $X^aY^{k-a}$ would suffice.. maybe it's because we have to consider the different combinations for the $a$ dice at the end?
There are different ways to choose which dice finish on the last round and which don't.

Quote:
 Originally Posted by proglote http://www.wolframalpha.com/input/?i...n-1%29%29^3%29 How can those infinite sums be computed by hand?
Expand and use Faulhaber's formula, perhaps?

August 7th, 2011, 12:27 PM   #10
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Re: Tossing dice

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 Originally Posted by CRGreathouse Expand and use Faulhaber's formula, perhaps?
I think not really take a look at the case k = 1 for instance, the expected value would be $\frac{1}{5}\cdot\sum_{n=1}^{\infty} n \cdot $$\frac{5}{6}$$^{n} = \frac{1}{5} \cdot ( 1 \cdot \frac{5}{6} + 2 \cdot $$\frac{5}{6}$$^{2} + ... )$ and Faulhaber should only serve for equal exponents I guess..

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