Probability of 2 sevens before 6 evens

[CAF123]

Exactly. Next question, what's the probability of throwing 6 evens before a 7? What's the chance of throwing one 7 and five evens followed by a 6th even?

P(6 evens before one seven) = $(3/4)^6 (1/4)$
P(1 seven and then 6 evens) = $(1/4) (3/4)^6$

 

[MarneMath]

P(1 seven and then 6 evens) wasn't the question. The question was P(5 evens and one 7 followed by another 6th at the very end.) I'm phrasing this as close as possible to how the initial math should look. :) *Hint how many ways can you arrange the first 6 rolls I'm asking for?

 

[CAF123]

P(1 seven and then 6 evens) wasn't the question. The question was P(5 evens and one 7 followed by another 6th at the very end.) I'm phrasing this as close as possible to how the initial math should look. :)

Ok, so P(5 evens and 1 seven followed by 1 even) = $(3/4)^5 (1/4) (3/4)$
And you can reorganise the first 6 rolls (that is with 5 evens and 1 seven) in 6! Different ways. So this should be multiplied by what I had.

So in the end, I have P(6evens and 1 seven) = $(3/4)^6 (1/4)$
P(5 evens and 1 seven followed by even) = $(3/4)^5 (1/4) (3/4) \cdot 6!$

 

[MarneMath]

I'm going to throw you a bone, you forgot the C(6,1) (Can you see why?) So this, in the end reduces to (3/2)(3/4)^6.I didn't notice something at first, but...P(6 evens before one seven) = (3/4)^6, the extra 1/4 would be the probability of P(6 evens and then one seven).

Ok, so now we have the two probabilities. These answers ( if you combine them) represents the ways we will lose the game, so can we rewrite this in such a way so that it actually represents a way we win? This should be the easy part.

Edit: I think you edited your post sometime after I hit 'advance reply', so some of my initial remarks look like they make less sense, but still you should be able to finish this now with the information I provided.

 

[CAF123]

I'm going to throw you a bone, you forgot the C(6,1) (Can you see why?) So this, in the end reduces to (3/2)(3/4)^6.


I didn't notice something at first, but...P(6 evens before one seven) = (3/4)^6, the extra 1/4 would be the probability of P(6 evens and then one seven).

Ok, so now we have the two probabilities. These answers ( if you combine them) represents the ways we will lose the game, so can we rewrite this in such a way so that it actually represents a way we win? This should be the easy part.

Edit: I think you edited your post sometime after I hit 'advance reply', so some of my initial remarks look like they make less sense, but still you should be able to finish this now with the information I provided.

So P(lose) = (3/4)^6 + (3/4)^5 (1/4) (3/4) C(6,1) (I wrote 6! At first, but I see why it is 6 choose 1 because we want to order where the seven may be chosen). Then P(win) = 1 - P(lose). I see why a reduced sample space makes this a whole lot easier. One last question:
Say I wanted to compute P(6 evens before 1 seven) mechanically (i try not to do this in practice, but i have a question)That is using P(6 evens)/(P(6 evens) + P(1 seven)) = (3/4)^6/((3/4)^6 + (1/4)) which is not equal to (3/4)^6. Why?

 

[MarneMath]

Honestly, I have no idea what you mean by your last question by mechanically, but, I have been up for the last 20 hours! Tomorrow, I'll recheck this thread and if no one else squared you away, I'll reenage the question.

Also, if you need/want me to show you how to do this problem without a reduced sample space and conditional probability I can do that too.

Lastly, if you want me to use Ray's method (what I think his method of setting up a system of 12 linear equations) I can do that too. If this is enough, that's fine too!

 

[CAF123]

Honestly, I have no idea what you mean by your last question by mechanically, but, I have been up for the last 20 hours! Tomorrow, I'll recheck this thread and if no one else squared you away, I'll reenage the question.

Also, if you need/want me to show you how to do this problem without a reduced sample space and conditional probability I can do that too.

Lastly, if you want me to use Ray's method (what I think his method of setting up a system of 12 linear equations) I can do that too. If this is enough, that's fine too!

I have in my book P(E before F) = P(E)/(P(E) + P(F)). Because we haven't really covered Markov chains etc.., It'd help if we could go over the conditional prob approach (I am interested in Ray's method, it is just that I am preparing for my exam and so I think it would be best if I left really understanding Ray's method until afterwards). Thanks!

 

[haruspex]

So P(lose) = (3/4)^6 + (3/4)^5 (1/4) (3/4) C(6,1)

I think it's easier to think in terms of P[win].
When you get the second 7, you will have 0 to 5 evens. In each case, the prior state was one seven and that many evens. But you do not care what route you took to that state.
Can you write down the formula for the number of ways of reaching one seven and v evens?
That done, can you write down the probability of getting to one seven and v evens?
That done, can you write down the probability of getting to two sevens and v evens?
Having done all that, all you have to do is sum over v.

 

[Ray Vickson]

I am not exactly sure why we need to deal with a reduced sample space in this problem. On anyone throw, the probability of getting a seven is 6/36 and for an even it is 18/36. Is it completely necessary to reduce the sample space? If so, why?

One need not reduce the sample space or use altered (conditional) probabilities. Basically, we have a finite-state Markov chain with two 'absorbing' states (W and L) and a 1-step transition probability matrix P = (P(i,j)), with P(i,j) = Pr{state i --> state j} in one step. If we let a(i) = probability that state W is ever reached, starting from state i, then standard equations are that
$$a(i) = P(i,W) + \sum_{j \neq W,L} P(i,j) a(j) \; \forall i \neq W, L.$$
Note that in general there will be some 'self-loops' i --> i whenever P(i,i) > 0; these correspond to tossing an odd non-7 in your problem. We can eliminate such self-loops, just by isolating a(i) on both sides of the equation:
$$(1-P(i,i) ) a(i) = P(i,W) + \sum_{j \neq i,W,L} P(i,j) a(j) \Longrightarrow a(i) = \frac{P(i,W)}{1-P(i,i)} + \sum_{j \neq i,W,L} \frac{P(i,j)}{1-P(i,i)} a(j).$$
This has the form
$$a(i) = Q(i,W) + \sum_{j \neq i,W,L} Q(i,j) a(j) \; \forall i \neq W,L,$$
where
$$Q(i,k) = \frac{P(i,k)}{1-P(i,i)} = \frac{P(i,j)} {\sum_{m \neq i} P(i,m)},$$ which is just the conditional probability that i --> k, given that i ≠ i (that is, given that we move). The Q (i,k) give the probabilities of the next state reached from i, given that the system undergoes a genuine change of state. Note that $\sum_{k \neq i} Q(i,k) = 1,$ as it must (for when the system leaves state i it must go somewhere).

 

[CAF123]

@haruspex
I haven't really defined a probability 'state' as such in my class. Does your approach use Markov chains?
@RGV I really appreciate all your answers, I just feel I can't really follow at the moment since I haven't done anything in my course about states, Markov chains etc., I will certainly refer back to this thread later though.

 

[haruspex]

@haruspex
I haven't really defined a probability 'state' as such in my class. Does your approach use Markov chains?

They are not probability states. There is a set of states having probabilities. There's nothing complicated about this. If you toss a coin 100 times, the numbers of heads and tails you have so far at some point in the sequence could be your 'state', and you can figure out the probabilities of being in various states after some number of tosses.
Yes, that's what Markov chains are about: states with probabilistic transitions between them. The present problem is a relatively simple one in that you can reduce the states (by ignoring odd throws other than 7) to the point where you cannot stay in or return to any given state. This makes the analysis much easier.
Do you want to use Markov chains here? Very well then, it uses Markov chains. Do you want to avoid using Markov chains here? In that case, no, it just uses some elementary reasoning.
The problem describes a system which is just the sort of thing Markov chains describe. So whatever method you use to solve it could reasonably be interpreted as 'using' Markov chains.

 

[CAF123]

They are not probability states. There is a set of states having probabilities. There's nothing complicated about this. If you toss a coin 100 times, the numbers of heads and tails you have so far at some point in the sequence could be your 'state', and you can figure out the probabilities of being in various states after some number of tosses.
Yes, that's what Markov chains are about: states with probabilistic transitions between them. The present problem is a relatively simple one in that you can reduce the states (by ignoring odd throws other than 7) to the point where you cannot stay in or return to any given state. This makes the analysis much easier.
Do you want to use Markov chains here? Very well then, it uses Markov chains. Do you want to avoid using Markov chains here? In that case, no, it just uses some elementary reasoning.
The problem describes a system which is just the sort of thing Markov chains describe. So whatever method you use to solve it could reasonably be interpreted as 'using' Markov chains.

Using your method I have the sum:
$$\sum_{i=0}^{5} (\frac{3}{4})^i (\frac{1}{4})^2 (i+1),$$ which gives the correct answer. This is another method using the reduced sample space.

 

[Ray Vickson]

Using your method I have the sum:
$$\sum_{i=0}^{5} (\frac{3}{4})^i (\frac{1}{4})^2 (i+1),$$ which gives the correct answer. This is another method using the reduced sample space.

I don't get the correct answer when I evaluate this. Your sum is 4547/2048 ≈ 2.22021, while the correct probability of two sevens before 6 evens is 4547/8192 ≈ 0.55505. I got these by just evaluating the y(i,j) backwards, starting from y(1,j) = p*(1+q+...+q^(j-1)) for j = 0,1,2,...,5, then using y(2,j) = p_o *y (2,j) + p_7*y(1,j) + p_e*y(2,j-1), with y(2,1) = p^2, where p = p_7/(p_7+p_e) and q = 1-p. I used the standard dice probabilities to get p_7 = 1/6, p_e = 1/2 and p_o = 1/3, giving p = 1/4, q = 3/4.

 

[D H]

I don't get the correct answer when I evaluate this. Your sum is 4547/2048 ≈ 2.22021

His sum correctly evaluates to 4547/8192. Wolfram evaluation: sum ((3/4)^i * (1/4)^2 * (i+1)) from i=0 to 5.

You must have omitted a factor of 1/4.Here's a way to look at CAF123's sum.
Define a "seven or even" roll as a roll of a pair of dice such that the sum of the dice is either a seven or an even number. The probability of a "seven or even" roll yielding a sum of seven is 1/4.

What's the probability of a sequence of i+2 such "seven or even" rolls where
- The last roll is a seven and
- There is exactly one roll of a seven in the first i+1 rolls?

This probability is (3/4)ⁱ(1/4)²(i+1).

Summing over i=0 to i=5 yields the total probability that one will roll a pair of sevens will before rolling six evens.

 

[Ray Vickson]

His sum correctly evaluates to 4547/8192. Wolfram evaluation: sum ((3/4)^i * (1/4)^2 * (i+1)) from i=0 to 5.

You must have omitted a factor of 1/4.

Your're right: I had (1/2)^2 in the sum instead of (1/2)^2. Sorry.