Probability of coin and dice

A fair coin is flipped until the first head is reached, at which point the coin flips stop. Suppose that for each coin flip (including the last head) there is an associated independent roll of a fair six-sided die.


a) Find the probability that the number of coin flips is X= 2 given that the sum of the associated die rolls is S= 4?


The combinations I found were (Flip,roll) = (T,1)(H,3), (T,2)(H,2), (T3,H1) ... and each of this occurs at (1/2)x(1/6) of getting each.
So (1/12)(1/2) x 3 possible combinations.
P= 3/144

b) Find the probability that the sum of the die rolls is S= 4 given that the number of coin flips X is even.
X=2; same as part a
X=4; one way and thats to get (T,1)x3 and (H,1)

so 3/144 x (1/12)^4 x 1 combinations is the answer?


c) Find the probability that the number of coin flips is X= 2 given that (i) the sum of the die rolls is S= 4 and also (ii) the first die roll showed 1?

No idea.


(d) Compute the unconditional expected value of the sum S



geometric expected value? 1/p = 1/(1/12) = 12

mussi said:

A fair coin is flipped until the first head is reached, at which point the coin flips stop. Suppose that for each coin flip (including the last head) there is an associated independent roll of a fair six-sided die.


a) Find the probability that the number of coin flips is X= 2 given that the sum of the associated die rolls is S= 4?


The combinations I found were (Flip,roll) = (T,1)(H,3), (T,2)(H,2), (T3,H1) ... and each of this occurs at (1/2)x(1/6) of getting each.
So (1/12)(1/2) x 3 possible combinations.
P= 3/144

Click to expand...

There is at least one difficulties with what you did. You are asked for a conditional probability, but you seem to have calculated an absolute. The question can be rephrased, "Of all the ways to get S=4, what fraction of the probabilities has X=2?" You have listed the three combinations with X=2. You must also consider (H,4) and three ways to add 1+1+2.
The joint probabilities of S=4 and various X are
P(X=1 and S=4) = (1/2)(1/6) = 1/12
P(X=2 and S=4) = [(1/2)(1/6)]^2 × 3 = 3/144
P(X=3 and S=4) = [(1/2)(1/6)]^3 × 3 = 3/1728
P(X=4 and S=4) = [(1/2)(1/6)]^4 = 1/20736
Take the sum of those four to get P(S=4)
Then the conditional probability is P(X=2 | S=4) = P(X=2 and S=4) / P(S=4)

b) Find the probability that the sum of the die rolls is S= 4 given that the number of coin flips X is even.
X=2; same as part a
X=4; one way and thats to get (T,1)x3 and (H,1)

so 3/144 x (1/12)^4 x 1 combinations is the answer?

Click to expand...

I am sure you want to ADD those two probabilities, NOT MULTIPLY.
P(X=even and S=4) = 3/144 + (1/12)^4
That is only one step, however. To get the conditional probability asked for, you have to compare to the total probability that X is even. That will be
P(X=even) = P(X=2) + P(X=4) + . . . = (1/4) + (1/16) + (1/64) + . . . = sum of a geometric series.
Of all the cases with even X, the conditional probabity that S=4 is
P(S=4 | X=even) = P(X=even and S=4) / P(X=even)

c) Find the probability that the number of coin flips is X= 2 given that (i) the sum of the die rolls is S

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= 4 and also (ii) the first die roll showed 1?

No idea.

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If the first roll is known to be 1, there are fewer ways to add up to 4:
X=2: 1 + 3
X=3: 1 + 1 + 2 or 1 + 2 + 1
X=4: 1 + 1 + 1 + 1

(d) Compute the unconditional expected value of the sum S

geometric expected value? 1/p = 1/(1/12) = 12

Click to expand...

For every independent roll of the dice, the expectation value is 3.5.
Find the expectation value for number of rolls, X, using the geometric series.

E[X × S] = E[X] × E

DrPhil said:

There is at least one difficulties with what you did. You are asked for a conditional probability, but you seem to have calculated an absolute. The question can be rephrased, "Of all the ways to get S=4, what fraction of the probabilities has X=2?" You have listed the three combinations with X=2. You must also consider (H,4) and three ways to add 1+1+2.
The joint probabilities of S=4 and various X are
P(X=1 and S=4) = (1/2)(1/6) = 1/12
P(X=2 and S=4) = [(1/2)(1/6)]^2 × 3 = 3/144
P(X=3 and S=4) = [(1/2)(1/6)]^3 × 3 = 3/1728
P(X=4 and S=4) = [(1/2)(1/6)]^4 = 1/20736
Take the sum of those four to get P(S=4)
Then the conditional probability is P(X=2 | S=4) = P(X=2 and S=4) / P(S=4)

Click to expand...

Why are you considering rolling 4 times?

I am sure you want to ADD those two probabilities, NOT MULTIPLY.
P(X=even and S=4) = 3/144 + (1/12)^4
That is only one step, however. To get the conditional probability asked for, you have to compare to the total probability that X is even. That will be
P(X=even) = P(X=2) + P(X=4) + . . . = (1/4) + (1/16) + (1/64) + . . . = sum of a geometric series.

Click to expand...

Sum = 1 ???

Of all the cases with even X, the conditional probabity that S=4 is
P(S=4 | X=even) = P(X=even and S=4) / P(X=even)

If the first roll is known to be 1, there are fewer ways to add up to 4:
X=2: 1 + 3
X=3: 1 + 1 + 2 or 1 + 2 + 1
X=4: 1 + 1 + 1 + 1

Click to expand...

P(X=2 and S=4) = [(1/2)(1/6)]^2 × 1 = 1/144
P(X=3 and S=4) = [(1/2)(1/6)]^3 × 2 = 2/1728
P(X=4 and S=4) = [(1/2)(1/6)]^4 = 1/20736
P(X=2 | S=4) = P(X=2 and S=4) / P(S=4) ????

For every independent roll of the dice, the expectation value is 3.5.
Find the expectation value for number of rolls, X, using the geometric series.

E[X × S] = E[X] × E

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Why isn't it .. E[XS] = E[X] + E ??
Also how did you get 3.5? Still don't quite get this one.

mussi said:

Why are you considering rolling 4 times?

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Because you have to add up ALL the ways you could get S=4:
X=1: 4
X=2: 1+3, 2+2, 3+1
X=3: 1+1+2. 1+2+1. 2+1+1
X=4: 1+1+1+1

For the conditional probability, take the ratio of the X=2 probability to the sum of the S=4 probabilities.

Sum = 1 ???

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The sum of the series including both odd and even X is 1, but there is twice as much probability of odd as of even. (Write out the odd terms and the even terms and compare.)
Thus....P(X=even) = 1/3

P(X=2 and S=4) = [(1/2)(1/6)]^2 × 1 = 1/144
P(X=3 and S=4) = [(1/2)(1/6)]^3 × 2 = 2/1728
P(X=4 and S=4) = [(1/2)(1/6)]^4 = 1/20736
P(X=2 | S=4) = P(X=2 and S=4) / P(S=4) ????

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Good. Now add up the three probabilities to get P(S=4), and calculate the ratio. Turns out that if you know the first turn was (T,1), and you also know S=4, there is a high (conditional) probability that X=2.

Why isn't it .. E[XS] = E[X] + E ??
Also how did you get 3.5? Still don't quite get this on

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The average of one throw of a fair die is (1+2+3+4+5+6)/6 = 3.5

The average for the sum of X throws is X*(3.5)
Multiply E[X] times 3.5 to get the expected total.

P(x=even) = 1/3?
1/4 + 1/8 ... > 1/3?
not quite sure why it isn't 1/2 then?

For E what formula are u using? It's not the standard np, etc but the average formula of rolling a dice once over infinite rolls (individual). E[x] = 1/p = 1/(1/6)) ? This is because you are expected to roll 6 times to get that average value of 3.5... Thus E[XS] = 3.5 x 6

This has been a very complex problem to work on, so I have assumed your class is at a pretty high level and that you have learned "all about" conditional probabilities. I am hopeful that your are understanding all the steps I have left out - at least you are having fewer questions!

mussi said:

P(x=even) = 1/3?
1/4 + 1/8 ... > 1/3?....it's not +1/8, but +1/16
not quite sure why it isn't 1/2 then?

Click to expand...

\(\displaystyle \displaystyle \sum f(x) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \cdot \cdot \cdot \to 1\)

\(\displaystyle \displaystyle \sum f(x=\text{odd}) = \frac{1}{2} +\frac{1}{8} + \frac{1}{32} + \cdot \cdot \cdot \)

\(\displaystyle \displaystyle \sum f(x=\text{even}) = \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \cdot \cdot \cdot\)

For every "odd" term, the corresponding "even" term is half. Thus the full sum of evens is half of the sum of odds.
When you apply the formula for sum of a geometric series, note that \(\displaystyle r = 1/4\).

For E what formula are u using? It's not the standard np, etc but the average formula of rolling a dice once over infinite rolls (individual). E[x] = 1/p = 1/(1/6)) ? This is because you are expected to roll 6 times to get that average value of 3.5... Thus E[XS] = 3.5 x 6

Click to expand...

The definition of expectation value is the sum of (probability)×(value) for all possible outcomes. For a fair six-sided die, there are six outcomes and the probability of each is (1/6). Using lower-case \(\displaystyle s\) for a single throw,

\(\displaystyle \displaystyle \mathrm E = \sum_{s=1}^6 s\ p(s) = 1\ (1/6) + 2\ (1/6) + 3\ (1/6) + 4\ (1/6) + 5\ (1/6) + 6\ (1/6) = 3.5\)

If you throw \(\displaystyle X\) times, the sum is \(\displaystyle X\) times a single throw. The expectation value is the sum over all possible \(\displaystyle X\) of (probability)×(value). In the geometric series, \(\displaystyle p(X) = (1/2)^X\).

\(\displaystyle \displaystyle \mathrm E = \sum_{X=1}^\infty X\ p(X)\ \mathrm E = 3.5\ \sum_{X=1}^\infty \dfrac{X}{2^X} = 3.5\ \left(\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \cdot \cdot \cdot \right)\)

That is not an easy sum to find in a handbook, but if you write out the first 6 or so partial sums, you should see what the limit is headed for (a simple number!).

DrPhil said:

There is at least one difficulties with what you did. You are asked for a conditional probability, but you seem to have calculated an absolute. The question can be rephrased, "Of all the ways to get S=4, what fraction of the probabilities has X=2?" You have listed the three combinations with X=2. You must also consider (H,4) and three ways to add 1+1+2.
The joint probabilities of S=4 and various X are
P(X=1 and S=4) = (1/2)(1/6) = 1/12
P(X=2 and S=4) = [(1/2)(1/6)]^2 × 3 = 3/144
P(X=3 and S=4) = [(1/2)(1/6)]^3 × 3 = 3/1728
P(X=4 and S=4) = [(1/2)(1/6)]^4 = 1/20736
Take the sum of those four to get P(S=4)
Then the conditional probability is P(X=2 | S=4) = P(X=2 and S=4) / P(S=4)

Click to expand...

Reviewing, to get P(S=4), is it not just
P( S=4) = (1/6) ; 1 flip
P(S=4) = (1/6)^2 × 3 ;2 flip
P( S=4) = (1/6)^3 × 3 ; 3 flip
P( S=4) = (1/6)^4 ; 4 flip
added? (minus the flipping of coin chance)

I ask this because you dont consider this for the P(x=even and S=4) case found below?

I am sure you want to ADD those two probabilities, NOT MULTIPLY.
P(X=even and S=4) = 3/144 + (1/12)^4
That is only one step, however. To get the conditional probability asked for, you have to compare to the total probability that X is even. That will be
P(X=even) = P(X=2) + P(X=4) + . . . = (1/4) + (1/16) + (1/64) + . . . = sum of a geometric series.
Of all the cases with even X, the conditional probabity that S=4 is
P(S=4 | X=even) = P(X=even and S=4) / P(X=even)

Click to expand...

mussi said:

Reviewing, to get P(S=4), is it not just
P( S=4) = (1/6) ; 1 flip
P(S=4) = (1/6)^2 × 3 ;2 flip
P( S=4) = (1/6)^3 × 3 ; 3 flip
P( S=4) = (1/6)^4 ; 4 flip
added? (minus the flipping of coin chance)

Click to expand...

When you write
"P( S=4) = (1/6) ; 1 flip"
you are saying that if there is 1 flip, then P( S=4) = (1/6).
Thus you have a conditional probability. I think that is what you mean by "; 1 flip". I use the notation "| X=1" to say the same thing:
P(S=4 | X=1) = (1/6)
P(S=4 | X=2) = (1/6)^2 × 3
. . .
P(S=4 | X>4) = 0

You are asked to find P(X=2 | S=4), but what you have so far is P(S=4 | X=2)
You can get the joint probability from either conditional probability by multiplying by the probability of the condition:

P(S=4 and X=2) = P(X=2 | S=4) P(S=4) = P(S=4 | X=2) P(X=2)

Solve for what you are asked to find:

P(X=2 | S=4) = P(S=4 and X=2) / P(S=4) = [P(S=4 | X=2) P(X=2)]/P(S=4)

In other words, the conditional probability is the ratio of the joint probability to the sum of all possible ways to have S=4. Your present question is how to determine P(S=4)

Now consider a tree diagram for the coin flips. For heads branch left and for tails branch right. Each branch has a probability of 1/2. If the first toss is H (with probability 1/2), then X=1 and P(S=4) = (1/2)(1/6). If you branch right, there is another toss. Probabilities multiply as you move down the branches: 1/4, 1/8, . . .. Adding up the four branches that may possibly give S=4,

\(\displaystyle \displaystyle P(S\!=\!4) = \sum_{i=1}^4 P(S\!=\!4\ |\ X\!=\!i)\ P(X\!=\!i) \)

I ask this because you dont consider this for the P(x=even and S=4) case found below?

Click to expand...

This part only looks different, because a different conditional question has been asked. The "Condition" is not "S=4," but rather "X=even." Using the same logic as above [called Bayses' Theorem, BTW],

P(S=4 | X=even) = P(S=4 and X=even) / P(X=even)

The shape of the tree diagram is the same, but to find all even X we don't have to ask anything about S. To get the conditional probability, we take the ratio of even X that do produce S=4, to the sum of all even X.

Hope this helps - like I said, it is a complex question. I should have asked if you have studied Bayses' Theorem?