Throwing three different dices - probability of getting a six

dr.trovacek

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Throwing three different dice - probability of getting a six

This is a fourth grade high school problem given in the lesson formula of total probability and Bayes' formula:

We have three dice. One of them is normal, the second has a six on three sides and the third has a six on all six sides. We throw a selected die two times and both times we get a six. We throw that die once again.
What is the probability that we will get a six again?

Solution in the textbook: \(\displaystyle \frac{61}{69}\)

Here is my (wrong) attempt at this problem.

I indexed the dice in the order mentioned in the text of the problem.
So probabilities of getting a six throwing a die two times for each of the dice are:
\(\displaystyle \frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}\) for the first die
\(\displaystyle \frac{3}{6} \cdot \frac{3}{6}=\frac{9}{36}\) for the second die
1 for the third die

Then I was thinking that this should be some kind of union of events: we get a six for the third time given that we selected the first die OR we get a six for the third time given that we selected the second die OR we get a six for the third time given that we selected the third die.

Probabilities for getting a six on a third throw, because every throw is an independent event, are:
\(\displaystyle \frac{1}{216}\) for the first die
\(\displaystyle \frac{27}{216} \) for the second die
1 for the third die

Since the probability for selecting each one of the dice is \(\displaystyle \frac{1}{3}\) final probability of getting a six when throwing one of the selected die for the third time is:
(probability that we selected the first die AND probability we got a six for the third time given that we selected the first die) OR (probability that we selected the second die AND probability we got a six for the third time given that we selected the second die) OR (probability that we selected the third die AND probability we got a six for the third time given that we selected the third die).

So this is my reasoning, which is, obviously wrong. I get the 61 in the numerator but i get \(\displaystyle \frac{61}{162} \) which is 37,65%. Even intuitively it seems It should be more than that, since we have one die which has 50% chance of getting a 6 and one having 100% of getting a six.

I would love to find out what are the mistakes in my reasoning. I came to this thought process trying to think in form of a probability three diagram, which I have used before solving probability problems in general.

Any help is much appreciated
 
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This is a fourth grade high school problem given in the lesson formula of total probability and Bayes' formula:
We have three dices. One of them is normal, second has a six on three sides and third has a six on all six sides.
We throw a selected dice two times and both times we got a six. We throw that dice once again.
What is the probability that we will get a six again?
Solution in the textbook: \(\displaystyle \frac{61}{69}\)
Comment: There are three dice but one die.
Say N denotes the normal die, H denotes the die with half sixes and A denotes all sixes.
Because of the notation nightmare, I will use \(\displaystyle PQ\) for \(\displaystyle P\cap Q\)

\(\displaystyle \mathcal{P}(S_3|S_1 S_2)=\dfrac{\mathcal{P}(S_3 S_1 S_2)}{\mathcal{P}(S_1 S_2)}\)
Because the events \(\displaystyle N,~H,~\&~A\) partition the space
\(\displaystyle \mathcal{P}(S_3 S_1 S_2)=\mathcal{P}(S_3 S_1 S_2 N)+\mathcal{P}(S_3 S_1 S_2H)+\mathcal{P}(S_3 S_1 S_2A)\)
There is a similar statement for \(\displaystyle \mathcal{P}(S_1 S_2)\)
So
\(\displaystyle \mathcal{P}(S_3|S_1 S_2)=\dfrac{\mathcal{P}(S_3 S_1 S_2)}{\mathcal{P}(S_1 S_2)}=\dfrac{6^{-3}+2^{-3}+1}{6^{-2}+2^{-2}+1}\)
NOW SEE HERE.
 
This is a fourth grade high school problem given in the lesson formula of total probability and Bayes' formula:

We have three dice. One of them is normal, second has a six on three sides and third has a six on all six sides.
We throw a selected die two times and both times we got a six. We throw that die once again.
What is the probability that we will get a six again?

Solution in the textbook: \(\displaystyle \frac{61}{69}\)

Here is my (wrong) attempt at this problem.

I indexed the dice in the order mentioned in the text of the problem.
So probabilities of getting a six throwing a die two times for each of the dice are:
\(\displaystyle \frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}\) for the first die
\(\displaystyle \frac{3}{6} \cdot \frac{3}{6}=\frac{9}{36}\) for the second die
1 for the third die

Then I was thinking that this should be some kind of union of events: we get a six for the third time given that we selected the first die OR we get a six for the third time given that we selected the second die OR we get a six for the third time given that we selected the third die.

Probabilities for getting a six on all three throws, because every throw is an independent event, are:
\(\displaystyle \frac{1}{216}\) for the first die
\(\displaystyle \frac{27}{216} \) for the second die
1 for the third dice

Since the probability for selecting each one of the dice is \(\displaystyle \frac{1}{3}\), the final probability of getting a six when throwing the selected die for the third time is:
(probability that we selected the first die AND we got a six for the third time given that we selected the first die) + (probability that we selected the second die AND we got a six for the third time given that we selected the second die) + (probability that we selected the third die AND we got a six for the third time given that we selected the third die).

So this is my reasoning, which is, obviously wrong. I get the 61 in the numerator but I get \(\displaystyle \frac{61}{162} \) which is 37,65%. Even intuitively it seems it should be more than that, since we have one die which has 50% chance of getting a 6 and one having 100% of getting a six.

I would love to find out what are the mistakes in my reasoning. I came to this thought process trying to think in form of a probability three diagram, which I have used before solving probability problems in general.

Any help is much appreciated

I'm trying to follow your reasoning in order to see where you went wrong, but I'm not sure exactly what you did. Can you state the actual calculation you did?

Above I have reworded some things to make it easier for me to read; I hope I haven't changed the meaning. One was removing the word "probability" where it didn't belong, because you confused events with their probabilities.

I suspect that the problem is that your probabilities of selecting a given die were a priori, not taking into account the known fact that you rolled two sixes, which makes some more likely than others to have been chosen.
 
my calculation

I noticed that in probability every word in the problem is especially important for getting the right solution (even more than other math areas I've encountered so far).
So event though I often struggle with the way the problem is stated and how to state the solution when I'm using my native language, is likely I just haven't expressed myself accurately enough in English. I apologise for that, I'll try to analyse in more depth what I may wrote wrong. I expect my English will improve now that I started to use it more often again


This is my calculation:
\(\displaystyle \frac{1}{3} \cdot \frac{1}{36} \cdot \frac{1}{6} +\frac{1}{3} \cdot \frac{9}{36} \cdot \frac{3}{6} + \frac{1}{3} \cdot 1 = \frac{61}{162}\)
 
I noticed that in probability every word in the problem is especially important for getting the right solution (even more than other math areas I've encountered so far).
So event though I often struggle with the way the problem is stated and how to state the solution when I'm using my native language, is likely I just haven't expressed myself accurately enough in English. I apologise for that, I'll try to analyse in more depth what I may wrote wrong. I expect my English will improve now that I started to use it more often again


This is my calculation:
\(\displaystyle \frac{1}{3} \cdot \frac{1}{36} \cdot \frac{1}{6} +\frac{1}{3} \cdot \frac{9}{36} \cdot \frac{3}{6} + \frac{1}{3} \cdot 1 = \frac{61}{162}\)

Yes, the exact wording of a problem is important here, because the problems themselves are highly sensitive to slight changes. It's not primarily an issue of language, but language usage can certainly make it easier to miss a detail.

Your calculation gives the probability that, if you randomly choose a die, you will get three sixes. That can be part of a solution, but is not what you are looking for.

I think my comment that you took the probability for each die to be 1/3 rather than taking into account that the all-sixes die is more likely to have been chosen, given that two sixes have been rolled, is accurate.
 
Modifying PKA's notation:

\(\displaystyle N \text { is the event of selecting the normal die;}\\

H \text { is the event of selecting the die with three 6's;}\\

A \text { is the event of selecting the die with all 6's;}\\

X \text{ is the event of rolling 6 on both the first and second rolls, and}\\

Y \text { is the event of rolling 6 on the third roll.}\)

Now you have a useful notation. (Thanks PKA.) Use it.

You are asked to find:

\(\displaystyle P(Y|X).\)

BY DEFINITION, \(\displaystyle P(Y|X) = \dfrac{P(Y \& X)
}{P(X)}.\)

So we have to find P(Y&X) and P(X). Things are simpler if you remember definitions.

Because rolls of a given die are independent, it is trivial to find

\(\displaystyle P(Y|N) = \dfrac{1}{6},\ P(X|N) = \dfrac{1}{36}, P(Y \& X |N) = \dfrac{1}{216},\\

P(Y|H) = \dfrac{1}{2},\ P(X|H) = \dfrac{1}{4},\ P(Y \& X|H) = \dfrac{1}{8},\\

P(Y|A) = 1, P(X|A) = 1, \text { and } P(Y \& X|A)= 1.\)

A theorem: the probability of one of of n mutually exclusive, cumulatively exhaustive, and equiprobable events is 1/n. And N, H, and A are mutually exclusive, cumulatively exhaustive, and equiprobable. So

\(\displaystyle P(N) = P(H) = P(A) = \dfrac{1}{3}\)

\(\displaystyle P(X) = P(X \& N) + P(X \& H) + P(X \& A) \text { and}\\

P(Y \& X) = P(Y \& X \& N) + P(Y \& X \& H) + P(Y \& X \& A)\)

because N, H, and A are cumulatively exhaustive.

Rolls of a die are independent so

\(\displaystyle P(X \& N) = P(X|N) * P(N) = \left (\dfrac{1}{6} \right )^2 * \dfrac {1}{3} = \dfrac{1}{36} * \dfrac{1}{3},\\

P(X \& H) = \dfrac{1}{4} * \dfrac{1}{3}. \text { and } P(X \& A) = 1 * \dfrac{1}{3} \implies \\

P(X) = \dfrac{1}{3} * \left ( \dfrac{1}{36} + \dfrac{1}{4} + 1 \right ) = \dfrac{1}{3} * \dfrac{1 + 9 + 36}{36} = \dfrac{1}{3} * \dfrac{46}{36} = \dfrac{23}{54}.\)

Similarly,

\(\displaystyle P(Y \& X \& N) = P(Y \& X|N) * P(N) = \left (\dfrac{1}{6} \right )^3 * \dfrac {1}{3} = \dfrac{1}{216} * \dfrac{1}{3},\\

P(Y \& X \& H) = \dfrac{1}{8} * \dfrac{1}{3}, \text { and } P(X \& A) = 1 * \dfrac{1}{3} \implies \\

P(Y \& X) = \dfrac{1}{3} * \left ( \dfrac{1}{216} + \dfrac{1}{8} + 1 \right ) = \dfrac{1}{3} * \dfrac{1 + 27 + 216}{216} = \dfrac{1}{3} * \dfrac{244}{216} = \dfrac{61}{162}.\)

\(\displaystyle \therefore P(Y|X) = \dfrac{\dfrac{61}{162}}{\dfrac{23}{54}} = \dfrac{61}{162} * \dfrac{54}{23} = \dfrac{61 * 2 * 3^3}{23 * 2 * 3^4} = \dfrac{61}{23 * 3} = \dfrac{61}{69}.\)

Let the theorems and definitions do your thinking for you.
 
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My second attempt

I think I see the mistakes I made in my reasoning and also the difference between what is required and what I have calculated.The part of this problem which gave me headaches is the last question for the probability of getting a six, that confused me for some reason. Obviously I still have a problem grasping relatively complex problems like this one.

I wanted to post my solution earlier but I could not connect to the forum.

Here it is:

\(\displaystyle
S_i=\mbox{{we get a six in the i-th throw}}, i=1,2,3 \\

D_i=\mbox{{we get a six throwing the i-th die}}, i =1,2,3
\)

The die denoted with 1 is the normal die, with 2 the die with half sixes and with 3 the one with all sixes.

So \(\displaystyle P(S_3|S_1S_2)\) is probability of getting a six in the third throw, given we got a six in the first and the second throw

\(\displaystyle P(S_3|S_1S_2) = \dfrac{P(S_3 S_1 S_2)}{P(S_1 S_2)} \)

So we have unions of intersections because each one of the intersections can occur

\(\displaystyle P(S_3S_1 S_2)=P(S_3 S_1 S_2 D_1)+P(S_3 S_1 S_2D_2)+P(S_3 S_1 S_2D_3)\\
P(S_1S_2)= (S_1S_2D_1) +(S_1S_2D_2)+(S_1S_2D_3) \)

Probability of getting a six throwing
  • the first die three times in a row: \(\displaystyle P(S_3 S_1 S_2 D_1)= \left( \dfrac{1}{6} \right)^3\),
  • the second die three times in a row: \(\displaystyle P(S_3 S_1 S_2 D_2)= \left( \dfrac{1}{2} \right)^3\),
  • the third die three times in a row: \(\displaystyle P(S_3 S_1 S_2 D_3)= 1^3\)

It's quite similar with the event \(\displaystyle (S_1S_2) \)

Accordingly, the final result is \(\displaystyle P(S_3|S_1 S_2) = \dfrac{P(S_3 S_1S_2)}{P(S_1 S_2)} = \dfrac{6^{-3} + 2^{-3} + 1}{6^{-2}+2^{-2} + 1} = \frac{61}{69}\)

-

Is it ok to define my events like this, from the reasoning aspect?

I'm also not sure if "i-th" I used in the initial events is grammatically correct. We use something similar in my native tongue and it feels much easier to read than having a lot of differents letters, it seems more associative.

Now I'm digging into JeffM's solution.

Thanks again to everyone who responded to the thread!
 
I think I see the mistakes I made in my reasoning and also the difference between what is required and what I have calculated.The part of this problem which gave me headaches is the last question for the probability of getting a six, that confused me for some reason. Obviously I still have a problem grasping relatively complex problems like this one.

I wanted to post my solution earlier but I could not connect to the forum.

Here it is:

\(\displaystyle
S_i=\mbox{{we get a six in the i-th throw}}, i=1,2,3 \\

D_i=\mbox{{we get a six throwing the i-th die}}, i =1,2,3
\)

The die denoted with 1 is the normal die, with 2 the die with half sixes and with 3 the one with all sixes.

So \(\displaystyle P(S_3|S_1S_2)\) is probability of getting a six in the third throw, given we got a six in the first and the second throw

\(\displaystyle P(S_3|S_1S_2) = \dfrac{P(S_3 S_1 S_2)}{P(S_1 S_2)} \)

So we have unions of intersections because each one of the intersections can occur

\(\displaystyle P(S_3S_1 S_2)=P(S_3 S_1 S_2 D_1)+P(S_3 S_1 S_2D_2)+P(S_3 S_1 S_2D_3)\\
P(S_1S_2)= (S_1S_2D_1) +(S_1S_2D_2)+(S_1S_2D_3) \)

Probability of getting a six throwing
  • the first die three times in a row: \(\displaystyle P(S_3 S_1 S_2 D_1)= \left( \dfrac{1}{6} \right)^3\),
  • the second die three times in a row: \(\displaystyle P(S_3 S_1 S_2 D_2)= \left( \dfrac{1}{2} \right)^3\),
  • the third die three times in a row: \(\displaystyle P(S_3 S_1 S_2 D_3)= 1^3\)

It's quite similar with the event \(\displaystyle (S_1S_2) \)

Accordingly, the final result is \(\displaystyle P(S_3|S_1 S_2) = \dfrac{P(S_3 S_1S_2)}{P(S_1 S_2)} = \dfrac{6^{-3} + 2^{-3} + 1}{6^{-2}+2^{-2} + 1} = \frac{61}{69}\)

-

Is it ok to define my events like this, from the reasoning aspect?

I'm also not sure if "i-th" I used in the initial events is grammatically correct. We use something similar in my native tongue and it feels much easier to read than having a lot of differents letters, it seems more associative.

Now I'm digging into JeffM's solution.

Thanks again to everyone who responded to the thread!
Yes now it is correct. In Jeff's solution, he used a factor of one third in each term. O.K. but they all divide anyway.
 
Yes now it is correct. In Jeff's solution, he used a factor of one third in each term. O.K. but they all divide anyway.
I meant to say, but somehow forgot, that my method was not that different from PKA's. I was more interested in tying it to basic definitions and theorems than in really doing something logically different. The multiplication by 1/3 of numerator and denominator has no effect on the result because it cancels out. It just relates back to the formula linking joint probability and conditional probability.
 
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