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There are 3 dice in a bag: 6-sided, 12-sided and 20-sided. We pick one die at random, roll it, and it gives us a number N. What's the probability that when we pick another die and roll it, we'll get a number smaller than N, if N = 4?
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Do you put the first die back in the bag?
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There are 3 dice in a bag: 6-sided, 12-sided and 20-sided. We pick one die at random, roll it, and it gives us a number N. What's the probability that when we pick another die and roll it, we'll get a number smaller than N, if N = 4?

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OK, I'll take a crack at it....

Possible scenarios.

1. pull 6 side first which shows the 4, leaves 32 (12+20) possible outcomes in the bag.
2. pull 12 side first which shows the 4, leaves 26 (6+20) possible outcomes in the bag.
3. pull 20 side first which shows the 4, leaves 18 (6+12) possible outcomes in the bag.


So there a universe of 76 total outcomes.

With two dice left in the bag, there are three outcomes less than 4 on each dice, so 6 potential outcomes less than 4.

With three scenarios, each having six potential outcomes less than 4, gives 18 total outcomes less than 4.

18/76 = 0.236842
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There are 3 dice in a bag: 6-sided, 12-sided and 20-sided. We pick one die at random, roll it, and it gives us a number N. What's the probability that when we pick another die and roll it, we'll get a number smaller than N, if N = 4?

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Thinking about this some more, I'll take another crack at it. Probably this is wrong too but I enjoy the exercise.

Possible scenarios.

1. pull 6 side first which shows the 4, leaves 32 (12+20) possible outcomes in the bag, six of which are less than 4. Probability of less than 4 in this scenario is 6/32 = 0.1875

2. pull 12 side first which shows the 4, leaves 26 (6+20) possible outcomes in the bag, six of which are less than 4. Probability of less than 4 in this scenario is 6/26 = 0.230769

3. pull 20 side first which shows the 4, leaves 18 (6+12) possible outcomes in the bag, six of which are less than 4. Probability of less than 4 in this scenario is 6/18 = 0.333333

Each scenario has a 1/3 chance of being the chosen scenario, so overall probability of less than 4 is (1/3)*0.1875 + (1/3)*0.230769 + (1/3)*(0.333333) = 0.250334
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There are 3 dice in a bag: 6-sided, 12-sided and 20-sided. We pick one die at random, roll it, and it gives us a number N. What's the probability that when we pick another die and roll it, we'll get a number smaller than N, if N = 4?

-------------------

Thinking about this some more, I'll take another crack at it. Probably this is wrong too but I enjoy the exercise.

Possible scenarios.

1. pull 6 side first which shows the 4, leaves 32 (12+20) possible outcomes in the bag, six of which are less than 4. Probability of less than 4 in this scenario is 6/32 = 0.1875

2. pull 12 side first which shows the 4, leaves 26 (6+20) possible outcomes in the bag, six of which are less than 4. Probability of less than 4 in this scenario is 6/26 = 0.230769

3. pull 20 side first which shows the 4, leaves 18 (6+12) possible outcomes in the bag, six of which are less than 4. Probability of less than 4 in this scenario is 6/18 = 0.333333

Each scenario has a 1/3 chance of being the chosen scenario, so overall probability of less than 4 is (1/3)*0.1875 + (1/3)*0.230769 + (1/3)*(0.333333) = 0.250334


I think you ended correctly, but missed part of each of steps 1-3.

1a. pull 6 side first which shows the 4. Pick 12 side with 3 possibilities of less than 4 out of 12 = 0.25.

1b. pull 6 side first which shows the 4. Pick 20 side with 3 possibilities of less than 4 out of 20 - 0.15.

2a. pull 12 side first which shows the 4. Pick 6 side with 3 possibilities of less than 4 out of 6 - 0.50.

2b. pull 12 side first which shows the 4. Pick 20 side with 3 possibilities of less than 4 out of 20 - 0.15.

3a. pull 20 side first which shows the 4. Pick 6 side with 3 possibilities of less than 4 our of 6 - 0.50.

3b. pull 20 side first which shows the 4. Pick 12 side with 3 possibilities of less than 4 out of 12 - 0.25.

The odds of finding your self in any one of steps 1a - 3a is the same - 1 out of 6. So the overall odds are the average of the odds of the steps - 1/6*(.25+.15+.5+.15+.5+.25) = .30

Ira
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Mike,

Possible scenarios.

1. pull 6 side first which shows the 4, leaves 32 (12+20) possible outcomes in the bag.
2. pull 12 side first which shows the 4, leaves 26 (6+20) possible outcomes in the bag.
3. pull 20 side first which shows the 4, leaves 18 (6+12) possible outcomes in the bag.


So there a universe of 76 total outcomes.

With two dice left in the bag, there are three outcomes less than 4 on each dice, so 6 potential outcomes less than 4.

With three scenarios, each having six potential outcomes less than 4, gives 18 total outcomes less than 4.

Unfortunately, you did not take into account the fact that those outcomes are not equally probable.

Norm.
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Bob,

There are 3 dice in a bag: 6-sided, 12-sided and 20-sided. We pick one die at random, roll it, and it gives us a number N. What's the probability that when we pick another die and roll it, we'll get a number smaller than N, if N = 4?

This problem actually is much simpler than it appears. Regardless of which die is pulled first and whether it's replaced or not, there's a 1/3 probability of picking each die on the second selection. Each die has 3 faces that are less than 4 (values 1, 2, and 3). Thus, the probability is

P = 1/3 x 3/6 + 1/3 x 3/12 + 1/3 x 3/20 = 1/6 + 1/12 + 1/20 = (10+5+3)/60 = 18/60 = 3/10.

Norm.
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adonsant2 writes:

Do you put the first die back in the bag?

I reply:

No, you don't replace the first die you drew. --Bob
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None of the answers I've seen so far in this thread are correct.

Here's an easy way to see it can't be the case that, if you know your first roll is a 4, each die is equally likely to have been your first pull from the bag. Let P(x) be the probability that your first pull was the six-sided die, given that your first roll was x. Obviously if x > 6, P(x) = 0. But the total probability that the first pull was the six-sided die must be 1/3, and it also must be some weighted average of all P(x). That's not possible if P(x) = 1/3 for all x <= 6. --Bob
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First consider how likely it is that each of the three type of dice were the one first picked. Getting a 4 is 1/6 of a chance for the 6 sided die, 1/12 chance for the 12 sided die, 1/20 for the 20 sided.

So the chance that a 6 sided die was picked is 1/6 / (1/6+1/12+1/20) = 0.556. Similar for the others

Chance it was 6 sided .556
chance it was 12 sided .278
Chance it was 20 sided .167.


Next given each of these one at a time let's work out what are the chances of getting a 3 or less. There is a 50% chance of picking each of the other 2 dice.

From the start of the 6 sider, there is a 3/12 chance of getting a 1, 2 or 3 if the 12 sider is the second die, 3/20 if the 20 sider. Each of these multiplied by their 50% chance and added together gets a probability of 1/4, if it was the 6 sider picked first. Multiply that by the likelihood of the 6 sider being picked (0.556 chance) and we get 0.111 chance that the 6 sider was picked and of a 3 or less on the second die.

Similarly the probability of the 12 sider being picked first and a 3 or less off the second die is 0.09.

And with the 20 sider first and 3 or less second time 0.063.

Total probability 0.264.
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On Pick 1 the probability of getting any particular die is 1/3.

If Die 1 = 6,

P(x = 4 | die_1 = 6) = 1/6
Thus P(die_1 = 6 & x = 4) = P(die_1 = 6) * P(x = 4 | die_1 = 6)= 1/3*1/6 

Similarly, the probability of getting either of the remaining die on pick 2 is 1/2 
Thus P(y < 4 | die_2 = 12) = 3/12
and P(y < 4 | die_2 = 20) = 3/20

thus P(y < 4 | die_1 = 6 & x = 4) = 1/2*3/12 + 1/2*3/20

Therefore P(die_1 = 6 & x = 4 & y < 4) = P(die_1 = 6 & x = 4) * P(y < 4 | die_1 = 6 & x = 4) = 1/3*1/6*[1/2*(3/12+3/20)] = 1/90

Similarly,
P(die_1 = 12 & x = 4 & y < 4) = 1/3*1/12*[1/2*(3/6+3/20)] = 13/1440
P(die_1 = 20 & x = 4 & y < 4) = 1/3*1/20*[1/2*(3/6+3/12)] = 1/160

Thus the P(x=4 & y <4) = 1/90 + 13/1440 + 1/160 = 38/1440 ~ .0269
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MrPlunger writes:

First consider how likely it is that each of the three type of dice were the one first picked. Getting a 4 is 1/6 of a chance for the 6 sided die, 1/12 chance for the 12 sided die, 1/20 for the 20 sided.

So the chance that a 6 sided die was picked is 1/6 / (1/6+1/12+1/20) = 0.556. Similar for the others

Chance it was 6 sided .556
chance it was 12 sided .278
Chance it was 20 sided .167.


Next given each of these one at a time let's work out what are the chances of getting a 3 or less. There is a 50% chance of picking each of the other 2 dice.

From the start of the 6 sider, there is a 3/12 chance of getting a 1, 2 or 3 if the 12 sider is the second die, 3/20 if the 20 sider. Each of these multiplied by their 50% chance and added together gets a probability of 1/4, if it was the 6 sider picked first. Multiply that by the likelihood of the 6 sider being picked (0.556 chance) and we get 0.111 chance that the 6 sider was picked and of a 3 or less on the second die.

Similarly the probability of the 12 sider being picked first and a 3 or less off the second die is 0.09.

And with the 20 sider first and 3 or less second time 0.063.

Total probability 0.264.


I reply:

Ding! Ding! Ding! We have a winner!

This is a disguised Bayes Theorem problem. If you know your first roll was a 4, it's a lot more likely that you picked the six-sided die in the first place. The total probability works out to be exactly 19/72. --Bob
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We pick one die at random, roll it, and it gives us a number N.

Maybe it's just the wording of the problem. Normally, when something is "picked at random" all valid items have the same chance to be picked. Nothing wrong with the math.
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5761796E65 writes:

Maybe it's just the wording of the problem. Normally, when something is "picked at random" all valid items have the same chance to be picked. Nothing wrong with the math.

I reply:

At the beginning of the problem, before you learned which number was rolled, each die was equally likely. Learning that the roll was a 4 necessarily affects the a priori probabilities. That's easy to see if instead of a 4, the roll was a 17. --Bob
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Wayne,

Normally, when something is "picked at random" all valid items have the same chance to be picked.

That's right. There's a probability of 1/3 of initially choosing the six-sided die, a 1/3 probability of choosing the 12-sided die, and a 1/3 probability of choosing the 20-sided die.

But here's the catch: there are 38 possible outcomes of the first roll -- 1 through 6 on the 6-sided die, 1 through 12 on the twelve-sided die, and 1 through 20 on the twenty-sided die. The outcomes for each die are equiprobable, but that does not make all 38 outcomes equiprobable. In particular, the actual probability of each outcome on the six-sided die is 1/18 (1/3 x 1/6), the probability of each outcome on the twelve-sided die is 1/36 (1/3 x 1/12), and the probability of each outcome on the twenty-sided die is 1/60 (1/3 x 1/20).

Now, the knowledge that the initial roll is a 4 means that we have to throw out all outcomes that don't have a 4 as the rolled value -- that is, the other five outcomes on the six-sided die, the other eleven outcomes on the twelve-sided die, and the other nineteen outcomes on the twenty-sized die -- and scale the probabilities of the remaining cases (that is, the cases that do produce a four) so they add up to ONE. Now,

1/18 + 1/36 + 1/60 = 10/180 + 5/180 + 3/180 = 18/180 = 1/10

so we multiply the probabilities of the remaining cases by 10 -- and the initial roll of 4 came from the six-sided die 10/18 = 5/9 of the time, from the twelve-sided die 10/36 = 5/18 of the time, and from the twenty-sided die 10/60 = 1/6 of the time.

This is what's known as "conditional probability" -- the knowledge that the initial roll was a 4 alters the probability each die was chosen.

Note that the impact would have been very different if the initial roll had been 7 or more. If it were in the range of 7 through 12, the first die could not have been the six-sided die because the six-sided die does not have those values -- and 1/36 + 1/60 = 5/180 + 3/180 = 8/180 = 2/45, so it would come from the twelve-sided die 5/8 of the time and from the twenty-sided die 3/8 of the time. And an initial roll of 13 or higher could come only from the 20-sided die.

Norm.
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nice problem. Two related problems:

GIVEN: a bag with three die a 6-sided a 12-sided and a 20-sided in it:

1) pick one dice from the bag and roll it. What is the chance you get a 4? A 10? A 15?

2) No matter what you got on the first roll, pick one of the two remaining dice from the bag and roll it. What is the chance you get a 4? A 10? A 15?

Cheers,
R:)
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I guessed 30% chance of rolling a 3 or less on the remaining two dice. I see.26 as the answer. Since I'm rounding up because there is no .6 of a die roll, I'm sticking with 30%.

30% chance of rolling less than a 4 on the remaining two dice.
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Steershifter writes (in part):

guessed 30% chance of rolling a 3 or less on the remaining two dice. I see.26 as the answer. Since I'm rounding up because there is no .6 of a die roll, I'm sticking with 30%.

I reply:

How do you manage 0.3 of a die roll? --Bob
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