*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?

- Bob78164
- Bored Moderator
**Posts:**19397**Joined:**Mon Oct 08, 2007 12:02 pm**Location:**By the phone

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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- jarnon
**Posts:**5252**Joined:**Tue Oct 09, 2007 9:52 pm**Location:**Merion, Pa.

First die is a cube:

probability 5/9

if second die is a dodecahedron, probability of <4 is 1/4

if second die is an icosahedron, probability of <4 is 3/20

First die is a dodecahedron:

probability 5/18

if second die is a cube, probability of <4 is 1/2

if second die is an icosahedron, probability of <4 is 3/20

First die is an icosahedron:

probability 1/6

if second die is a cube, probability of <4 is 1/2

if second die is a dodecahedron, probability of <4 is 1/4

Probability of <4 is (5/9) x ((1/2) x (1/4) + (1/2) x 3/20)) + (5/18) x ((1/2) x (1/2) + (1/2) x (3/20)) + (1/6) x ((1/2) x (1/2) + (1/2) x (1/4)) = 19/72 = 0.263888...

probability 5/9

if second die is a dodecahedron, probability of <4 is 1/4

if second die is an icosahedron, probability of <4 is 3/20

First die is a dodecahedron:

probability 5/18

if second die is a cube, probability of <4 is 1/2

if second die is an icosahedron, probability of <4 is 3/20

First die is an icosahedron:

probability 1/6

if second die is a cube, probability of <4 is 1/2

if second die is a dodecahedron, probability of <4 is 1/4

Probability of <4 is (5/9) x ((1/2) x (1/4) + (1/2) x 3/20)) + (5/18) x ((1/2) x (1/2) + (1/2) x (3/20)) + (1/6) x ((1/2) x (1/2) + (1/2) x (1/4)) = 19/72 = 0.263888...

- MarleysGh0st
**Posts:**27844**Joined:**Mon Oct 08, 2007 10:55 am**Location:**Elsewhere

I like jarnon's calculations, except that he seems to be calculating the probability of which die was picked first based on the probability of that die rolling a 4. The problem clearly states that the first die was chosen randomly; therefore, those probabilities in the final calculation are all 1/3. Obviously, if we replayed this experiment and got certain different rolls--say, 17*--we could infer more about which die was rolled first.

*And in that particular case, the final answer would be 1.

*And in that particular case, the final answer would be 1.

- littlebeast13
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From a realistic standpoint, if you look at the first die to see that it rolled 4, it's going to be obvious from sight which die it is, which would completely change the calculations...

lb13

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- Vandal
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I will go on the record stating that this math problem is neither cute nor little.

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- Bob78164
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Jarnon's solution is correct. Once you know that the first roll is a 4, that makes it more likely that your random choice was the six-sided die.MarleysGh0st wrote: ↑Tue Sep 08, 2020 7:41 amI like jarnon's calculations, except that he seems to be calculating the probability of which die was picked first based on the probability of that die rolling a 4. The problem clearly states that the first die was chosen randomly; therefore, those probabilities in the final calculation are all 1/3. Obviously, if we replayed this experiment and got certain different rolls--say, 17*--we could infer more about which die was rolled first.

*And in that particular case, the final answer would be 1.

The easy way to see that 1/3, 1/3, 1/3 can't be right is to realize that if you average the probabilities that your first die was the six-sided die across all numbers from 1 to 20, the answer for the six-sided die has to be 1/3. But that probability is clearly 0 for any number greater than 6, so it has to be greater than 1/3 for the numbers 1 through 6, or the average will end up less than 1/3. --Bob

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- MarleysGh0st
**Posts:**27844**Joined:**Mon Oct 08, 2007 10:55 am**Location:**Elsewhere

I withdraw my objection. Without realizing it, my counter-example was pointing at what you just explained: while the choice of the first die is random, the value of N infers, to a variable extent, which die was chosen.Bob78164 wrote: ↑Tue Sep 08, 2020 11:38 amJarnon's solution is correct. Once you know that the first roll is a 4, that makes it more likely that your random choice was the six-sided die.MarleysGh0st wrote: ↑Tue Sep 08, 2020 7:41 amI like jarnon's calculations, except that he seems to be calculating the probability of which die was picked first based on the probability of that die rolling a 4. The problem clearly states that the first die was chosen randomly; therefore, those probabilities in the final calculation are all 1/3. Obviously, if we replayed this experiment and got certain different rolls--say, 17*--we could infer more about which die was rolled first.

*And in that particular case, the final answer would be 1.

The easy way to see that 1/3, 1/3, 1/3 can't be right is to realize that if you average the probabilities that your first die was the six-sided die across all numbers from 1 to 20, the answer for the six-sided die has to be 1/3. But that probability is clearly 0 for any number greater than 6, so it has to be greater than 1/3 for the numbers 1 through 6, or the average will end up less than 1/3. --Bob

And lb13 has a valid point, if

- Bob Juch
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I had to take my neurons back in time 50 years, but I came up with Jarnon's solution.

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Si fractum non sit, noli id reficere.

Teach a child to be polite and courteous in the home and, when he grows up, he'll never be able to drive in New Jersey.