Probability of Even Score with 2 Tetrahedral Dice

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In summary, the conversation discusses a question involving two tetrahedral dice and finding the probability that the score is even, given that at least one die lands on a 3. The formula P(A|B) = P(A&B) / P(B) is used, where A represents an even number and B represents the first die landing on a 3. The final answer is 3/7.
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scolty
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Hi,

im trying to figure out how to solve the following question and any hints would be greatly appreciated.

Question: Two tetrahedral dice, with faces labelled 1,2,3,4 are thrown and the number on which each lands is noted. The score is the sum of the two faces. Find the probability that the score is even, given that at least one die lands on a 3.

The final answer is suppose to be 3/7.

Below is as far as i had managed to get.

P(A|B) = P(A&B)/P(B)

where:
A is the sum = an even number, ie 2nd dice = 1 or 3
B the first dice is 3.

if the above is correct (ie P(B) isn't the probability that a single dice is showing 3) then is 1/4, otherwise i think its 3/16 (that may be wrong.)

im lost from here on. As i mentioned earlier, any assistance would be appreciated.
 
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  • #2
i managed to solve this by writing out the sequence of dice rolls as follows:
(sure 99% of the ppl using this forum know more about stats than i do, so this is for the 1%)

Cg2wZ.jpg


green are the possible scenarios where one die is a 3.

red are the corresponding scenarios where the total is an even number

P(A|B) = P(A&B) / P(B)

where P(B) = 7/16 (note I am not entirely sure why its not 8 since 3,3 isn't counted as two possibilities.)
P(A&B) = 3/16 (See the combinations which are both green and red)

answer = 3/7
 

FAQ: Probability of Even Score with 2 Tetrahedral Dice

What is the probability of getting an even score when rolling two tetrahedral dice?

The probability of getting an even score when rolling two tetrahedral dice is 1/2 or 50%. This is because there are 4 possible outcomes for each die (1, 2, 3, or 4), and when you roll two dice, the total number of outcomes is 4x4=16. Out of those 16 outcomes, 8 of them will result in an even score (2, 4, 6, or 8). Therefore, the probability is 8/16 or 1/2.

How do you calculate the probability of getting an even score with two tetrahedral dice?

To calculate the probability of getting an even score with two tetrahedral dice, you need to first determine the total number of outcomes when rolling two dice. This is found by multiplying the number of outcomes for each die (4) by itself (4x4=16). Then, determine the number of outcomes that result in an even score (8). Finally, divide the number of outcomes that result in an even score by the total number of outcomes (8/16) to get the probability, which is 1/2 or 50%.

Is the probability of getting an even score with two tetrahedral dice affected by the order of the dice?

No, the probability of getting an even score with two tetrahedral dice is not affected by the order of the dice. This is because each die is independent of the other, meaning the outcome of one die does not affect the outcome of the other. Therefore, the probability of getting an even score remains the same regardless of the order in which the dice are rolled.

How does the probability of getting an even score change if one of the dice is loaded?

If one of the dice is loaded, the probability of getting an even score with two tetrahedral dice will change. This is because the loaded die will have a higher chance of landing on certain numbers, which will affect the overall probability of getting an even score. The exact change in probability will depend on the degree to which the loaded die is biased.

Can you use the probability of getting an even score with two tetrahedral dice to predict the outcome of a single roll?

No, the probability of getting an even score with two tetrahedral dice cannot be used to predict the outcome of a single roll. While the probability does give an idea of the likelihood of getting an even score in the long run, it does not guarantee that the next roll will result in an even score. Each roll is independent and the outcome is ultimately determined by chance.

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