I swear I leaned this in school at some point but its been bothering me .
If you throw a six sided die, the odds of rolling a six are 1/6.
But what are the odds of rolling at least one six if you throw two dice at the same time? I thought it was 1/6 + 1/6 or 2/6, but suspect I might be wrong.
This is the idea that has me stuck. If you throw 6 dice at the same time what are the odds of rolling at least one 6 ? It can’t be 6/6 or 100%. The odds of rolling at least one 6 are certainly high but absolutely not 100%, so the logic I used for the two dice can’t be correct. There must be a formula for this but I’m having trouble searching for it . Thanks !
In: 12
The chance is essentially 1/6 for each dice and each throw (even if you do them simultaneously) is independent.
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It is equal to 1 minus the probability of you rolling no sixes.
For an individual die that’s 5/6 since any number between 1 and 5 works. The probability of rolling 6 dice with no sixes is (5/6)**6. The probability of rolling at least one 6 is therefore 1-(5/6)**6
There are two ways to solve this: either you can work out all possible permutations of six dice (there are 46,656 of them), count which ones have one or more sixes, and divide that by the total.
Or, you can calculate the odds that you roll zero sixes and take the opposite. This is the easier method: the odds of not rolling a six are 5/6. To find the probability that two independent events will occur, you multiply them. So if you have six dice, you should multiply 5/6 by itself six times, i.e. raise it to the sixth power. This is 15625/46656, or about 33%. That is the odds of rolling zero sixes, so the odds of rolling at least one six are 67%.
The problem with the 1/6 + 1/6 approach (for two dice) is that the particular situation where you roll a pair of sixes is counted twice. You counted it in the first 1/6 and also in the second 1/6. This smidgen of probability needs to be removed from one of the 1/6’s to get an accurate total probability. (Picture a Venn diagram with two intersecting circles and you want the total area; you must not count the center area twice.) With many dice there are even more of these multiple six situations that were counted multiple times and must be removed. That makes your additive approach complicated. Feasible but complicated. It happens to be easier to figure out everything that doesn’t involve any sixes and then take 1 minus that.
The logic for 2 dice is as follows. When you roll 2 dice, lets say the first once lands on 1, there are 6 possibilities for the second die. That is also true for all 6 possibilities of the first die. You can get 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2…. 6,4 6,5 6,6 or 36 possible pairs.
Now we just count all the pairs that contain a 6
We get 1,6 2,6 3,6 4,6 5,6 6,6
There is also 6,1 6,2 6,3 6,4 6,5 and we already counted 6,6 for a total of 11 of 36 possibilities or a 30.6% chance.
You can also picture a grid.
1 2 3 4 5 6
1 x
2 x
3 x
4 x
5 x
6 x x x x x x
From the visual we can also see a 5×5 grid or 25 pairs that don’t contain a 6 or a 69.4% of no sixes. The same logic can be expanded out for more dice. 5/6 * 5/6 * 5/6… ect which is better than counting pairs.
Let’s think of odds as a portion of all the possible outcomes. For this example lets use coins because manually listing the possible outcomes of dice is a PITA.
Flipping a coin 4 times, every possibility is: HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT
We can calculate the total amount of outcomes by taking the amount of possible outcomes of a single “turn,” (in this case 2) and taking it to the power of the amount of “turns,” we have (in this case 4). 2^(4) = 16 possible outcomes.
Which odds we get depends on which questions we ask.
**What are the odds we get at least 3 tails?** Well, we count up the amount of outcomes that have three tails in them (4) and take that out of the total outcomes (16). 4/16 or simplified down we have a 1/4 chance of getting at least 3 tails.
**What are the odds we get 3 tails in a row?** There are only 3 outcomes that have 3 in a row, meaning our odds are 3/16.
**What are the odds we get the same number of heads as tails?** 6 outcomes have this property, so our odds are 6/16 or 3/8.
You can do the same thing with dice. If we listed out all the possible outcomes of rolling a dice twice (same as rolling 2 dice at once), with each chance having 6 possible outcomes. 6^(2) = 36 possible outcomes total. Each roll has a possibility of landing on a 6 once, which means 2 out of the 36 possibilities, or in other words we have a 2/36 = 1/13 chance or rolling a 6 at least once.
Give it a shot yourself: What do you think the odds will be of getting both dice to land on a 6?
There are definitely formulas to work this out, but to me the easiest way is to do the opposite. What are the chances none of them are 6?
5/6 chance that any individual dice is not a 6. So the chance none are 6 is (5/6)^6 =0.335. therefore the chance at least 1 is a six is 1-0.335 = 0.665. or 66.5% chance.
Which obviously isn’t the same as 100%, which may not be what many people would first think!
You can repeat with as many dice as you want. Formula would still be 1-(5/6)^x where X is number of dice (or number of rolls, if you just reroll a single dice). You’ll note that as x gets larger the answer gets larger, but will never reach 1 as it’s still not a guarantee that you’ll ever get a 6.
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Rolling at least one 6, is the exact opposite of rolling exactly zero 6. For the first dice you have 5/6 chance to not roll a 6, AND you need the second dice to not roll a six either, AND the third dice, AND the fourth, and so on
In probability, when you hear AND you should think multiplication. Therefore the answer is ⅚ × ⅚ × …
6 times in total, which is (⅚)^6 ≈ 0.33
This means you have around ~33% chance of rolling zero six when throwing six dices. Therefore in ~67% of the cases, there will not be zero 6 (so either one two three four five or six 6)
And when you hear OR, in probability, you should think +. So for example “What is the probability of hitting 6 OR 1 in 6 throws”, is either 6 OR 1 for the first, AND 6 OR 1 for the second, AND 6 OR 1 for the third, and so on → (⅙+⅙)^6
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