If I throw 6 dice, what is the chance that at least one "1" shows up?

Question

I'm amazed at how many people choose to comment/answer when they have absolutley no clue. The correct answer is .665

Answer 1

1/6 chances

Answer 2

1 in 6 (actually 6 combinations out of 36 total combinations of faces fits the criteria, which reduces to 1 in 6.)

It does not matter how many you throw at once. For a single die, the odds are a simple 1 in 6. Throwing more dice in a group does not reduce the odds of scoring a 1 showing as each individual die still has a 1 in 6 probability. You are increasing the number of times you try to get a 1 to show is all. As with coins, flipping multiple coins does not change the odds on the next toss being heads to anything other than the original 1 in 2 of the 1st tossed coin.

Consider instead of throwing all 6 at once, you threw a single die 6 times. Has this changed the odds? No, still 1 in 6. You are just throwing 6 at once instead of 6 in sequence. Someone said the odds change if in sequence. Not so. Each individual toss has a 1 in 6 probability. Just because my 1st toss did not score a 1 does not raise or lower the probability of the 2nd toss, which is still 1 in 6. Consider the coins again. If I toss 99 coins and each time the coin came up heads, what are the odds the 100th toss will be heads? Wrong, still 50-50. (A lot of people say heads is more likely. A mechanical coin flipper CAN consistently flip sequential coins to come up the same but only if started the same It's like the bread sliding off the table and ALWAYS landing butter side down if it was butter side up while on the table. I'm talking you and your hands, not a device.) History of tosses does not affect future random events. Previous tosses of dice do not change the next die toss from 1 in 6.

Is it possible I have this bassackwards and history CAN change the future random event?

EDITED: the answer following me is wrong. Vegas would go out of business with his 100% reasoning. His 6 out 6 is flawed since there are 36 individual faces which can come up, with 6 of each which means 6 out of 36 meets the criteria. Vegas lives off people with this 100% kind of reasoning.

Answer 3

well, if you throw 1 die, then the chance of a "1" is of course 1 in 6

if you threw all the dice individually, one at a time, each one would have a one in 6 chance of coming up 1

on average, with many throwings, you would therefore get one "1" per throw of six dice

for a single throw, the easiest way to look at it is to calculate the odds that you don't get a "1"

5/6*5/6*5/6*5/6*5/6*5/6=.335

so the odds of getting a "1" with a single throw, is .665, or there is a 66.5% chance that any throw will have a one

Answer 4

1/6 it will be a one

Answer 5

There is a 5/6 chance for each seperate one to not show a "1"
So for 6 die, the odds would be 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6.
This results in a 33.5% chance to not roll any ones, or a 66.5% chance to roll at least one.

Answer 6

Probability (roll a 1)= 1/6

Answer 7

Well, if you are throwing 6 dice once, the probability of any one number showing up is 6/36 or 1/6.

There are a total of 36 sides and and there are 6 of each number.

If, however, you are throwing 1 die 6 times, the probability is 1/36. (1/6)(1/6)(1/6)(1/6)(1/6)(1/6)

Answer 8

1:6 [all six dice are tossed on every throw]

It doesn't matter how many dice are thrown, if you're looking for only one "1."

Now, if you're looking for two or more "1's" to appear, the probability changes. Don't really know what it is, but my guess would be (with all six dice tossed on every throw):

Two "1's" in 12 throws
Three "1's" in 18 throws
Four "1's" in 24 throws
Five "1's" in 30 throws
Six "1's" in 36 throws

Just a guess . . .

Answer 9

1/(6*6)

Answer 10

these answers are ridiculous. the chance of atleast one 1 showing up is 100%. each die has a one in six chance of a one showing, and you have six dice. that is six chances in six....100%. that does NOT mean you will definitely get a one showing on every roll. just that the odds are 6 in 6. you people who do not undertand statitics should not answer these questions