This should be simple but apparently its not.

[gamblore99]

Hey guys, This probability question is really confusing me, ive asked a friend and my stats proffesor (I don't think she understood my question), and both couldn't help me. Here it is....please help

If I have one die, and want to role a 6, the probability of me getting a 6 is 1/6. now say I have 2 die, and role them at the same time, they are mutually exclusive so I can find the probability of getting a six on either dice (getting a 6 on both counts), so I can add the sum of there probabilities to get 2/6.

Now say I have 7 dice, and I roll them at the same time and want to get a 6. Again mutually exclusive, so i add up the individual probabilites and get 7/6 as the chance of rolling a 6. But probability cannot exceed 1. This does not make sense. I am also aware that 7/6 is the expected amount of 6s I would get when rolling 7 dice. BUt it cant be the probability as probability cannot be greater than one, and in this case it is, and I know it is very possible to roll 7 dice together and not get a one. Therefore the probabilities of 2 mutually exclusive events do not add. But they Do. Don't they?

Someone please help me make sense of this conundrum. Thank you

[snowlarbear]

nope, for the 2 dice example, it is:

chance die1 rolls 6: 1/6
chance die2 rolls 6: 1/6

you don't add them. you mentioned you want a 6 on BOTH dice, so its (1/6) * (1/6) = 1/36.

alternately, if you are just looking for the prob. you will get a 6 on EITHER or BOTH dice, you have to take into account all possibilites of getting a 6 on one die and coming up blank on the other (5 possible outcomes), plus the same situation for the other die (another 5), plus the probability they will both be a 6 (1/36).

(5/36) + (5/36) + (1/36) = 11/36 and thats the probability you'll get a 6 on either or both dice.

i think that is right, it has been a while since math class so i don't remember the general formula which would give you the answer for the 7 dice situation. maybe its 7 choose 1 or something? anyway, you definately don't add both the probabilities.

[udontknowmickey]

No, they do not add.

for your problem you can do it as the previous poster responded for two dice, but as you add dice it gets progressively more complicated.

There is an easy method for computing higher numbers of dice . Take the probability that you will NOT get a 6 on any of the 7 dice. This is 5/6 for each die, and since you want it to be not a 6 on each die, its 5/6*5/6*...(7 times) *5/6 or (5/6)^7. This is your probability of not ever getting a 6, so the probability of getting at least 1 "6" is 1-(5/6)^7

[pzhon]

[ QUOTE ]
now say I have 2 die, and role them at the same time, they are mutually exclusive so I can find the probability of getting a six on either dice (getting a 6 on both counts), so I can add the sum of there probabilities to get 2/6.

[/ QUOTE ]
I think you are confused about the meaning of mutually exclusive. If rolling a 6 on the first die and on the second die were mutually exclusive, it would mean that it is impossible to roll a 6 on both dice at the same time.

Since these events are not mutually exclusive, you can't add the probabilities to find the probability of their union.

Perhaps you are confusing the term "independent" with mutually exclusive. The events you can describe in terms of the first die alone are independent of the events you can describe in terms of the second die alone.

[gamblore99]

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[ QUOTE ]
now say I have 2 die, and role them at the same time, they are mutually exclusive so I can find the probability of getting a six on either dice (getting a 6 on both counts), so I can add the sum of there probabilities to get 2/6.

[/ QUOTE ]
I think you are confused about the meaning of mutually exclusive. If rolling a 6 on the first die and on the second die were mutually exclusive, it would mean that it is impossible to roll a 6 on both dice at the same time.

Since these events are not mutually exclusive, you can't add the probabilities to find the probability of their union.

Perhaps you are confusing the term "independent" with mutually exclusive. The events you can describe in terms of the first die alone are independent of the events you can describe in terms of the second die alone.

[/ QUOTE ]

thank you. this post has really helped me understand my mistake. I Will hang your picture on my kitchen wall.

Thank you to everyone else, your responses were also helpful, though not kitchen wall worthy.

[Cerril]

Good enough... this is just intuitive but another way to look at it (which has next to nothing to do with probability) is that the 'average' number of times you roll a six is 1/6, or one out of every six rolls.

With two dice it's 2/6, or every three rolls you average a 'six'.

With seven dice the 7/6 means that you're going to average 7 'sixes' every six rolls, or that you're likely to see more than one six come up every time (well, a fraction more than one).

That doesn't help with the probability of rolling 'at least one six on seven dice rolled once' of (5/6)^6 or ~72%

[BrettK]

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With seven dice the 7/6 means that you're going to average 7 'sixes' every six rolls, or that you're likely to see more than one six come up every time (well, a fraction more than one).

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Averaging seven sixes when only six are possible doesn't seem counter-intuitive?

You have a 1/6 chance of rolling each of six numbers.
If you roll six dice, the highest probability lies with that you will see one of each number.

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That doesn't help with the probability of rolling 'at least one six on seven dice rolled once' of (5/6)^6 or ~72%

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Seventy-two percent is correct, but the equation is 1-[(5/6)^7]

Brett

[mmbt0ne]

I think he means you should see 7 sixes every 6 trials, where each trial involves rolling 7 dice.

[gamblore99]

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With seven dice the 7/6 means that you're going to average 7 'sixes' every six rolls, or that you're likely to see more than one six come up every time (well, a fraction more than one).


[/ QUOTE ]

you will not see 7 success on 6 roles. you will see 7/6 successes. I think thats what you mean, and anyways 7/6 Expected Value is correct.

[Cerril]

There are actually seven possible sixes every roll, since you're rolling seven dice, so 42 possible sixes every seven rolls (if you're keeping a tally of every time that face shows up).

And uh, yeah, right... I do the (1-x) part in my head whenever I see probability stuff, so my 'showing my work' part sometimes looks a little silly.