3
$\begingroup$

I play a tabletop game and we roll dice for game outcomes. The dice have 8 sides and a certain result (lets call it a success) appears on 2 sides, so 25% of success.

If I roll 4 dice, the chance of having at least one success is about 70% (binomial distribution for 4 dice).

Say I'm rolling 4 dice, and then I'm rolling another 4 dice. Statistically, this is the same as rolling 8 dice, right? Therefore, I can apply the binomial distribution to 8 dice. The chance of 5 or more successes in 8 dice is about 3%.

Say I roll 4 dice and get 4 successes. I can conclude that the chance of having at least one other success in the next 4 dice is about 3%.

However, a lot of other players disagree and think that it is not 3% but 70%, because I'm rolling 4 dice again, and it doesn't matter what I rolled before.

I would appreciate if you could either confirm I am right, or explain to me why I'm wrong. thanks!

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Let's consider a simpler version of the same question, where you will flip a fair coin twice. By assuming the chance of each flip of being heads is 50%, that's a 25% chance of two heads. You flip the coin once and happen to see heads. Analogously with your question, it seems you would maintain that the chance of getting one more heads is still 25%. How do you square that with your original assumption that the chance of a heads on the second flip is 50%? $\endgroup$
    – whuber ♦
    Commented Jun 17, 2016 at 21:40
  • $\begingroup$ the way I see it is that the difference in probability predictions is due to the fact that you're predicting two different events. 25% is the chance for 2 heads in 2 flips. 50% is the chance of 1 head in 1 flip. they are both correct, as they are different events, but one prediction uses more knowledge so you can use that for better accuracy. $\endgroup$
    – XBear
    Commented Jun 18, 2016 at 2:19
  • 2
    $\begingroup$ But how does that resolve the contradiction in your reasoning, where you assert that the second flip has both 25% and 50% chance of heads? Note that there is no ambiguity about this event: in concerns the second flip only. The point is that your original calculation that yielded 25% assumed the outcome of the first flip gives no information about the second flip (that's what independence means and it was the assumption of independence that allowed you to multiply $1/2$ by $1/2$ to get 25% in the first place). Thus there is no possibility of more "knowledge." $\endgroup$
    – whuber ♦
    Commented Jun 18, 2016 at 14:16

1 Answer 1

Reset to default
1
$\begingroup$

Unfortunately, you are wrong, and the other players are right. The chance of you getting at least 1 success out of 4 dice is still 70%, because previous events don't affect future events.

Think about it this way - you have a coin that you know is fair (50/50). You flip it 1,000 times, and it lands on heads every time. What are the odds that it'll be a head the next time you flip it? It might seem like the coin has 'used' up all of its luck already in getting heads, but it's still 50/50 because previous outcomes don't affect future ones. Coins aren't affected by previous outcomes - otherwise, you'd need to get a brand new coin each time you wanted to flip a coin fairly. In the same manner, your 4 dice rolls aren't affected in any way by your previous rolls.

In more mathematical terms (this isn't the proper notation but hopefully you get the gist)

$P(success = 1$ out of $4) = 0.70$

$P(success = 5$ out of $8) = 0.03$

However, you've already rolled 4 dice, so you need to account for that. The real probability you're looking at is:

$P(success = 5$ out of $8 | success = 4$ out of $4)$

which is equal to 0.7.

(All of this assumes that your dice are actually fair - if they're rigged that's another story.)

Improve this answer
$\endgroup$
6
  • $\begingroup$ what I don't understand about your answer is that it seems you're saying that rolling 4 dice first, followed by 4 dice, is different than rolling 8 dice all together. $\endgroup$
    – XBear
    Commented Jun 17, 2016 at 21:15
  • 2
    $\begingroup$ Rolling 4 dice and then 4 dice and rolling 8 dice are the same. But in your case, as Kontorus points out, you are looking for the probability of rolling 1 success out of 4, so 0.70. This is the case because in the example you describe, you already roll 4 successes out of 4 with the first four dice (this is the outcome with a low probability). $\endgroup$
    – Antoine Vernet
    Commented Jun 17, 2016 at 21:24
  • $\begingroup$ I still don't understand. the dice don't know you looked at them midway. you roll 8 dice, the chance of having 5 or more successes is 3%. if the first 4 dice are all successes, why the chance of having at least another one in the next 4 is 70% instead of 3%? $\endgroup$
    – XBear
    Commented Jun 17, 2016 at 21:59
  • 1
    $\begingroup$ Think about this idea - if you roll 8 dice, the chance of having 5 or more successes is 3%. What if you roll the first 4 dice and get NO successes. What's the chance of getting 5 or more successes out of the 8 dice rolls now? It's 0 - you can't possibly get 5 or more successes because the maximum number of successes will be 4. The probability has changed, because some of the dice are now fixed in value and aren't up to chance. $\endgroup$
    – Duncan
    Commented Jun 17, 2016 at 22:12
  • 1
    $\begingroup$ @XBear Rolling 4 dice and then looking at them changes the probabiltiy of the overall outcome after the 8th toss (5 successes in 8) because looking changes the information from what it was when you started. If you get 4 successes in the first four tosses you know you only need one more success in the remaining 4 tosses, which is much easier to achieve than if you only had say 1 success in the first four tosses (where you'd need 4 successes in the remainder). $\endgroup$
    – Glen_b
    Commented Jun 18, 2016 at 3:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.