# A (simple ?) probability question.
**URL:** https://boards.straightdope.com/t/a-simple-probability-question/656348
**Category:** Factual Questions
**Created:** [April 21, 2013, 11:34pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348 "2013-04-21T23:34:53Z")
**Posts on this page:** 20
**Page:** 2
### Author:  [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)
#### Post date: [April 22, 2013, 5:10am UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/21 "2013-04-22T05:10:31Z")
> [@Jragon](#):
>
> It doesn’t matter whether you roll the dice at the same time, or one and then wait to see if you should roll the other. The calculations to get there are different, but the result is the same. If you roll both at the same time – Great Antibob’s set method is correct, if you roll them one after the other, RadicalPi is correct.
>
> The reason for this is that the roll of the second die is dependent on the first roll. Specifically, you have a 5/6 chance that you will make the second roll. This is basically what RadicalPi was showing.
That makes sense, I was thinking in terms of opportunity before the dice were ever rolled ignoring the fact that one dice would have to be rolled before the other.
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### Author:  [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)
#### Post date: [April 22, 2013, 7:46am UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/22 "2013-04-22T07:46:27Z")
> [@HoneyBadgerDC](#):
>
> I never said that, the odds suggest you would be but odds don’t mean anything will actually happen.
Then what do you mean when you say the odds are 1/1?
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### Author:  [@septimus](https://boards.straightdope.com/u/septimus)
#### Post date: [April 22, 2013, 9:46am UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/23 "2013-04-22T09:46:18Z")
**Great Antibob** should have closed the thread (or ask for a move to BBQ Pit) after his correct answer in #2.
But as long as we’re continuing, it might be noted that there is a sense in which number of sixes after three rolls is 3/6 = 0.5, or after six rolls is 1.0. Those are _the (mean) expected number of sixes_. Sometimes you get fewer than 1 six after 6 rolls, because sometimes you get more than 1.
That k/6 gives the _expected count_ of sixes in k rolls may be very obvious, but the same idea provides a sometimes-overlooked shortcut in harder problems.
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### Author:  [@mcgato](https://boards.straightdope.com/u/mcgato)
#### Post date: [April 22, 2013, 12:42pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/24 "2013-04-22T12:42:21Z")
I’m always amused when people who obviously know nothing about a subject make posts as though they have great knowledge about the subject. HoneyBadgerDC has amused me.
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### Author:  [@C\_K\_Dexter\_Haven](https://boards.straightdope.com/u/C_K_Dexter_Haven)
#### Post date: [April 22, 2013, 1:30pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/25 "2013-04-22T13:30:11Z")
Another way of verifying: what are the chances of LOSING? Not getting a 6 on die #1 (probability 5/6) AND not getting a 6 on die #2 (prob 5/6) Basic probability theory: multiply when you need both results, so 5/6 x 5/6 = 25/36. Thus the chance of winning (without a 4) is 1 - 25/36 = 11/36. QED.
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### Author:  [@Colibri](https://boards.straightdope.com/u/Colibri)
#### Post date: [April 22, 2013, 3:03pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/26 "2013-04-22T15:03:45Z")
> [@HoneyBadgerDC](#):
>
> That makes sense, I was thinking in terms of opportunity before the dice were ever rolled ignoring the fact that one dice would have to be rolled before the other.
**Moderating**
**HoneyBadgerDC** , it’s not particularly helpful to keep posting answers on a subject that you are not informed about, especially when others have already posted correct information. Let’s refrain from this in the future.
Colibri
General Questions Moderator
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### Author:  [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)
#### Post date: [April 22, 2013, 3:51pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/27 "2013-04-22T15:51:07Z")
> [@septimus](#):
>
> **Great Antibob** should have closed the thread (or ask for a move to BBQ Pit) after his correct answer in #2.
>
> But as long as we’re continuing, it might be noted that there is a sense in which number of sixes after three rolls is 3/6 = 0.5, or after six rolls is 1.0. Those are _the (mean) expected number of sixes_. Sometimes you get fewer than 1 six after 6 rolls, because sometimes you get more than 1.
>
> That k/6 gives the _expected count_ of sixes in k rolls may be very obvious, but the same idea provides a sometimes-overlooked shortcut in harder problems.
A sidetrack on expected values.
Imagine you have a bunch of cups, numbered in sequence from one to however many cups you have. And you have an equal number of balls, numbered the same way. And you have a machine that shakes up the balls and randomly drops one ball in each cup.
How many balls should you expect to find in a cup with the same number?
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [April 22, 2013, 4:08pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/28 "2013-04-22T16:08:18Z")
If the game is to roll at least one six on a pair of dice, then your probability of winning is 11/36, or slightly more than 30%. If you’re allowed to reroll once if you get a four, then your probability of winning goes up to 55/144, or slightly better than 38%. How much value is there in getting more chances to reroll?
Suppose I let you reroll up to two times. Then your probability of winning is 11/36 + 9/36(11/36 + 9/36 \* 11/36), which works out to be 77/192, or about 40.2%. That’s still a pretty good increase, so what if you get more rerolls than that?
Let’s now look at the most extreme case. Suppose that you can keep rerolling as long as you get a four. In that case, the probability of winning _p_ is given by the solution to the equation _p_ = 11/36 + 9/36_p_. This is an easy computation, and we find that _p_ = 11/27, or about 40.7%. There’s very limited value in having more than two rerolls–in fact, having three gets you all the way up to 40.5%, which is essentially the same probability unless you’re doing a lot of gambling.
But we can be a little bit more precise in quantifying the declining marginal value of rerolls. Let _p_(_n_) denote the probability of winning when you’re given _n_ reroll attempts. We can pretty easily see that _p_(0) = 11/36, and _p_(_n_) = 11/36 + 9/36_p_(_n_ - 1) for _n_ \> 0. This is easy to solve with your favorite computer algebra system, and it gives us the solution of _p_(_n_) = 11/27(1 - 1/4[sup]n + 1[/sup]). From this, we see that the additional value of each roll is going to zero roughly as fast as 1/4[sup]n[/sup].
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### Author:  [@Great\_Antibob](https://boards.straightdope.com/u/Great_Antibob)
#### Post date: [April 22, 2013, 4:09pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/29 "2013-04-22T16:09:11Z")
> [@Little\_Nemo](#):
>
> How many balls should you expect to find in a cup with the same number?
Via wiki, [random permutations](http://en.wikipedia.org/wiki/Random_permutation).
With sufficient number of balls, the distribution of what are called “fixed points” approaches Poisson with mean 1.
So, the expected number of matches is 1 with sufficiently large number of cups/balls. The ‘mass’ of the distribution is about 37% with no matches, about 37% with 1 match, and roughly 26% with more than 1 match.
As an aside, the probability of absolutely zero matches approaches 1/e or roughly 37% and is called a [derangement](http://en.wikipedia.org/wiki/Derangement) (one of my favorite combinatorics terms). 1/e is also roughly the probability of exactly 1 match in this case.
Taking it back to the original question and my response: I didn’t realize there’d be quite this much contention in this thread.
When explaining a combinatorics problem with only a few outcomes, I like to go ahead and list outcomes rather than rely solely on computation. For students, it helps reinforce the idea that the numbers should bear some resemblance to reality (what we like to call a ‘sanity check’) and helps build up intuition in both numeracy and combinatorics.
I guess that backfired a bit here. But all the alternate methods for getting the same numbers is another good way of showing there’s never just one way of thinking about any problem.
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### Author:  [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)
#### Post date: [April 22, 2013, 4:28pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/30 "2013-04-22T16:28:43Z")
I don’t think the correct answer, which is 374/900, has been posted. The chances of at least one 6 are, as stated many times is 11/36. No argument. The chances of at least one 4, in the absence of at least one 6, are 9/25. Then you roll again and those chances are 11/36. Thus the total is
11/36 + (9/25)(11/36) = 374/900
(if my arithmetic is correct).
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### Author:  [@Great\_Antibob](https://boards.straightdope.com/u/Great_Antibob)
#### Post date: [April 22, 2013, 4:46pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/31 "2013-04-22T16:46:39Z")
> [@Hari\_Seldon](#):
>
> The chances of at least one 4, in the absence of at least one 6, are 9/25.
This part doesn’t seem correct. The probability should be 9/36.
And I’ll list all outcomes to make it clear:
1-1 1-2 1-3 1-4 1-5 1-6
2-1 2-2 2-3 2-4 2-5 2-6
3-1 3-2 3-3 3-4 3-5 3-6
4-1 4-2 4-3 4-4 4-5 4-6
5-1 5-2 5-3 5-4 5-5 5-6
6-1 6-2 6-3 6-4 6-5 6-6
Of these, 11 have at least one six. 9 have a 4 but no 6. And the remaining 16 have neither.
I think your error comes from a misunderstanding. Given there are no 6’s, the probability of a 4 is indeed 9/25.
But the probability of no 6’s is 25/36, which you must also include:
Prob(4s and no 6s) = Prob(4 given no 6s)_Prob(no 6s)
= (9/25)_(25/36)
= 9/36
In a textbook, you might see it given as:
```
P(A and B) = P(A|B)*P(B)
```
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### Author:  [@Learjeff](https://boards.straightdope.com/u/Learjeff)
#### Post date: [April 22, 2013, 7:37pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/32 "2013-04-22T19:37:40Z")
> [@Great\_Antibob](#):
>
> Via wiki, [random permutations](http://en.wikipedia.org/wiki/Random_permutation).
>
> With sufficient number of balls, the distribution of what are called “fixed points” approaches Poisson with mean 1.
>
> So, the expected number of matches is 1 with sufficiently large number of cups/balls. The ‘mass’ of the distribution is about 37% with no matches, about 37% with 1 match, and roughly 26% with more than 1 match.
>
> As an aside, the probability of absolutely zero matches approaches 1/e or roughly 37% and is called a [derangement](http://en.wikipedia.org/wiki/Derangement) (one of my favorite combinatorics terms). 1/e is also roughly the probability of exactly 1 match in this case.
> Taking it back to the original question and my response: I didn’t realize there’d be quite this much contention in this thread.
>
> When explaining a combinatorics problem with only a few outcomes, I like to go ahead and list outcomes rather than rely solely on computation. For students, it helps reinforce the idea that the numbers should bear some resemblance to reality (what we like to call a ‘sanity check’) and helps build up intuition in both numeracy and combinatorics.
>
> I guess that backfired a bit here. But all the alternate methods for getting the same numbers is another good way of showing there’s never just one way of thinking about any problem.
That sounds right to me.
Amusingly, taking the case of 2 cups, there are two possibilites. Either both balls are in their cups, or neither is. So, the expectation is 1, yet 1 is the answer you’ll never get. Isn’t probability fun!
With 3 cups, there are 6 orders. Of 6 cases, in one we get 3 balls correct; in 3 more we get 1 ball correct. The average is (3 + 3) / 6 = 1. Odd that it matches the case for 2 balls, but odd things can happen near the endpoints.
For 4 balls, we get a total of 18 hits in 24 cases, so the probability drops. I guess it climbs from there?
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### Author:  [@Learjeff](https://boards.straightdope.com/u/Learjeff)
#### Post date: [April 22, 2013, 8:20pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/33 "2013-04-22T20:20:33Z")
counting failure. For 4 balls, we get 24 of 24. The number of hits always matches the number of permutations exactly, and the answer is (as The Great Antibob says) 1.
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [April 22, 2013, 8:22pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/34 "2013-04-22T20:22:26Z")
Here’s one of my favorites:
You roll six standard six-sided dice, and you count the number of unique faces you get. So if you roll (5, 6, 1, 5, 2, 5), that’s four, and if you roll (1, 3, 6, 5, 2, 2), that’s five.
What’s the probability that you get four?
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### Author:  [@Learjeff](https://boards.straightdope.com/u/Learjeff)
#### Post date: [April 22, 2013, 8:23pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/35 "2013-04-22T20:23:47Z")
So the expected number of matches is 1 with _any_ number of balls. It’s not asymptotic, as I understood Bob’s comment to mean.
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### Author:  [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)
#### Post date: [April 22, 2013, 8:28pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/36 "2013-04-22T20:28:03Z")
> [@Learjeff](#):
>
> For 4 balls, we get a total of 18 hits in 24 cases, so the probability drops. I guess it climbs from there?
No, it’s 24 hits in 24 cases. The expected value remains one. It’s always one, regardless of how many pairs of cups and balls there are.
eta: ninja’d by Learjeff himself.
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### Author:  [@President\_Johnny\_Gentle](https://boards.straightdope.com/u/President_Johnny_Gentle)
#### Post date: [April 22, 2013, 9:38pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/37 "2013-04-22T21:38:20Z")
> [@ultrafilter](#):
>
> Here’s one of my favorites:
>
> You roll six standard six-sided dice, and you count the number of unique faces you get. So if you roll (5, 6, 1, 5, 2, 5), that’s four, and if you roll (1, 3, 6, 5, 2, 2), that’s five.
>
> What’s the probability that you get four?
I’m coming up with 23400/46656 = 325/648, or about 50.2%, but I’m not coming up with a clever way of counting the answers.
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### Author:  [@Cartoonacy](https://boards.straightdope.com/u/Cartoonacy)
#### Post date: [April 22, 2013, 10:08pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/38 "2013-04-22T22:08:37Z")
> [@Little\_Nemo](#):
>
> No, it’s 24 hits in 24 cases. The expected value remains one. It’s always one, regardless of how many pairs of cups and balls there are.
I didn’t get what you were saying until I tried listing all the permutations. A “hit” is when a ball lands in a cup with the same number. So if each ball lands in the same-numbered cup, that counts as four hits, not one. Now it makes sense.
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### Author:  [@Jragon](https://boards.straightdope.com/u/Jragon)
#### Post date: [April 22, 2013, 11:09pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/39 "2013-04-22T23:09:53Z")
> [@ultrafilter](#):
>
> Here’s one of my favorites:
>
> You roll six standard six-sided dice, and you count the number of unique faces you get. So if you roll (5, 6, 1, 5, 2, 5), that’s four, and if you roll (1, 3, 6, 5, 2, 2), that’s five.
>
> What’s the probability that you get four?
Hmm, that’s a tricky one. I’m not going to hash it out now, but it strikes me that this is **similar** to a scenario without replacement, but couched in a medium with replacement, if that makes sense.
In that way, the chance of getting one unique is 1, in that your first roll is always unique, the chance of getting 6 uniques is 1 \* 5/6 \* 4/6 \* 3/6 \* 2/6 \* 1/6 = 6!/(6^6). The tricky part of any number in “the middle” (2-5 uniques) is that you need to explicitly count out the probabilities of getting a further unique. The other thing that’s going to make it tricky, and the calculation messier, is the branching nature – that is, if you fail on roll 4 to get your fourth unique, you can still succeed on 5. However, if you succeed on 5 you must fail on 6, but if you fail on 5 (and so on).
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [April 22, 2013, 11:17pm UTC](https://boards.straightdope.com/t/a-simple-probability-question/656348/40 "2013-04-22T23:17:41Z")
Might want to check your math on getting one unique.
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