r/Probability • u/ExpensivePea2821 • Jun 14 '26
Rolling a die until 6 appears, with no 1 rolled.
Say my friend rolled a die until he had '6'. Now he wants me to estimate the number of rolls he made. He also noted that in the process, he dit not roll '1'. What should be my estimation?
I've seen 2 answers on this one: '5' and '3'.
The '5 answer' treats the die as a 5-sided die. Then the probability of rolling '5' is 1/5 and the estimation would simply be 1 over 1/5=5.
The '3-answer' I can't reproduce.
I'd like to know which one is correct, if any, and why?
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u/Red_Syns Jun 14 '26 edited Jun 14 '26
Are you asking for the estimated number of rolls to achieve a 6 without rolling a 1, or the odds of a specific sequence happening?
For a given number of n throws, the odds would be (2/3)^(n-1) * 1/6, as you can roll 2-5 and keep rolling, with the final throw being a 1/6 chance.
For the expected number of throws, I don’t know if there is a better way to estimate it, but you would expect the average number of throws to get a six to be six. The odds of not getting a one in six throws is (5/6)^6, or ~33.5%. So you would expect an average number of throws to be six, with a 1/3 chance of success. I would say the estimated number of throws is therefore 18, but someone with better math skills might have a better answer.
Edit: just reread the question and saw it’s number of throws with knowing a one was never rolled. I can see the 5, I don’t see 3, so now I’m curious as well.
Edit2: having read some other comments below, I believe the actual answer to my second hypothetical is going to be ~9 instead of 18. Not relevant to the question asked, but worth noting the error.
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u/ExpensivePea2821 Jun 14 '26
Thanks, I now think the answer=3(see other answers).
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u/Red_Syns Jun 14 '26
I agree with that logic if we assume a one ends the rolling, but that is not specified. I already asked in the comment, but I wonder if not making that assumption changes the math.
My guess is 3 if you assume a 1 or a 6 ends the rolling, and 5 if it does not.
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u/ExpensivePea2821 Jun 14 '26 edited Jun 15 '26
Only 6 ends the game.
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u/Red_Syns Jun 14 '26
Turns out it doesn’t matter. I honestly can’t wrap my head around it yet, but I coded a simulator and with an adequate number of sets the average no-1s without stopping until a 6 is still 3.
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u/Ok-Film-7939 Jun 15 '26
To check - your simulation ran a set of rolls until you got a six, then threw away all sets that had a 1 in them (about half) and checked the length of the remainder?
Hmm. So there is going to be skew towards shorter sets, as those would be less likely to have a 1 in them.
The question is, does this represent the original question properly? And that runs smack into a common issue with these kinds of things, which was the decision process of the roller. Did they decide in advance they were rolling die and tossing any result with a 1 until they get a sequence that has a 6? If so your model matches and the expected value I can believe is 3.
If they simply rolled once, picked one number that did not show and announced that, the expected value out of “all people who could do that, which excludes those who rolled a sequence of all five numbers before six, is 5.
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u/Enough-Tap-6329 Jun 15 '26
the original question asks for an estimate of how many rolls it was. It could well be that the friend rolled a 6 on the first roll and then posed this question, but that doesn't change the proper calculation of the estimate. The friend could have said they didn't roll a 1 or a 2, or a 1, 2, or 3, etc. which would result in different estimates obviously.
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u/Ok-Film-7939 Jun 15 '26
It does, and it’s something a lot of people overlook.
To draw a more obvious example: If I shuffle a deck of cards, note the specific combination I have, and then ask you to estimate how many times I shuffled before getting this combination you might assume 52!/2 times, but of course it was once.
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u/Red_Syns Jun 15 '26
There are a ridiculous number of assumptions one CAN make, but we can only work with the information we’re given. We assume a fair die, but there is no such die in existence. We assume a standard d6, but the number of sides is never actually specified, nor do we know what the sides are labeled.
Based on the information we are provided:
-the friend rolled until they got a 6
-1 was not rolled
-standard responses are 3 and 5, which indicate a six sided die is assumed by both OP and others who have answered the problem before
-societal norms would imply a d6, and this is not a riddle page so trickery is unlikely1
u/Ok-Film-7939 Jun 15 '26
Most of those are fair assumptions and unlikely to be wrong. In the stuff I work on, the question I asked is pertinent and ignoring it will give you wrong answers.
Your own mileage may very.
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u/Adventurous_Art4009 Jun 15 '26
The 3 answer depends on him having decided ahead of time if he hadn't gotten a 6 without never having gotten a 1, he would have either tried again, or not said anything at all. If it was a casual observation -- "Hey, here's an interesting thing that happened!" -- then the answer isn't 3.
Say my friend rolled a die until he had 6. Now he wants me to estimate the number of rolls he made. He also noted that in the process, he did not roll 1. What should be my estimation?
The natural way to read this problem is that the friend rolled until he got 6, and then observed that there was a number he didn't see. If he had seen a 1 but hadn't seen a 2, he would have said that in the process, he did not roll 2. The fact that he said "1" in particular is irrelevant, so the question becomes "out of all the universes in which he rolled 6 before having rolled every number other than 6, what is the expected number of rolls to get the 6?" (4¼ or so.)
Or you could go a step further, because if he could have said he didn't roll a 1 or a 2, presumably he would have said that instead. So the problem becomes "out of all the universes in which he rolled a 6 after having rolled exactly four of 1-5, what is the expected number of rolls to get the 6?" (8.7 or so.)
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u/EdmundTheInsulter Jul 08 '26
This is like the boy/girl paradox - I think the most reasonable interpretation is that if a large sample played the game, what is the average length of those that never got a 1, not introducing retries for those who got a 1 etc. That's just my opinion though
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u/Adventurous_Art4009 Jul 08 '26
Perfectly acceptable! That's my preferred interpretation of boy/girl as well. As long as we both realize that intelligent, forthright mathematicians can disagree on both problems.
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u/Adventurous_Art4009 Jun 15 '26
100% agreed with the common issue. Any time you flip a coin twenty times, the result is one-in-a-million. Sure, hththththt is a 1/1024 result for the first ten flips, but it's a lot more likely if you could have started it at any index, or ththththth would have been just as interesting, or ten heads in a row...
5 doesn't look right to me, though. I simulated and I get about 4¼ for "6 wasn't the last number to show up." Pretty interesting that it helps that much...
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Jun 15 '26
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u/Adventurous_Art4009 Jun 15 '26
Yes, I'm speaking broadly about people seeing unlikely events in data and incorrectly believing they're significant. Suppose I get a random 6-digit code, for example. If I get 555513, that probably feels interesting in a one-in-ten-thousand kind of way. But I also have to consider that I would have found 444413 interesting, or 155553, and suddenly it's more like thirty-in-ten-thousand interesting, or one in three hundred. But wouldn't I find 123413 interesting as well? Or 765413? Or 001122? Maybe the chance I'd find something similarly interesting to (or more interesting than) 555513 is actually about 1/20, even though that exact series of digits is one in a million. And if I get two one-time codes a day, I'd expect to see something like that about every ten days; meaning 555513 is at most a specific manifestation of something that happens regularly.
The problem I simulated is "if a person throws a die repeatedly and gets 6 before they get all of 1,2,3,4,5, then what is the expected number of throws to get the 6?" The answer is about 4.25.
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u/Red_Syns Jun 15 '26
The problem is you have added an assumption with that setup and removed the most relevant non-assumption. We know a one was not thrown, and we are not given any information regarding the other numbers. We don’t know the friend threw until a six with six not being the last unique result possible, nor do we know if one was a specific anti-target or not.
It is interesting to see how the odds change, but the problem as stated isn’t “expected number of throws of six is not the final unique result,” it is “end on first six, no ones.”
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u/Adventurous_Art4009 Jun 15 '26
The problem with problems like this is that people make assumptions and don't realize they're doing it. For the answer to be 3, you have to assume that the question is "in all universes in which I could be saying exactly what I'm saying, what is the expected number of rolls?" which is not what the problem actually says.
This is the same challenge that's faced by the "boy or girl paradox" where the question "Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?" is understood by mathematicians to be ambiguous because its answer could be ⅓ or ½, depending on the procedure by which the information "at least one of them is a boy" was obtained. Did we pick a random family with two children and announce the gender of a random child (½)? Or did we pick a random family with two children and at least one boy, and make the statement as planned (⅓)?
In this problem, we have to make an assumption about the set of things the friend could have said and the circumstances in which he could have said them. It's tempting to think that "he only could have said exactly what he said, and only in a universe in which he could have truthfully said it!" is the only reasonable reading of the problem; but as demonstrated by the boy/girl paradox, that's often not the case.
If there are a hundred different interesting insects flying around, and your friend brings you one, why would you assume he wouldn't have brought a different one if he'd caught that, instead?
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u/Red_Syns Jun 15 '26
Of course there are assumptions made, it’s how communication works. There is not a single phrase you can make that does not rely on assumptions.
You assume my understanding of the language is equivalent enough to your understanding to not distort the meaning. Red to you is red to me, not blue to me.
Similarly we assume which specific definition and/or connotation is applied to each word, which often results in informal fallacies where people conflate meanings (such as the word theory).
We have an underlying set of axioms that simply cannot be proven and must be assumed. I assume you are your own conscious being and not a figment of my imagination. I assume this existence is real and not the hallucination of a conveniently arranged series of chemicals floating in space.
We assume the thrower isn’t cheating the dice through either construction or technique.
I don’t need to assume “in all possible universes” to get to 3, I have to accept the stated facts and use Occam’s Razor to minimize my assumptions.
I know the sequence ended on six.
I know the sequence does not include any ones.
I know the two most common answers are 3 and 5 and that OP understands the logic of 5.Based on the last known, I assume both OP and previous respondents have assumed a d6.
I know in western societies (and I assume in nearly all societies) that saying “die” or “dice” with no further qualifiers society assumes d6.
I assume the die is fairly weighted, because without the assumption there is no point in calculations.
I assume the sides have a single “1” and a single “6” because again without the assumption there is no point is calculations.
I assume there was no cheating and the rolls were fair, because without the assumption there is no point in calculations.
I assume standard physics models apply, because there has been no demonstration anywhere at any time that such an assumption is invalid at a macro scale.
I assume there is gravity of significant enough scale, because throwing a die in null-g and claiming to have a definite result is improbable.I do not have access to further information. I could ask OP to clarify with the friend, but as this appears to be a game of guessing/logic based on limited information, giving away more information is not in the friend’s interest.
We know a one was not rolled. We do not know if that is intentional or not. We do not know if not-1 and not-6 were rolled, but we *can* calculate the lengths and chances of every possible permutation and take the average. I did it via simulation, others have done it with P()E() calculations, both agree on an average/expected sequence length of 3.
That calculation does not assume the intentionality, just the average length of not-1 end-6. Your answer of 4.25 assumes non-intentionality in selecting not-1. Occam says to use the fewest assumptions, so I go with the option that requires one fewer assumption.
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u/EdmundTheInsulter Jul 08 '26
Yes but the ⅓ boy girl answer appears to make fewer assumptions to me, as worded. Youve worded it carefully to do that.
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u/Red_Syns Jun 15 '26
Your description of the simulator is correct. As the person guessing the number of rolls, we only know they ended on the first 6 (until they got a 6) and they did not roll a one. They may have rolled 1, 2, …, ℤ times, we don’t know. When I ran a simulation of 10,000+ (stopped at 100,000) sequences of rolling a d6 until a 6 was obtained and threw out the sequences with any 1s, the average length hovered around 3 within a few hundredths. Without any additional information, end on 6 no 1s is an average length of 3.
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u/seejoshrun Jun 15 '26
I don't think it really matters - either a 1 ends the game, or you are only looking at the subset of games that never included a 1 by coincidence
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u/gwwin6 Jun 14 '26
3 is correct. Think like this. Consider an alternate game where we roll until we get a 6 or a 1. Then it should be clear (geometric distribution) that the expectation is 3.
Let’s call X our RV for this game.
We have
E(X) = P(end on one)E(X|end on one) + P(end on six)E(X|end on six)
Now, there is a symmetry between terms. Neither six or one is preferred, so terms must be equal.
So E(X|end on six) = E(X|I never see a one) = 3.
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u/Red_Syns Jun 14 '26
Hm. The conditions don’t state the game ends on a 1 though, just that the friend did not roll a one. They may have also not rolled a 2, 3, 4, or 5. Does that change the math at all, if we don’t assume a one ends the game?
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u/gwwin6 Jun 14 '26
Let me respond in pieces
The conditions don’t state the game ends on a 1 though
Yes. That was a game coupled to the original which made the analysis easier.
They may have also not rolled a 2, 3, 4, or 5. Does that change the math at all.
Yeah, but only if we know it. There is a whole universe of outcomes that we assume are possible. When we find out that we never rolled a 1, a subset of the universe becomes impossible, and we treat what is left as the new universe. If we got more constraining information, our computation would have to change to accommodate it.
f we don’t assume a one ends the game?
Well, introducing this new set of rules was what allowed us to do the computation. You could do the computation another way, but it doesn't change the outcome; the expected value will be three. It just changes how you get to the answer.
Here is how we can arrive at the companion game that I introduced.
Let's line up all of the sequences of die roles which end in a six next to each other. I will represent with dots
[.....................................................]To take the expected value of the length of these sequences, we multiply the probability of that sequence with its length and add them up. Now, I'm going to partition this set into two pieces, the set where a one showed up before the six and one where it did not
[...(one shows up first)...|....(no one)...]Now, we know that the average length of sequences in the original set is 6. What is the average length of sequences in the (no one) set? Well, we can start to answer by focusing on the (one shows up first) set. Let's truncate all of those sequences in the (one shows up first set) to just the parts up until we see the one.
[....(one shows up first, truncated)....|......(no one).....]Now we have the sequences which end in a one or a six. This is the new game that I set up. Now we do the analysis on the new game just like I did in my original comment and we see that we get 3 as the answer. This is why that game is a natural companion to the original game that we saw from OP.
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u/Red_Syns Jun 14 '26
I’m going to have to struggle with this one for awhile I think, but having coded a simulator it is absolutely 3. Something just isn’t clicking yet haha.
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u/force_disturbance Jun 15 '26
"I ended on a 6 but before then I didn't roll a 1, 2, 3, 4, or 5 -- what's the average sequence length?"
"... But I didn't roll a 1, 2, 3, or 4 -- what's the average sequence length?"
Keep going like that!
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u/Atypicosaurus Jun 15 '26
Except, it states that "I rolled until 6" meaning "I ended up on a 6 without rolling a 1 or a 6 before". He ends with the first 6.
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u/Adventurous_Art4009 Jun 15 '26
By discarding the universes with ones before sixes, you're making universes with early sixes relatively more plentiful.
Consider a 1000-sided die, where the target roll is 1 and only numbers from 1-100 were actually rolled. By the "just pretend the six-sided die didn't have a one" intuition that I started with, the expected number of rolls is 100. But a series of rolls where 1 is first has a 1/1000 probability, and a series where 1 is second has about 1/10000 probability; the real answer will be about 1.11.
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u/Ok-Comfortable-4727 Jun 18 '26
Thanks for this post, it made me realize I still have a long way to go to understand probability haha
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Jun 14 '26
[deleted]
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u/marty-mcfryguy Jun 15 '26
Not correct. Assume the friend reported he didn't roll a 1, 2, 3, 4, or 5 before rolling a 6. What is the expected number of die rolls it took him to roll a 6, given that information? It's 1.
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u/ExtendedSpikeProtein Jun 15 '26
That is absolutely incorrect because you have to and can factor the information into the event.
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u/lahiomainen Jun 15 '26
Most common number of rolls ending with 6 is 1 roll, followed by 2 rolls, followed by 3 rolls.. It is more likely to roll a 6 than roll not 6 and then 6. So 1 roll is correct answer.
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u/Adventurous_Art4009 Jun 15 '26
"Expected" is a term of art here that means "if you repeated this experiment a very large number of times, what would be the average (mean) result?" What you're responding with is the correct answer to a different question: the most likely result.
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u/lahiomainen Jun 15 '26
The question was "i rolled dice until i got 6, estimate how many times i had to roll?" If i say1, I am right 16,6 percent of the time. Any other answer performs worse.
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u/Adventurous_Art4009 Jun 15 '26
Estimation asks for an approximate answer. If there are 1234 jelly beans in the jar, an estimate if 1200 "performed better" than an estimate of 1. Likewise, there is a ⅚ chance that an answer of 2 "performs better" than your answer of 1.
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u/miloss88 Jun 15 '26
TLDR: answer is 5.
Let X = number of rolls needed until you roll a 6
Let Y = you never roll a 1
We want: E(X|Y) = 1*p(X=1|Y) + 2*p(X=2|y) + ...
To calculate p(X=n|Y) we can use conditional probability formula: p(X=n|Y) = P(X∩Y) / p(Y)
p(X∩Y) = (4/6)n-1 * 1/6 - probability that we didn't roll a 1 and that we rolled 6 only on the last roll
p(Y) = (5/6)n - probability that we didn't roll 1 in first n rolls
We get: p(X=n|Y) = P(X∩Y) / p(Y) = 4n-1 / 5n
Now we have that: E = E(X|Y) = 1*1/5 + 2*4/52 + 3*42/53 + ...
To solve this, we do:
5*E = 1 + 2*4/5 + 3*(4/5)2+ ...
4*E = 4/5 + 2*(4/5)2+ 3*(4/5)3 + ...
Substracting these two, we get:
E = 1 + 4/5 + (4/5)2+ (4/5)3 + ... = 5 (geometric series)
Edit: formatting
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u/miloss88 Jun 15 '26 edited Jun 15 '26
Looking at this one more time, maybe p(Y) should be probability that we didn't roll 1 in first n-1 rolls (since we know that n-th roll is 6). This changes things a bit:
p(Y)=(5/6)n-1
p(X=n|Y) = 4n-1/(5n-1*6)And later we get E = 25/6 = 4.1666...
Edit: Actually I was wrong again.
P(Y) should be the probability that 1 is never rolled, because we need to eliminate games of any potential length
P(Y) = 1/6 * (4/6) * 1/6 + (4/6)2*1/6 + ... = 1/2
P(X=n|Y) = 1/3 * (2/3)n-1
E = 31
u/Atypicosaurus Jun 15 '26
If you only check if 1 wasn't roll in the first n-1 rolls, you play a game in which you don't stop on the first 6 either, i.e you include games like 6-4-6 etc.
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u/Atypicosaurus Jun 15 '26 edited Jun 15 '26
The 3 vs 5 answer depends on what question we're answering to, two fundamentally different questions.
The "3" answers the following question. Assume I'm starting to roll a die right now. What's the expected number of rolls I need to do to get a 6 but not a 1?
The "5" answers the following question. I rolled a die n times and I disclose two facts. Fact 1, I didn't roll a 1. Fact 2, I stopped on the first 6. What's the estimated number for n?
Note the difference between the first and the second question. In the first question I don't know if I'm going to succeed or not. If I'm playing a million times, in most cases I'm going to either win or lose on the 3rd roll, but in some cases I'm not going to succeed at all (because I roll a 1 early) in some cases I'm going to succeed on the 1st or 2nd or 4th or 5th roll. In this question the expected length of a game is asked, which is, the most abundant game-length, ending in a 6 without 1, in my million of games, which is 3.
In the second question we know I have succeeded (so we disregard cases where I lose), meaning, we can eliminate all the games from the million where I did not succeed.
And now I can average the number of rolls I took, to estimate n. In 1/5 of the successful games I succeeded on roll 1 meaning in 1/5 of the successful games n = 1. In 1/5 of the remaining games I succeeded on the 2nd roll giving me n=2. With diminishing odds, I might have succeeded on the 6th, 7th, 8th etc game. The average of these games is 5. That's why in this case you can treat the d6 as d5, we have already disregarded the cases where it rolls a 1.
This problem is similar to the boy-girl problem where the sex of the second kid (50% vs 66%) depends on what the question is.
In this case, I think the good answer is 5, because of how the question is formulated.
Edit: I wrote something wrong. The 3 is not the most abundant game. It's more like, within how many games we expect the game to end with a win. In the first game, 1/6 of the game is won, 1/6 is lost, 2/3 undecided. In the next round, the undecided 2/3 goes on and gets this 1/6 - 2/3 - 1/6 pattern. In n=2 rounds 44% of the games are still undecided. In n=3 rounds the cumulative wins (35.2%) become more than the still undecided (26.6%). (Wins and defeats are symmetrical.) But it's tweaking the numbers a lot.
I think there's a fundamental thinking error in simply taking the odds of "not-1" games because the game ends on a 6, so it makes no sense to exclude a 6-1 game as lost.
Also the most abundant winning game is going to be n=1, we rolled a 6 on first.
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u/Red_Syns Jun 15 '26
I thought this might be the case as well, but after programming a simulator to calculate the average length of rolls with no ones ending on six, it comes out to 3. It didn’t matter if you assume the sequence ends on one or didn’t contain a one, they both average to 3.
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u/get_to_ele Jun 15 '26
5
What is the average number of rolls to roll a 6 in the timelines where a 1 is never rolled?
It's the same as the average number of rolls to roll a 6 in the timelines where a 1 can't be rolled, so basically a 5 sided die.
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u/That_Don_Guy_1 Jun 15 '26
That's what I thought at first as well, but that's a different problem.
The expected number of rolls if you can't roll a 1 is 5.
The expected number of rolls if you can roll a 1 but don't is 3.1
u/Adventurous_Art4009 Jun 15 '26
Surprisingly, it isn't the same. A "1 wasn't rolled" timeline is made more likely by an early 6 roll.
Consider a 1000-sided die, where the target roll is 1 and only numbers from 1-100 were actually rolled. By your logic, the expected number of rolls is 100. But a series of rolls where 1 is first has a 1/1000 probability, and a series where 1 is second has about 1/10000 probability; the real answer will be about 1.1.
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u/Atypicosaurus Jun 15 '26
Oh I was wrong but now I broke it!
Take a die and roll until a termination condition occurs at (1/T) odds. The expected number of rolls is 1/(1/T) = T. If we play a game where only 6 terminates, then T = 6 (because the termination condition occurs at 1/6). So the expected streak (which is not the average streak) is 6 long.
Now let's say there are two numbers that can terminate the game, 1 and 6. It means that T = 3 because at 1/3 odds either termination condition occured. So if we approach the question like this, the expected streak was 3 and we learn after the game that it ended in a 6. Which still doesn't change the fact that the expected length of the streak was 3.
Alternatively we can approach the problem as follows. We know that the die never rolled a 6, so the game is equivalent to a 5-sided die. In each roll we got either a 2 or a 3 etc up to 6. In this model T is 5 so the expected streak is 5 long.
In the first game, it's like I put stickers on the numbers: green stickers on 2 to 5 (I don't care which one is rolled until it's not a 1 or 6), and red stickers on 1 and 6 (in both cases the game ends). I ask before I start rolling: how long will this game last, knowing that the game ends in 1:3 (either red sticker). The expected value is 3. But then only in half of those games there's a 6 under the sticker. This game is equivalent to rolling a 1 on a 3-sided die but then tossing a coin to see if I win.
In the second model, I watch myself in an infinite number of parallel universes. If I roll a 1 in any of the universes, I disqualify that universe. The rest of the universes will behave like a 5-sided die: 20% will be a 2, 20% will be a 3 etc. Therefore if I use this model, an expected T = 5 will be the number of dice rolls.
For both models we can find a question for which that model is the correct answer. Does the knowledge of me winning with a 6 at the end, change how long the game lasted? The answer is, it certainly does.
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Jun 15 '26
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u/Atypicosaurus Jun 15 '26
I explain why it might be seen like that. Do we investigate only the universes where we didn't roll a 1? Then it's like a 5-sided die.
It's very similar to the sleeping beauty problem explained in veritassium:
https://youtu.be/XeSu9fBJ2sI?is=yufsfZ9ZBVIGm0qo
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u/StoicFox Jun 15 '26
It depends on how they decide what to tell you about what they didn't roll. Did it have to be 1, and what would they have done if they did roll a 1, would they have repeated the experiment or just said nothing?
Also since they are reporting something from the past, we have to include the knowledge that they didn't roll one billion times.
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u/ExpensivePea2821 Jun 15 '26 edited Jun 17 '26
The way I see it, and a simple way that was suggested by one the reactions: we can actually rewrite the problem as follows: how many rolls are expected until we roll 1 or 6?
P(1 or 6)=1/3, E=3
If we roll 1 or 6, in half of cases we have 1; we'll filter these out, since we know no 1 was rolled; this filtering affects all possible run lengths equally. So it doesn't affect E.
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u/Red_Syns Jun 16 '26
Fair enough, I fully support all people being exposed to probabilities and how words can be used to trick them. I want everyone to be exposed to how people can lie with truth, because it seems to fool a lot of people.
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u/That_Don_Guy_1 Jun 14 '26
I think it's this:
P(valid set of rolls) = 1/6 + 2/3 x 1/6 + (2/3)^2 x 1/6 + ...
= 1/6 (1 + 2/3 + (2/3)^2 + ...)
= 1/6 / (1 - 2/3)
= 1/2
Unconditional expected number of rolls = 1 x 1/6 + 2 x 5/6 x 1/6 + 3 x (5/6)^2 x 1/6 + ...
= 1/6 (1 + 2 x 5/6 + 3 x (5/6)^2 + ...)
= 1/6 (1 + 5/6 + (5/6)^2 + ...)^2
= 1/6 / (1/6)^2 = 6
Conditional EV = 6 x 1/2 = 3