What is the expected number of rolls till a 1 comes up on a standard 6-sided die, conditional on only odd numbers coming up?
First, without the conditional, the answer would be 6 rolls. (This is the expected value of a Geometric distribution, if you want to get very technical, which of course we do.)
What about with the conditional? One way you might reason is like so: Only getting odd rolls is like having a 3-sided die, with just the 3 odd numbers: 1, 3, 5. So for the same reason the answer is 6 with with 6-sided die, it's 3 with this hypothetical 3-sided die.
I would've fallen for that but now believe it's wrong. We can discuss further in the comments.
This market resolves to whatever a proof plus a simulation both agree on.
Source: Stan Wagon
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For a given k >= 1, the probability p_k that there are exactly k rolls and all rolls are odd is 1/6 (1/3)^(k-1). The probability all rolls are odd is sum p_k = 1/4, so the expected number of rolls is 4 sum kp_k = 3/2.
Solution modified from https://www.caltechmathmeet.org/_files/ugd/006423_aa7143be4f534217bd7a089051d4e07f.pdf problem 4 which is easily seen to be equivalent.
I can't see a way that it isn't 3 rolls, even though that's very much the vote. What are the expected rolls for a 3 coming up, under the same conditional? And for a 5? The die doesn't know the difference between a 1, 3 or 5, and I can't see how it would be 1.5 rolls for all of them (which is the current market favourite) and still work out? What am I missing here?
Maybe I'm somehow not understanding the framing of the question - is there some reason that the odds of a 1 on each roll isn't 1/3rd, and each roll is independent? How else are you modeling this to get a different answer?
Ah, well I think I see a different answer, depending on what you mean by "conditional on only odd numbers coming up". If you mean you've got a standard 6-sided dice, but it's only got odd numbers on it (replace the 2 with a 1, the 4 with a 3, and the 6 with a 5 - or just blank them out, via the conditioning magic), then I think you - and any simulation - would find it takes on average 3 rolls to get a 1. I suppose this is what you would call "conditioning by subtraction".
You mean something else, I think - you're just using a full die with 1-6 on it, but starting the count again whenever there's an even number rolled (edited that - I'd wrongly said restart on odd!). Which I think is a different problem.
@CliveFreeman Yeah, I was confused about this for a quite a while. To simulate this, we roll the D6 until a 1 appears and count how many rolls that took. Except if an even number is rolled, start over. So the easiest way to have a sequence of odd rolls ending in 1 is to just roll the 1 off the bat, which I think is part of the intuition here.
@dreev OK well I wrote the code - it's simpler to code when you think of it as generating the sequence backwards. Each sample starts with a 1 - those are the sequences you're counting. Then keep generating rand(1,6) as long as you get a 3 or a 5. Each one of those adds one to the length of the sequence. If you get any other number, go round the sampling loop again.
And of course, sure enough, you get an average length of 1.5. Thinking of it backwards in this way also helped me with the proof - you're summing a geometric series with a ratio of 1/3 each time, so that's 1/1-p for p=1/3, which is just 3/2.
@dreev hmmm wait that's close to 2? It's gotta be something really close to 1 right? By far the most likely case is they just rolled 344, not that they rolled 820 then 344, isn't it? (the answer to the problem in the post is 3/2 I'm just offering intuition for the fact that the answer is so low)
@dreev Guess the answer generally is n/(n-k) where n is # of sides k is # of options other than the one you are going for thus 1000/999 in my question and 3/2 = 6/4 in the problem.
@NateWatson the answer has to be the same whoever happened to win so the answer is just the number of rolls needed to avoid hitting 820 in general which is 1/(999/1000).
To make these intuitive, it helps to imagine a plausible scenario where you learn exactly the information given in conditional, rather than just positing by fiat that it's the only information you have.
You are in a conference in a ballroom with 1000 people and at some point everyone is told to look under their chairs for a distinct number 1-1000 (you have 344). They raffle some prizes by rolling 3d10 to generate a number and calling it out the number. But at the start you hear the speaker say to the person rolling the dice "if you ever get a duplicate number just reroll".
You can't see the person rolling the dice because there's a piller blocking your view. The first number called is 820. Second called is your number, 344! As you're walking up to get your prize, what's your belief about the number of times the dice were rolled to generate the second number?
(conditional on only 344 and 820 being rolled, what is the expected number of rolls taken to get 344?)
@NateWatson Is the answer 1+1000/999?
(This is my 3rd attempt to submit this comment. Hope they don't all end up appearing!)
Notice that when a 1 is attained, when its previous number is 1, 2, 4, or 6, it means that the 1 was attained in just 1 roll, meaning there's a 2/3 probability of generating a 1 in just 1 roll. This is assuming that the rolls are reset back to 0 when a 2, 4, or 6 is attained. Therefore, the probability of having a sequence of length n is just 2/3((1/3)^(n-1)). The expected value, therefore, is just 2/3(sum(n * 1/3^(n-1) from n = 1 to inf), which can just be evaluated as (2/3)(9/4) = 3/2.
Simulation for C++ program is shown here: https://ideone.com/Cfa24x
I think there'll be a nice intuition pump for this by considering a more extreme version of the problem, but I haven't found it yet. I think @jacksonpolack has a good intuition thinking in terms of a game tree but that's not intuitive for me yet either.
the extreme version would be how long it takes to pick 1 out of 10, vs how long it takes to pick 1 out of 10B, conditioning on all rolls <= 10. In the first case it takes ten, but in the second case your first roll is almost always conditioned away. If your first roll is 1 (.1) then you get 1. If your first roll is 2-10 then you do another roll and p=1 - 10^-9 you are conditioned away. so it's almost always 1!
the 'game trees' correspond for both cases, but the nodes of the trees have different weights because nodes that are deeper are more likely to be dropped by future even numbers?