[Guest]

I've decided that rolling cube-shaped dice is too old-fashioned, and have instead painted the numbers 1 to 6 evenly around a wheel. Every time a dice roll is required I toss a coin. If it lands on heads, I turn the wheel one-sixth of a turn clockwise, if it lands on tails, I turn the wheel one-sixth of a turn anticlockwise. I then call out the number at the top. This gives me a fun and useful random number generator. Before commencing I will give the wheel a good spin so the number I start from will be random.

What is the average number of turns it will take to produce a 6?

EDIT (for clarity): Each turn consists of a coin toss followed by 1/6 rotation depending on the coin toss. The spin happens only once, before we start making turns, to give us a random starting point. The result of the spin doesn't count as a turn in itself, so if the spin lands on 6 we haven't done a 6 in zero turns.

[Prime]

If I understood the conditions right. Each turn consists of a random spin + one coin toss.

Then ...

It takes on average 6 turns to produce "6". It is still a random pick of one from 6.

[Guest]

If I understood the conditions right. Each turn consists of a random spin + one coin toss.

Then ...

It takes on average 6 turns to produce "6". It is still a random pick of one from 6.

No sorry, perhaps I wasn't clear, the spin is just to create a random place to start from. No spinning after that. I'll amend the OP.

[Guest]

I've decided that rolling cube-shaped dice is too old-fashioned, and have instead painted the numbers 1 to 6 evenly around a wheel. Every time a dice roll is required I toss a coin. If it lands on heads, I turn the wheel one-sixth of a turn clockwise, if it lands on tails, I turn the wheel one-sixth of a turn anticlockwise. I then call out the number at the top. This gives me a fun and useful random number generator. Before commencing I will give the wheel a good spin so the number I start from will be random.

What is the average number of turns it will take to produce a 6?

EDIT (for clarity): Each turn consists of a coin toss followed by 1/6 rotation depending on the coin toss. The spin happens only once, before we start making turns, to give us a random starting point. The result of the spin doesn't count as a turn in itself, so if the spin lands on 6 we haven't done a 6 in zero turns.

Let me know if the think path is correct

if n =1, if means that we either (a) have 5 on the top, and we get a tail, 1/6 anticlockwise will get '6' or (b) have 1 on the top, and we get a head, 1/6 clockwise will get 6.

so for n = 1, p(n=1) = 1/6*1/2 + 1/6*1/2 = 1/6

for n =2, is either at (a), have 4 on top, 2 times get a tail, and 1/6 anticlockwise twice, will get us '6', (b) have 2 on the top, 2 time get head, 1/6 clcokwise twice, or ©, have 6 on top, either head then tail or tail then head will bring us back to 6.

hence, n = 2, p(n=2) = 1/6 * 1/2* 1/2 + 1/6 * 1/2 * 1/2 + 1/6 * (1/2*1/2 + 1/2 * 1/2)

p(n=2) = 1/24 + 1/24 + 1/12 = 1/6

and so on...... ?

[Guest]

This gets complicated quickly because of the fact that you can go around either direction. Let me see if there's an easier way to think about it.

[Guest]

Let me know if the think path is correct

if n =1, if means that we either (a) have 5 on the top, and we get a tail, 1/6 anticlockwise will get '6' or (b) have 1 on the top, and we get a head, 1/6 clockwise will get 6.

so for n = 1, p(n=1) = 1/6*1/2 + 1/6*1/2 = 1/6

for n =2, is either at (a), have 4 on top, 2 times get a tail, and 1/6 anticlockwise twice, will get us '6', (b) have 2 on the top, 2 time get head, 1/6 clcokwise twice, or ©, have 6 on top, either head then tail or tail then head will bring us back to 6.

hence, n = 2, p(n=2) = 1/6 * 1/2* 1/2 + 1/6 * 1/2 * 1/2 + 1/6 * (1/2*1/2 + 1/2 * 1/2)

p(n=2) = 1/24 + 1/24 + 1/12 = 1/6

and so on...... ?

You're not wrong, but that looks like hard work.

[Guest]

You're not wrong, but that looks like hard work.

I have counted to n=4

p = 1/6 no matter what n is ??? :huh

[Guest]

I have counted to n=4

p = 1/6 no matter what n is ??? :huh

I think it changes at n=3:

either:

1 to 2 to 1 to 6

3 to 2 to 1 to 6

3 to 4 to 5 to 6

5 to 4 to 5 to 6

(1/6*1/2*1/2*1/2) * 4 = 1/12

[Guest]

6 times.

From the wheel, there is a 2/6 (1/3) chance that 6 is next to the current number on the wheel.

Then there is a 1/2 chance the coin will throw you the right way.

1/3 * 1/2 = 1/6

Therefore on average you need 6 spins.

[Guest]

6 times.

From the wheel, there is a 2/6 (1/3) chance that 6 is next to the current number on the wheel.

Then there is a 1/2 chance the coin will throw you the right way.

1/3 * 1/2 = 1/6

Therefore on average you need 6 spins.

The inspiration for this puzzle was to provide a counterexample to Bonanova's assertion (

here) that if you have 6 equally probable events you'd have to wait, on average 6 times for the first appearance of any one of them. Seems pretty obvious, and it may be that I have done my sums wrong, but I don't think that's the case here.

[Guest]

Alright, I think the easiest way to calculate this will be to break it down by the original spin.

Obviously, if 6 is spun, 0 flips are required

We can make it a little simpler: 5 will be the same as 1, and 4 will be the same as 2. Even better, 2 will be the same as 1 half the time, and the same as 3 the other half. This leaves basically two calculations: expected flips for a spin of 1, and expected flips for a spin of 3.

This will still be a lot of work, but I'll get it started on 3.

P(3 spun)=1/6

To get to 6 you need to go either 4-5-6 or 2-1-6. Basically, this means that you need a string of flips which looks like this:

[H=2+T]H or [T=2+H]T. The brackets indicate that there could be any number of flips in there, as long as the difference is two. For instance, the following all land on 6: TTT, HHH, HHTHH, THTTT, THTHHTHHH, etc. Note that the number of flips must be odd.

The probability of getting to 6 in exactly n flips will be:

_(n-1)C_((n-1)/2+1)*.5ⁿ

Then double that because you can go either direction. And make sure that you don't double count when working with higher ns.

So yeah, it's still pretty hard.

[Guest]

So to start with you have a 1 in 6 chance of spinning a 6. Let's say you spin a 5. How do you resolve with the random nature of fliping a coin the fact that you could flip heads, tails, heads, tails, heads, etc an infinite number of times and combinations and never reach the 6 on the wheel? You could start at 5 and flip 4 (heads or tails depending on which way the wheel is numbered) in a row and be on 1 then start flipping the opposite side of the coin and get farther away from the 6 again.

[Guest]

The inspiration for this puzzle was to provide a counterexample to Bonanova's assertion (here) that if you have 6 equally probable events you'd have to wait, on average 6 times for the first appearance of any one of them. Seems pretty obvious, and it may be that I have done my sums wrong, but I don't think that's the case here.

I'm going to have to disagree with that one... because we have two different stages to the process (and one of them - the initial spin - only happens once) I don't think this is a valid counterexample. Once you've done the spin, the events are not equally probable, and so it may well take > 6 'moves' to get where you're heading.

A similar 'counterexample' would be: There are six different paths. At the end of each path is a number from one to six. It takes 5000 paces to get to the end of the path to find out which number is at the end.

You select which path you walk down by rolling a fair dice: what is the average number of steps you will take before you meet a six?

Then the answer is 30,000 steps... but I don't think its fair to say you had to do the thing 30,000 times before you got a six

[Guest]

Okay, I read the spoiler from the original author but if you flip a coin to determine direction of rotation and the coin having two sides has a probability of 1/2 for either heads or tails won't you be constantly going back and forth? I know that's not how coin flipping works in reality but it very easily could factor in to the idea of something having a 1/6 possibility of ever occurring. You're not just doing probability for the wheel and its numbers but you're also doing probability for clicking the wheel one way and then the other by way of a coin toss which will push you either way equally as often. We should all do it and publish our results and then find the average.

[Guest]

I think it changes at n=3:

either:

1 to 2 to 1 to 6

3 to 2 to 1 to 6

3 to 4 to 5 to 6

5 to 4 to 5 to 6

(1/6*1/2*1/2*1/2) * 4 = 1/12

If is 1: 1 to 2 to 1 to 6 (that's your answer) and 1 to 6 to 5 to 6 (this another one I found)

If is 3: as per you stated,

If is 5: 5 to 4 to 5 to 6 (that's your answer) and 5 to 6 to 1 to 6 (this another one I found)

But again those extra ones that I found are not really what you missed. Because we have already achieved 6 for n = 1 and it shouldn't continue already.

Ok, I need to check again my list.

[Guest]

OK, I'll answer all the posts on this page...

Chuck Rampart:

Just to be clear, if the spin lands on 6 we haven't got a 6 yet. Each turn is a coin toss then a move, then you read the number. The spin is just a randomised starting position, not a turn in itself.

As for your calculations, you started well.

TDRAdam:

Yes you could do an infinite number of turns and never get 6. But then you could roll a die infinitely many times and never get 6 too. That's all part of the fun.

armcie:

Let's be clear about what the spin does. If you were to repeat the process a large number of times, enter into it at some random point, and start counting from there, you'd be equally likely to start from any number. The spin performs the same function. It gives us a starting point as if we had entered into the process at some random stage.

Once you have done the spin, every number is as likely (at every stage) as every other number (although obviously the sequence of numbers is more restricted)

A similar 'counterexample' would be: There are six different paths. At the end of each path is a number from one to six. It takes 5000 paces to get to the end of the path to find out which number is at the end.

You select which path you walk down by rolling a fair dice: what is the average number of steps you will take before you meet a six?

Then the answer is 30,000 steps... but I don't think its fair to say you had to do the thing 30,000 times before you got a six

That wouldn't be a counterexample. On average you'd walk down 6 paths in order to get to the 6, exactly as Bonanova was suggesting. Not sure what you mean about doing it 30,000 times.

billy:

I look forward to your test results!

woon:

As we've established, there could be an unlimited number of turns before you get a 6. You might need to factor that into your workings.

[Guest]

The inspiration for this puzzle was to provide a counterexample to Bonanova's assertion (here) that if you have 6 equally probable events you'd have to wait, on average 6 times for the first appearance of any one of them. Seems pretty obvious, and it may be that I have done my sums wrong, but I don't think that's the case here.

I fail to see how this represents a counterexample. In Bonanova's example, the probability is equally distributed on each roll. In this scheme, after the die is rolled, the probability of a 6 on the next coin toss is 0 if the initial roll was 2, 3, 4, or 6.

As JC would have said: alea iacta est! ;-)

[Guest]

I fail to see how this represents a counterexample. In Bonanova's example, the probability is equally distributed on each roll. In this scheme, after the die is rolled, the probability of a 6 on the next coin toss is 0 if the initial roll was 2, 3, 4, or 6.

As JC would have said: alea iacta est! ;-)

I was refuting a specific line of reasoning, namely that if you rolled a die 6 times the expected number of appearances of a 6 would be 1. Therefore the expected number of rolls to produce a 6 would be 6.

In this case the initial condition remains true but the conclusion does not.

[EventHorizon]

Let w be the average number of flips it takes to get a 6 when a 6 is initially rolled.

Let x be the average number of flips it takes to get a 6 when a 5 or 1 is initially rolled.

Let y be the average number of flips it takes to get a 6 when a 4 or 2 is initially rolled.

Let z be the average number of flips it takes to get a 6 when a 3 is initially rolled.

After 1 flip after starting with 3 you will be on either 4 or 2,

z = 1 + y

After 1 flip after starting on 4 or 2, you will be on either 3 (50%) or in the set {1,5} (50%),

y = 1 + (x/2 + z/2)

y = 1 + x/2 + (1+y)/2

y/2 = 1 + x/2 + 1/2

y = 2 + x + 1

y = x+3

Similarly for x,

x = 1 + (0/2 + y/2)

x = 1 + y/2

x = 1 + (x+3)/2

x/2 = 1 + 3/2

x = 2 + 3 = 5

Similarly for w,

w = 1 + (x/2 + x/2)

w = 1 + x

w = 6

so we have

w = 6

x = 5

y = 8

z = 9

Since each number is equally likely as a starting point, we have (6+5+8+9+8+5)/6 = 41/6 = 6 + 5/6

If zero flips are required when 6 is initially rolled, then simply set w to 0 to get 35/6 = 5 + 5/6

Anyone see an error in logic?

Edited October 7, 2008 by EventHorizon

[EventHorizon]

if 6 is an "absorbing state" (eg, you cannot move off of it), then that makes w = 1. This makes the average equal to (1+5+8+9+8+5)/6 = 36/6 = 6.

[Guest]

Let w be the average number of flips it takes to get a 6 when a 6 is initially rolled.

Let x be the average number of flips it takes to get a 6 when a 5 or 1 is initially rolled.

Let y be the average number of flips it takes to get a 6 when a 4 or 2 is initially rolled.

Let z be the average number of flips it takes to get a 6 when a 3 is initially rolled.

After 1 flip after starting with 3 you will be on either 4 or 2,

z = 1 + y

After 1 flip after starting on 4 or 2, you will be on either 3 (50%) or in the set {1,5} (50%),

y = 1 + (x/2 + z/2)

y = 1 + x/2 + (1+y)/2

y/2 = 1 + x/2 + 1/2

y = 2 + x + 1

y = x+3

Similarly for x,

x = 1 + (0/2 + y/2)

x = 1 + y/2

x = 1 + (x+3)/2

x/2 = 1 + 3/2

x = 2 + 3 = 5

Similarly for w,

w = 1 + (x/2 + x/2)

w = 1 + x

w = 6

so we have

w = 6

x = 5

y = 8

z = 9

Since each number is equally likely as a starting point, we have (6+5+8+9+8+5)/6 = 41/6 = 6 + 5/6

If zero flips are required when 6 is initially rolled, then simply set w to 0 to get 35/6 = 5 + 5/6

Anyone see an error in logic?

I want to see an error in the logic, but I can't find it. I think you've got it, and I definitely think that's the easiest way to find it.

I had assumed you could break it down a little, but would eventually need to actually go through some infinite series of Bernoulli trials.

Well played.

[Prime]

I was refuting a specific line of reasoning, namely that if you rolled a die 6 times the expected number of appearances of a 6 would be 1. Therefore the expected number of rolls to produce a 6 would be 6.

In this case the initial condition remains true but the conclusion does not.

Actually, on that one point I agree with Bonanova.

If an event occurs on average n times out of T, then you can say that the average number of trials for an event to occur first time is T/n.

The disagreement that I had with Bonanova in another third topic (his "dicey" problem) was about the assertion that if the average number of trials for an event to occur is n, then the probability for an event to occur sometime during the first n trials is 50%.

My disagreement with Bonanova in the average number of trials to roll "6" topic is essentially, that I believe Bonanova's proof to be recursive (circlular). It first assumes what the average is then calculates it based on the same assumption.

Your problem here is more interesting. It is a variation on a random walk problem. I agree with EventHorizon's solution (above), which is economical and precise.

Edited October 7, 2008 by Prime

[Guest]

The disagreement that I had with Bonanova in another third topic (his "dicey" problem) was about the assertion that if the average number of trials for an event to occur is n, then the probability for an event to occur sometime during the first n trials is 50%.

Oh. I don't remember that one, but I fell into sort of the opposite trap, computing the median (incorrectly anyway), and trying to call it the average.

It is easy to see that this is false by considering the simplest of all examples: tossing a fair coin. On average, it takes 2 tosses for a "heads" to occur, but at least one "heads" occurs with 75% probability in 2 tosses.

[Guest]

Ok, so this is kind of cheating and by no means eloquent, but it's been to long since I took Probability and Stats... so I brute forced this one by simulating it in an excel spreadsheet. Here's how I basically did it:

1) Random number generator 1-6 to represent the first spin

2)Random number generator 1-2 to represent heads or tails for each toss

3) (this was the fun part) Series of IF statements to either add or subtract from the number,to keep the numbers within 1-6 and lastly to recognize when six has been rolled and record how many times it took

4) average across 40,000 trials

A seperate sample size calculation showed me that roughly 16,000 trials would get me within +- 1% of the answer, with a 99% confidence interval.

From all this, the average I consistently get to the nearest hundreth is 3.5

I'm going to work on trying to get a more detailed description of my excel formulas so anyone can replicate it, since unfortunately it seems like I can't upload an excel file. Any ideas on how to share this info?

Let me know what you think!

[unreality]

Interesting that 3.5 is the average roll value of a single 6-sided dice. Are you sure you got the if-statements to work recursively for as many trials to get 6? Just checking Interesting result

[Guest]

here are ten possible outcomes starting at 1

Since I don't remember if it says this in the puzzle let H = +1 and T = -1

1 is the Start Point:

1, H2, H3, H4, H5, H6 = 5

1, H2, H3, H4, H5, T4, T3, T2, T1, T6 = 9

1, H2, H3, H4, T3, T2, T1, T6 = 7

1, H2, H3, T2, T1, T6 = 5

1, H2, T1, T6 = 3

1, H2, H3, H4, H5, T4, H5, H6 = 7

1, H2, H3, H4, H5, T4, T3, H4, H5, H6 =9

1, H2, H3, H4, H5, T4, T3, T2, H3, H4, H5, H6 = 11

1, H2, H3, H4, H5, T4, T3, T2, T1, H2, H3, H4, H5, H6 =13

1, T6 = 1

so far ten variations have produced the number 7 as an average... but the variations are actually infinite so I'll try with a real world model because you can add and subtract one forever.

The next ten results are from me arbitrarily picking numbers and then flipping a coin.

3. t,h, h, t, t, h, t, t, h, h, h, t, h, h, h 15 flips

5. t, h, h 3

2. h, t, t, t, 4

6. t, h, 2

4. t, h, h, t, h, h, 6

1. t 1

4. h, t, t, t, h, h, h, t, h, t, t, t, t, h, t, h, t, t 18

5. h, 1

3. h, t, t, t, t 5

4. t, h, t, h, t, h, h, t, t, h, h, t, h, h 14

so in this manner, I got 6.9 which is close enough to 7 I think. Unfortunately I don't have a spinning wheel but I think I chose starting numbers randomly enough with assistance of a receipt and my own fleeting thoughts. I'll tell you what though 3's and 4's are a royal pain. LOL!

Is there an equation that presents 7 in any of the answers?