Coinroll.it - Dice rolling game | Instant bets | Off-the-chain | 1% House edge

[indianplayers]

Wow amazing how the 11+ losses are appearing a lot lately. I guess it's my luck. 

[GCInc.]

Wow amazing how the 11+ losses are appearing a lot lately. I guess it's my luck. 

What's an 11+ loss?

[indianplayers]

Wow amazing how the 11+ losses are appearing a lot lately. I guess it's my luck. 

What's an 11+ loss?

11+ straight losses.

[Namworld]

Wow amazing how the 11+ losses are appearing a lot lately. I guess it's my luck. 

What odds are you playing at?

[indianplayers]

50% and only rolled couple thousand times, but since yesterday I've gone more than 5 times over 11 losses in a row. One time I think it was 14-15 times. Oh Well. 

[Namworld]

50% and only rolled couple thousand times, but since yesterday I've gone more than 5 times over 11 losses in a row. One time I think it was 14-15 times. Oh Well. 

Eh, well I don't know how many thousands times you rolled but 11 losses streaks are to be expected with so many rolls. Although if it was less than 10 thousand, I guess that's indeed unlucky.

[GCInc.]

50% and only rolled couple thousand times, but since yesterday I've gone more than 5 times over 11 losses in a row. One time I think it was 14-15 times. Oh Well. 

Certainly worse than expected luck. What are the odds of 6 times 11 losses in a row for 2000 rolls? For one such streak it is 1/2048... and that means one out of "2048 series of 11 rolls" (22000+ rolls in average for one such streak)?

[indianplayers]

50% and only rolled couple thousand times, but since yesterday I've gone more than 5 times over 11 losses in a row. One time I think it was 14-15 times. Oh Well. 

Certainly worse than expected luck. What are the odds of 6 times 11 losses in a row for 2000 rolls? For one such streak it is 1/2048... and that means one out of "2048 series of 11 rolls" (22000+ rolls in average for one such streak)?

I don't know, but I lost a lot of money that's for sure. 

[dooglus]

Certainly worse than expected luck. What are the odds of 6 times 11 losses in a row for 2000 rolls? For one such streak it is 1/2048... and that means one out of "2048 series of 11 rolls" (22000+ rolls in average for one such streak)?

No, it doesn't work like that.  You don't multiply 2048 by 11.

Because imagine if the last 5 of one set of 11 lost, and then the first 6 of the next set of 11 also lost.  You would count that as 11 losses in a row, but it wouldn't be any of your 2048 sets of 11.

I think the expected number of rolls to get a sequence of 11 losses at 50% chance of a single loss is 2048.  But then something comes to mind where organofconti corrected me on that, so I'm no longer sure.  I'll see if I can dig up his post.

Edit: here: https://bitcointalk.org/index.php?topic=80312.msg1042019#msg1042019

It's interesting, because the discussion starts when he makes exactly the same mistake you made, and I follow up with exactly the mistake I just made.  Eventually we reach a conclusion I think.

Edit2: It's an interesting read, but the conclusion is that it takes on average 2046 plays to get a sequence of 11 losses at 50% chance per loss.

[dooglus]

Certainly worse than expected luck. What are the odds of 6 times 11 losses in a row for 2000 rolls? For one such streak it is 1/2048... and that means one out of "2048 series of 11 rolls" (22000+ rolls in average for one such streak)?

No, it doesn't work like that.  You don't multiply 2048 by 11.

Because imagine if the last 5 of one set of 11 lost, and then the first 6 of the next set of 11 also lost.  You would count that as 11 losses in a row, but it wouldn't be any of your 2048 sets of 11.

I think the expected number of rolls to get a sequence of 11 losses at 50% chance of a single loss is 2048.  But then something comes to mind where organofcorti corrected me on that, so I'm no longer sure.  I'll see if I can dig up his post.

Edit: here: https://bitcointalk.org/index.php?topic=80312.msg1042019#msg1042019

It's interesting, because the discussion starts when he makes exactly the same mistake you made, and I follow up with exactly the mistake I just made.  Eventually we reach a conclusion I think.

Edit2: It's an interesting read, but the conclusion is that it takes on average 4094 plays to get a sequence of 11 losses at 50% chance per loss.

[Luke-Jr]

Bets are off-the-chain

Thank you!

[toddtervy]

Bets are off-the-chain

Thank you!

Hey, no problem.  Anything for you guy.  Primedice.com also has good game.

[GCInc.]

It's an interesting read, but the conclusion is that it takes on average 2046 plays to get a sequence of 11 losses at 50% chance per loss.

Interesting indeed. "On average" is understandable for the layman, but I find it vague. I gather it means exactly 50% odds (to get a sequence of 11 losses at 50% chance per loss). If you can say that way. I'd suppose graphically it comes from the bell-shaped distribution curve, exactly half of which is dissected at the 2046 plays.

That has bothered me for a long time. I actually figured it would be something like that, but honestly thought it could be just slightly more than 2048, like in one of your early replies in the Satoshidice thread you quoted.

I find those calculations thrilling, even have a little book that contains examples of counter-intuitive everyday probabilities. But I tend to always forget how to calculate them. For instance I can't instantly come up with a solution on how to calculate answer to the question "What are the odds of 6 times 11 losses in a row for 2000 rolls?". It is well solvable with the formulas in your other thread, and a bit of time to sit down.

Also, I find it unbelievable that getting exactly 2 heads in a row for a fair coin flip takes on average 6 flips! "On average" again is supposed to mean 50% probability. Need to twist my head around that for a while.

[Scrat Acorns]

Also, I find it unbelievable that getting exactly 2 heads in a row for a fair coin flip takes on average 6 flips! "On average" again is supposed to mean 50% probability. Need to twist my head around that for a while.

In statistics and Quantum Mechanics (which also uses statistics heavily) common sense is useless. The human brain simply fails at understanding probabilities and large numbers.

[dooglus]

Interesting indeed. "On average" is understandable for the layman, but I find it vague. I gather it means exactly 50% odds (to get a sequence of 11 losses at 50% chance per loss). If you can say that way. I'd suppose graphically it comes from the bell-shaped distribution curve, exactly half of which is dissected at the 2046 plays.

I'm sorry, I was being lazy.  I mean the expected number of rolls.  ie. the sum of the product of the number of rolls and the probability of it taking that many rolls.

Also, I find it unbelievable that getting exactly 2 heads in a row for a fair coin flip takes on average 6 flips! "On average" again is supposed to mean 50% probability. Need to twist my head around that for a while.

How's this for weird?  A plot of probability against number of rolls.  There's a 0.25 probability that it will take 2 rolls to get 2 heads, etc:



The probability of it taking n rolls is:

  fib(n-2) / 2^n

where fib() is the Fibonacci sequence!

fib(0) = fib(1) = 1
fib(2) = 2
fib(3) = 3
fib(4) = 5
fib(n) = fib(n-1) + fib(n-2)

And you thought Mr. Fibonacci was just for the wackos doing technical analysis of the Bitcoin price?  

Edit: weird because of the step in the graph.  The probabilities that it will take 3 or 4 tosses to get 2 heads in a row are the same:

Code:

2:  1/4
3:  1/8 (THH)
4:  2/16 (TTHH and HTHH)
5:  3/32 (HTTHH, HTTHH, TTTHH)
6:  5/64
7:  8/128
8: 13/256

[GCInc.]

I'm sorry, I was being lazy.  I mean the expected number of rolls.  ie. the sum of the product of the number of rolls and the probability of it taking that many rolls.

I have to think about that for a while. How does that "expected number of rolls" relate to the probability of a certain sequence, ie. "what is the probability of getting a sequence of at least 11 losses when tossing a fair coin 4094 times?"

I have a hunch it would be 50%, and it comes back to the bell-shaped distribution curve, but hunches in this area are often wrong.

but the conclusion is that it takes on average 4094 plays to get a sequence of 11 losses at 50% chance per loss.

This correction went unnoticed for me last round...so it's 4094 and not 2046.

Edit: weird because of the step in the graph.  The probabilities that it will take 3 or 4 tosses to get 2 heads in a row are the same:

Yes, there is something that occurs to me mystical at that particular anomalous step

I find it weird when I look at it from another point of view: The probability of heads-heads sequence is 0.25, so you would need on average 4 sequences of 2 tosses.

x x    first sequence
  x x  second sequence
    x x  third sequence
      x x  fourth sequence
12 34 5  toss number

4 sequences of 2 tosses can be obtained from 5 tosses total, not 6. Now, why is one of those sequences invalid so you need the extra toss?

I must digest your chart, Fibonacci and the Markov chains referred to in the other thread to get this clear.

In statistics and Quantum Mechanics (which also uses statistics heavily) common sense is useless. The human brain simply fails at understanding probabilities and large numbers.

That's very true. However with hard practise you can understand some of it up to very limited numbers (I'm attempting now the range 1 to 6 ). The counter-intuitiveness is fascinating. I would like to craft a game designed around this 2-heads-in-a-row thing when I can understand it. Most people think it's only 4 tosses you need. Without using mathematics too much and through the practical example, I'd say it's 5, but even that is too low.

[hasher87]

I feel like i wanna make a different script this time around

instead of going for 1,2,4,8+++ for every losses at 2x modifier, why not go for 1 only, but increasing number of modifier 2x,3x,4x,5x.....999x...

could have been awesome, because sometimes i got hit with < 500  

the odds events is

selectGame(event, 60000)

[pennypig]

Wow amazing how the 11+ losses are appearing a lot lately. I guess it's my luck. 

I'm responsible for more than 10% of the total bets at coinroll.it.  Did the martingdale strategy for quite some time and from my experience you need money for more than 14 losses in a row.

Pic related:
https://i.imgur.com/XpnrI6B.png

[iANDROID]

Stop using Windows 8 and you will win.

[pennypig]

Stop using Windows 8 and you will win.

I agree, but that's chromeOS...arguably a lot shittier