tucenaber
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June 06, 2014, 03:38:44 PM |
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EDIT: Can you explain this in more detail? I'm told that the expected number of rolls before you see a 28 loss streak is around 514 million, but I don't know why.
/.../ Also, using simulation, I tried to figure out the number of rolls you need to make to have a 50% chance of seeing a streak of length N. It looks to me like the number is about sqrt(2) times smaller than the above numbers, and so we need to have seen 410M/sqrt(2) = 290M rolls to have a 50% chance of seeing a streak of length 28. Since we've seen 357M rolls, we presumably have a greater than 50% chance of seeing such a streak. This is pretty low quality research I did using simulation. The sqrt(2) thing in particular is very dodgy. An exact formula can be found here: https://math.stackexchange.com/questions/59738/probability-for-the-length-of-the-longest-run-in-n-bernoulli-trialsand a more useful approximation here: http://mathforum.org/library/drmath/view/56637.htmlThe answer seems to be 284 million rolls.
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TwinWinNerD
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June 06, 2014, 03:43:50 PM |
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EDIT: Can you explain this in more detail? I'm told that the expected number of rolls before you see a 28 loss streak is around 514 million, but I don't know why.
/.../ Also, using simulation, I tried to figure out the number of rolls you need to make to have a 50% chance of seeing a streak of length N. It looks to me like the number is about sqrt(2) times smaller than the above numbers, and so we need to have seen 410M/sqrt(2) = 290M rolls to have a 50% chance of seeing a streak of length 28. Since we've seen 357M rolls, we presumably have a greater than 50% chance of seeing such a streak. This is pretty low quality research I did using simulation. The sqrt(2) thing in particular is very dodgy. An exact formula can be found here: https://math.stackexchange.com/questions/59738/probability-for-the-length-of-the-longest-run-in-n-bernoulli-trialsand a more useful approximation here: http://mathforum.org/library/drmath/view/56637.htmlThe answer seems to be 284 million rolls. Thank you! Very intresting read.
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dooglus (OP)
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June 06, 2014, 09:48:46 PM |
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The answer seems to be 284 million rolls.
The question isn't: "how many rolls do we need so that I have a >50% chance of seeing a streak of length L or longer", or "how many rolls do we need so that the expected longest streak is of length L". The question is: "we have seen N rolls, what is my chance of seeing a streak of length L or longer?" Which is kind of the same thing, but in the opposite direction, and apparently much harder to calculate.
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Just-Dice | ██ ██████████ ██████████████████ ██████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████ ██████████████ ██████ | Play or Invest | ██ ██████████ ██████████████████ ██████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████ ██████████████ ██████ | 1% House Edge |
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Azlan
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June 06, 2014, 10:00:33 PM |
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Consider rolling a 6 sided die until you roll a six. The expected number of rolls to get a six is 6.
True, but incomplete - according to the small sample bias - rare events (<0.5) go 'unseen' more often then expected when the number of observations is small. Im sure the links offered by others goes into this. Take your die: You want to roll a six. You roll 6 times. Whats the probability of not seeing the six? 5/6 ^ 6 = 33.48% So a third of the time (with six observations) you will not see the 6 at all. It gets better. Imagine a 1000 sided die. You bet against the outcome "1000" and you roll exactly 1000 times. The probability of getting away with it (not seeing the '1000' even once) is 0.999 ^ 999 = 36.8% Many of your players are implementing this logic with bets with 98% chance of winning. The small sample bias predicts that, in 100 rolls, they will lose less then expected. Since there is no free sandwich in life you can guess what happens the other 63.2% of the time you roll the 1000 sided die 1000 times..... You will see the rare outcome ('1000') one *or more* times. Back to a streak of 28 reds (p=0.5). If you expect it to happen 1 in 236M rolls, the small sample bias (relative to the probability) predicts you will see such a steak once *or more* 63% of the time in 236M rolls but wont see it at all ~36% of the time. M_acchi
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dooglus (OP)
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June 06, 2014, 10:26:26 PM |
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Consider rolling a 6 sided die until you roll a six. The expected number of rolls to get a six is 6.
True, but incomplete Only because you deleted the next line that I wrote: But if you only roll 4 times, you have about a 51% of seeing a six, even though you've rolled less than the expected number.
The whole point I am making is that there is a difference between "the expected number of rolls to see a 6" and "the number of rolls you need to have a 50% chance of seeing a 6". Take your die: You want to roll a six. You roll 6 times. Whats the probability of not seeing the six? 5/6 ^ 6 = 33.48% So a third of the time (with six observations) you will not see the 6 at all.
Read about what "expected value" means. It sounds like you think it means "most likely value", but it doesn't. Take your 1000 sided die. Let's play a game. If you roll a 6, you win a million dollars, but if you roll anything else, you lose a dollar. What's the expected value of rolling the die? Almost every time you roll (99.9% of the time), you lose a dollar. So you 'expect' to lose almost every time. But that's not what expected value means. Your expected value for each roll is +$999.001: (1 * 1e6 - 999 * 1) / 1000 = 999.001
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Azlan
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June 06, 2014, 11:08:24 PM |
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The whole point I am making is that there is a difference between "the expected number of rolls to see a 6" and "the number of rolls you need to have a 50% chance of seeing a 6".
Agreed, and its a point worth making. Read about what "expected value" means.
Just looked it up. Seems to mean the average. But expected value misses VERY important stuff. For example, the expected value of playing 6 rounds of russian roulette (1 bullet in 6 chambers and spinning the barrel EACH time...so its like the die) ... ... is that you are 1/6 dead. Not very useful info. Im saying something similar to you - that its more interesting that with 6 plays (1 bullet) you will live 33% of the time. You are asking about 50%. with 5 plays, you live 5/6 ^ 5 = 40% of the time and, as you pointed out, with 4 plays 5/6 ^ 4 = 48% of the time. But that's not what expected value means. Your expected value for each roll is +$999.001: (1 * 1e6 - 999 * 1) / 1000 = 999.001
yes, but thats the average and assumes infinite rolls. given finite rolls (like 1000 for example), the more interesting fact pertains to your question about 50% ..... and the answer is greater then 500 rolls (for 1/1000 event). ...its 692 rolls (.999 ^ 692)= 50% : the chance of seeing it one or more times
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iWin
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June 07, 2014, 01:09:32 AM |
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Not sure if it's been asked before but will there ever be an autobet on this site?
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vinboy
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June 07, 2014, 01:49:34 AM |
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hi https://just-dice.com/ seems to be down at the moment? or is it just me?
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BayAreaCoins
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June 07, 2014, 01:55:03 AM |
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TwinWinNerD
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June 07, 2014, 01:56:35 AM |
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vinboy
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June 07, 2014, 01:58:40 AM |
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ok.. tried to connect thru a VPS and it worked..
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dooglus (OP)
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June 07, 2014, 03:15:49 AM |
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Not sure if it's been asked before but will there ever be an autobet on this site?
I don't think it's a good idea. Why would you want a program to bet for you on a site where there's no winning strategy? If you dislike playing enough that you do want to actually do it, perhaps you should stop playing. You can just send your coins directly to my donation address in the FAQ if you want to get rid of them quickly.
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Just-Dice | ██ ██████████ ██████████████████ ██████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████ ██████████████ ██████ | Play or Invest | ██ ██████████ ██████████████████ ██████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████ ██████████████ ██████ | 1% House Edge |
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seuntjie
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June 07, 2014, 09:59:17 AM |
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Not sure if it's been asked before but will there ever be an autobet on this site?
Dooglus will probably never implement his own, but there are a few 3rd party bots you can use. Here is one: https://bitcointalk.org/index.php?topic=307425
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MoneyGod
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June 07, 2014, 05:16:31 PM |
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Not sure if it's been asked before but will there ever be an autobet on this site?
Dooglus will probably never implement his own, but there are a few 3rd party bots you can use. Here is one: https://bitcointalk.org/index.php?topic=307425But I don't think these bots are helpful for these games because these most of games depend on luck also
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chaosPT
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June 07, 2014, 05:53:25 PM |
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Not sure if it's been asked before but will there ever be an autobet on this site?
Dooglus will probably never implement his own, but there are a few 3rd party bots you can use. Here is one: https://bitcointalk.org/index.php?topic=307425But I don't think these bots are helpful for these games because these most of games depend on luck also The bot are for more gambler to come over , Do JD have any whale these day ?
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boumalo
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June 07, 2014, 08:00:53 PM |
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Not sure if it's been asked before but will there ever be an autobet on this site?
I don't think it's a good idea. Why would you want a program to bet for you on a site where there's no winning strategy? If you dislike playing enough that you do want to actually do it, perhaps you should stop playing. You can just send your coins directly to my donation address in the FAQ if you want to get rid of them quickly. Some players love to autobet, it is actually fun and casinos are about giving fun to their clients BTW I have been re-reading the first pages of this thread, it is fun, it brings us back
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freedomno1
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Learning the troll avoidance button :)
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June 08, 2014, 10:02:26 AM |
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That would explain why I thought a whale died or something when the bankroll went up 2.1% over the last 4 days Although if Theymos was actually gambling his donations that would be one way of giving back
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Believing in Bitcoins and it's ability to change the world
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boumalo
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June 08, 2014, 11:12:33 AM |
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That would explain why I thought a whale died or something when the bankroll went up 2.1% over the last 4 days Although if Theymos was actually gambling his donations that would be one way of giving back It is obviously not the real Theymos If he was betting he would use a nickname The bitcointalk cold wallet addresses can be checked for balance
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