I can't control what the dice do, they have no memory but over the long-run of my shoot, every 100 should result in 16.666 7's and if I've thrown the dice 300 times, I'd expect to see ~50 7's, but if I've seen 80, would it be fair to say that there's a statistically significant variance in my result?
I ask because under that scenario, isn't it better for me (and only me) to throw the dice X amount of times, see how many 7's, and if I've rolled enough 7's to get into "significant variance" that'd be the time to try and bet heavier knowing that I've seen "too many." Conversely, if I've thrown the dice 300 times and see 30 7's, would that be "too few" and rely on starting small with long rolls rather than big initial wagers?
My gut tells me there's something wrong w/ my logic - someone punch me so I can throw it up. :)
~N
Quote: foolshopeSo in 100 rolls, we should see 16.666 7's. If we saw 24, would that be "statistically significant?" What if we only saw 8?
I can't control what the dice do, they have no memory but over the long-run of my shoot, every 100 should result in 16.666 7's and if I've thrown the dice 300 times, I'd expect to see ~50 7's, but if I've seen 80, would it be fair to say that there's a statistically significant variance in my result?
I ask because under that scenario, isn't it better for me (and only me) to throw the dice X amount of times, see how many 7's, and if I've rolled enough 7's to get into "significant variance" that'd be the time to try and bet heavier knowing that I've seen "too many." Conversely, if I've thrown the dice 300 times and see 30 7's, would that be "too few" and rely on starting small with long rolls rather than big initial wagers?
My gut tells me there's something wrong w/ my logic - someone punch me so I can throw it up. :)
~N
Kapow!
Your timeline is too small. The 1/6 rolls will get into shape over the course of many more rolls than you or I will experience in a single session. Just because you witnessed only 30 7's in 300 rolls, doesn't mean that the previous 300 rolls weren't 7 rich, bringing the Universe back into balance. Previous events have no influence on the present, nor the future, from our limited perspective.
Quote: foolshopeSo part of the problem is that there are "all the 7's ever rolled in the entire universe" and the number of 7's *I've* rolled... and since I've never kept track of how many rolls of the dice I've made OR how many 7's I've rolled, I can't even make any sort of approximation or speculation, correct?
Pretty much. You can expect that your personal experience will approximate the bell shaped curve with the seven in the center. However, it's an error to assume you can use the past rolls to anticipate the next one, or that the rolls are "due" for a correction. I recently watched a table where they opened a brand new stick of dice, and the first shooter then shot 8 consecutive come out sevens. He then asked to change dice. Do you think that was the right thing to do? One school of thinking is that more sevens are coming, another would say all the sevens have been used up.
Unfortunately, the only way to see a trend is in hindsight.
Ah, but even if you had kept precise track ... its not just YOUR history of rolling 7s but all those people in the universe who simply roll more 7s. It can be said that eventually all results of all persons will indeed even out to 16.666 rolls of 7s but you only really care about that very next roll anyway. The only thing that can be said about that very next roll is "you pays your money and you takes your chance". If you want to view the result as an "extra" 7, fine.Quote: foolshopeand since I've never kept track of how many rolls of the dice I've made OR how many 7's I've rolled, I can't even make any sort of approximation or speculation, correct?
But even then, i'm not sure 100 or even 300 rolls is enough to prove anything.
Plus, you'd have the problem of every new shooter picking 2 dice out of the five available. There really is no accurate way to track it.
Ya, and tomorrow we come back and we cut offs your johnson!
Thanks guys. This should prove interesting for table conversation tonight.
~N
Quote: foolshopeSo in 100 rolls, we should see 16.666 7's. If we saw 24, would that be "statistically significant?" What if we only saw 8?
24, 8, ARE NOT "statistically significant?"
You can see the probabilities for any range in the data below.
The Law of Large Numbers is all about PERCENTAGES.
after 1,000 dice rolls the absolute number of 7s that rolled (in a sim for example) DIVIDED by 1000 will be close to 1/6
after 10,000 dice rolls still getting close to 1/6 even more than 1,000 dice rolls but at the same time one could be farther away from the EV of 10,000*(1/6)
After 1 million dice rolls we would be even closer to 1/6, than at 1k or 10k rolls, but could also be further away than 1,000,000*(1/6)
Example:see how the % are converging where the absolute values (the difference) is all over the place and at times larger or smaller than expectation. This was just a quick simulation in Excel
rolls | 7s | % | EV | Diff |
---|---|---|---|---|
10 | 2 | 20.000% | 1.666666667 | 0.333333333 |
100 | 18 | 18.000% | 16.66666667 | 1.333333333 |
1,000 | 171 | 17.100% | 166.6666667 | 4.333333333 |
10,000 | 1640 | 16.400% | 1666.666667 | -26.66666667 |
100,000 | 16883 | 16.883% | 16666.66667 | 216.3333333 |
1,000,000 | 166751 | 16.675% | 166666.6667 | 84.33333333 |
10,000,000 | 1666081 | 16.661% | 1666666.667 | -585.6666667 |
Distribution values below for 100 dice rolls.
x prob[X=x] prob[X<x] prob[X>=x] prob[X<=x] prob[X>x]
5 0.0002909031 0.0000940202 0.9999059798 0.0003849233 0.9996150767
6 0.0009211932 0.0003849233 0.9996150767 0.0013061165 0.9986938835
7 0.0024740617 0.0013061165 0.9986938835 0.0037801782 0.9962198218
8 0.0057521934 0.0037801782 0.9962198218 0.0095323716 0.9904676284
9 0.0117600399 0.0095323716 0.9904676284 0.0212924115 0.9787075885
10 0.0214032726 0.0212924115 0.9787075885 0.0426956842 0.9573043158
11 0.0350235371 0.0426956842 0.9573043158 0.0777192212 0.9222807788
12 0.0519515800 0.0777192212 0.9222807788 0.1296708012 0.8703291988
13 0.0703344467 0.1296708012 0.8703291988 0.2000052479 0.7999947521
14 0.0874156695 0.2000052479 0.7999947521 0.2874209174 0.7125790826
15 0.1002366344 0.2874209174 0.7125790826 0.3876575518 0.6123424482
16 0.1065014240 0.3876575518 0.6123424482 0.4941589758 0.5058410242
17 0.1052484661 0.4941589758 0.5058410242 0.5994074418 0.4005925582
18 0.0970624742 0.5994074418 0.4005925582 0.6964699161 0.3035300839
19 0.0837802409 0.6964699161 0.3035300839 0.7802501570 0.2197498430
20 0.0678619951 0.7802501570 0.2197498430 0.8481121521 0.1518878479
21 0.0517043773 0.8481121521 0.1518878479 0.8998165294 0.1001834706
22 0.0371331437 0.8998165294 0.1001834706 0.9369496731 0.0630503269
23 0.0251859583 0.9369496731 0.0630503269 0.9621356314 0.0378643686
24 0.0161609899 0.9621356314 0.0378643686 0.9782966213 0.0217033787
25 0.0098258819 0.9782966213 0.0217033787 0.9881225031 0.0118774969
26 0.0056687780 0.9881225031 0.0118774969 0.9937912811 0.0062087189
27 0.0031073302 0.9937912811 0.0062087189 0.9968986113 0.0031013887
28 0.0016202507 0.9968986113 0.0031013887 0.9985188620 0.0014811380
29 0.0008045383 0.9985188620 0.0014811380 0.9993234003 0.0006765997
30 0.0003808148 0.9993234003 0.0006765997 0.9997042151 0.0002957849
31 0.0001719809 0.9997042151 0.0002957849 0.9998761960 0.0001238040
24 or more 7s = 0.0378643686 (3rd column)
Know our win/loss/time limits, stick to'm, have fun, generally stay away from the center of the table, and when possible - max out our odds.
Another Q:
We play in the Seattle area and Tulalip has 10x odds... but we have different styles. I like to place the come and add odds, he likes to play $26/7 / 52/54 across with a $5 come. When a place bet is hit, he'll take the winnings and make it a come since it's rare to press from $54 across to $300 (all #s w/ $60). His reasoning is "I'm more likely to hit one number once than that same number a few times - I'd rather start with bases-loaded but "light" than loading the bases like me and immediately throwing 5X odds on a number.
Any thoughts on this? We typically end-up about the same - is that to be expected?
Quote: foolshopeSo the bottom line is (I can't wait to explain this to my friend tonight!) is even if we had 10,000,000 rolls and saw 585 less 7's than what is expected, that's no reason to "bet more" and in fact, there is *no* number of "too many 7's" that can even be relied upon, hence the term "gambling."
Know our win/loss/time limits, stick to'm, have fun, generally stay away from the center of the table, and when possible - max out our odds.
Yes, Sounds good to me.
If you understand ev and variance you can calculate the standard deviation and use it to see the appropriate ranges your results can fall into.
10,000,000 trials would be Sqrt(10,000,000*(1/6)*(5/6)) = 1178.5 = 1 sd (sqrt(N*P*(1-P))
Try the math yourself with just 6 dice rolls and see how many actual 6-roll distributions follow the theoretical 6-roll distributions.
You can do this simulation with 2 dice rolling on the floor ( that should not cost any $$$ to do).
The more # of short distributions you do all end up into one larger distribution that starts to approach it's percentages, NOT necessarily the absolute values.
answers below:
x prob[X=x] prob[X<x] prob[X>=x] prob[X<=x] prob[X>x]
0 0.3348979 0.0000000 1.0000000 0.3348979 0.6651020
1 0.4018775 0.3348979 0.6651020 0.7367755 0.2632244
2 0.2009387 0.7367755 0.2632244 0.9377143 0.0622856
3 0.0535836 0.9377143 0.0622856 0.9912980 0.0087019
4 0.0080375 0.9912980 0.0087019 0.9993355 0.0006644
5 0.0006430 0.9993355 0.0006644 0.9999785 0.0000214
6 0.0000214 0.9999785 0.0000214 1.0000000 0.0000000
0.6651020 (almost 2 out of 3) of the 6-roll distributions will contain (on average) 1 or more 7s. (col #3)
0.2632244 ( slight better than 1 in 4) will contain (on average) 2 or more 7s.
An exercise like this can show how expectations within ranges actually work. It only takes a few "hits" to bring the percentages into balance (+ or -) while the absolute numbers are still not "balanced". This is why in gambling nothing is really due, just to make up for something that has hit too much or not enough.
He is accepting a lower payout for selecting numbers but his inclusion of the outside bets is working against him and even his doing other than the six and eight is working against him.
Most nights he "should" come in a bit behind you.
Quote: foolshope
We play in the Seattle area and Tulalip has 10x odds... but we have different styles. I like to place the come and add odds, he likes to play $26/7 / 52/54 across with a $5 come. When a place bet is hit, he'll take the winnings and make it a come since it's rare to press from $54 across to $300 (all #s w/ $60). His reasoning is "I'm more likely to hit one number once than that same number a few times - I'd rather start with bases-loaded but "light" than loading the bases like me and immediately throwing 5X odds on a number.
Any thoughts on this? We typically end-up about the same - is that to be expected?
It sounds like you are both eventually getting the same $52 - $55 in action. So you eventually end up the same. His place bets wouldn't be in action coming out , but he does run the risk of losing more on the point/seven, one-and-done, shooter. Does he take the other place bets down once come bet point is established?
So if you toss the dice 100 times and see 36 or more 7s, then that would be cause for concern (but still possible).
Quote: s2dbaker36 or more 7s in a 100 toss session is outside of 5 standard deviations.
So if you toss the dice 100 times and see 36 or more 7s, then that would be cause for concern (but still possible).
Every distribution is relative to the number of possible attempts.
A quick 10 million player computer simulation shows a few hitting 36 or more.
1 in 402,926.3 to roll 36 or more; 1 in 609,527.3 to roll exactly 36 7s.
Now if this happens more than 2 times in a row, I would want some new dice.
# of 7s per 100 rolls | freq | % |
---|---|---|
0 | 0 | 0.00000% |
1 | 2 | 0.00002% |
2 | 28 | 0.00028% |
3 | 167 | 0.00167% |
4 | 795 | 0.00795% |
5 | 2930 | 0.02930% |
6 | 9305 | 0.09305% |
7 | 24699 | 0.24699% |
8 | 57188 | 0.57188% |
9 | 117232 | 1.17232% |
10 | 213329 | 2.13329% |
11 | 348533 | 3.48533% |
12 | 518394 | 5.18394% |
13 | 703164 | 7.03164% |
14 | 873949 | 8.73949% |
15 | 1001619 | 10.01619% |
16 | 1065409 | 10.65409% |
17 | 1052291 | 10.52291% |
18 | 969930 | 9.69930% |
19 | 837784 | 8.37784% |
20 | 680167 | 6.80167% |
21 | 517980 | 5.17980% |
22 | 373138 | 3.73138% |
23 | 251660 | 2.51660% |
24 | 162052 | 1.62052% |
25 | 99125 | 0.99125% |
26 | 56550 | 0.56550% |
27 | 31279 | 0.31279% |
28 | 16393 | 0.16393% |
29 | 8079 | 0.08079% |
30 | 3874 | 0.03874% |
31 | 1726 | 0.01726% |
32 | 722 | 0.00722% |
33 | 313 | 0.00313% |
34 | 113 | 0.00113% |
35 | 54 | 0.00054% |
36 | 14 | 0.00014% |
37 | 11 | 0.00011% |
38 | 1 | 0.00001% |
39 | 1 | 0.00001% |
Quote: s2dbakerZero 7s in 100 tosses is NOT outside of 5 standard deviations.
True, but it can never be less than 0. So there is a limit.
Much harder to do No 7s than 36 or more.
The probability of Zero 7s in 100 dice rolls is: 1 in 82,817,974.5
Excel = 1/ BINOMDIST(0,100,1/6,FALSE)
That is why it did not happen in my 10 million simulation.
I needed more players!
Quote: s2dbakerI was thinking of a result set with an even distribution. I looked at my meager 50K trials and saw 4 as my least number of 7s. I should go by the 5 Standard Deviations % rather than actual avg- 5*stddev.
Excellent point to bring up.
I often get tripped thinking normal when it is not normal etc. hehe
Here is a photo for all to see what this means.
Enjoy
Remember boys and girls, both curves do NOT touch the x-axis. The photo just makes it look that way.
If you're seeing less 7's then you should, BUY the 4/10 for $35 each (assuming same rules)