The Chances are One in Ten Trillion

Discussion in 'Physics & Math' started by Jozen-Bo, Oct 15, 2009.

1. Jozen-BoThe Wheel Spinning King!!!Registered Senior Member

Messages:
1,597
I reposted this under a new name, because when I tried to edit the name in the other post it didn't change...

I am going to calculate some statistics and see if I can get this right. I'll use two sets. Correct me if I am wrong.

We'll take 20 dice, each having 20 sides and divide them into two groups. We pit these dice against each other, the one's rolling the highest number winning. Statistically it should come out 50% of the time from both groups winning with higher rolls...in other words 50-50 or five from each group coming out on top.

If one group out rolls the other ALL ten times, what are the chances.

1 beating 1 is a 50% chance.
If one die beats 2 dice this is a 25% chance.

If 1 die beats 10 this is 50% to the square root of 10...

50
25
12.5
6.25
3.125
1.5625
0.78125
0.390625
0.1953125
0.09765625 at 1 die beating 10 dice, thus 1 dice has an approximate chance of 1 in a thousand of out rolling 10 other dice.

(2) for this to happen again for another dice we need square the final result of the first.... 0.09765625 Time to find an online calculator....

we get 0.00953674316

(3) to find the chance of another roll beating the 10 in succession we multiply the reminder by the 0.09765625

we get 0.00093132257

repeat to up to 10....

(4) 0.00009094947
(5) 8.88178e-6
(6) 8.6736e-7
(7) 8.47e-8
(8) 8.27e-9
(9) 8.1e-10
(10) 8.1e-11

or basically .000000000081

Thus the chances are 1 in a ten trillion....

now comes the tough calculations...

1 die beating 50...(the easy part)

and then 10 die beating 50...(the part that takes more time to calculate)

I'll get back to this soon, unless someone else beats me to the answers!

First I need to get some food in me!

3. Jozen-BoThe Wheel Spinning King!!!Registered Senior Member

Messages:
1,597
So, it appears no one is disputing my calculations. Why...because they are correct!!!

5. PinwheelBannedBanned

Messages:
2,424
There is an error......somewhere.

7. Jozen-BoThe Wheel Spinning King!!!Registered Senior Member

Messages:
1,597
If 1 die beats 10 this is 50% to the square root of 10...

OK...I misused the term square root here, I am tired with too much to do...excuse me. It is 50% divided by 2 10 times. Doesn't change the answer, it was just an oversight because I'm overworked...there is no square root in the process of this calculation. There, now it is correct!

Last edited: Oct 15, 2009
8. Jozen-BoThe Wheel Spinning King!!!Registered Senior Member

Messages:
1,597
If your referring to the my use of the word square root, yes. I just pointed this out, squaring the 50% has nothing to do with the process to calculate the answer. I got some food in me, my brain is fed. I was tired and hungry when writing this out.

If you are saying that the answer is incorrect (ignoring my misuse of the term square root), then show the proof or I will make fun of you for not backing up your statement with proof (it's only fair, since this has been done to me enough times)...

I would be happy if you did find the error and proved it, I want to be corrected if I am wrong. Off course, I'd be happier if my calculations are correct, which they are...

9. Jozen-BoThe Wheel Spinning King!!!Registered Senior Member

Messages:
1,597
Hmmmm....

It looks like I'll have to spread the thread in order to get some sort of response. So far no one has found any errors other then me catching my own grammar error...

1 in 10,000,000,000,000

10. PinwheelBannedBanned

Messages:
2,424
I think people are dying of sheer boredom before they get to the end, and thus are having difficulty in replying.

11. Jozen-BoThe Wheel Spinning King!!!Registered Senior Member

Messages:
1,597
Probably so....at this time...

This thread is not meant to entertain.

I figured I might get some help here...

or is this place solely about entertainment...

12. CptBorkValued Senior Member

Messages:
6,411
The probability of a die from one side beating 10 dice from the other side is not $0.5^{10}$ as far as I can tell. If you rolled 10 times on each side (including a redo when you get a tie), then $0.5^{10}$ would be correct, but this is not how it's done. You're rolling 1 die and then keeping that value fixed while you roll 10 others. If, for instance, your 1 die beats the first 5 rolls on the other side, then chances are this die has a relatively high value and is more than 50% likely to beat the next roll. For a more concrete example, you have a 5% chance of rolling a 20 on your one die, in which case it's guaranteed to tie or beat every single roll from the other side. Your chances of seeing this happen are thus at least 5%, and adding the probabilities to get this outcome with a roll of less than 20 will only raise this number. I'm pretty sure the correct way to calculate this probability involves a rather ugly-looking sum, maybe I'll plug it into Matlab later on.

13. rpennerFully WiredValued Senior Member

Messages:
4,833
For one (fair) 20-sided die against another , they tie 5% of the time.
So the odds are only $\frac{\frac{1}{2}(n)(n-1)}{n^2} = \frac{\frac{1}{2}(20)(19)}{20^2}$ 47.5% for the picked one to "beat" the other.

Likewise for a two-sided die (a coin) the odd are only 25%.

Now for one die to beat all of M n-sided dice, the calculation is:
$\frac{\sum_{i=1}^{n-1} i^M}{n^{M+1}} = \frac{S_M(n-1)}{n^{M+1}} = \frac{\sum_{i=0}^M { M + 1\choose{i}} B_i (n-1)^{M+1-i}}{(M+1)n^{M+1}} = \frac{\sum_{i=0}^M \frac{(M+1)...(i+1)}{(M+1-i)...(1)} B_i (n-1)^{M+1-i}}{(M+1)n^{M+1}}$

Where $B_i$ are the fabulous Bernoulli numbers.
The go (starting with i=0): 1, 1/2, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66 and get weirder from there.

For M = 1 we have: $\frac{\sum_{i=0}^1 \frac{(2)...(i+1)}{(2-i)...(1)} B_i (n-1)^{2-i}}{2n^2} = \frac{\frac{(2)...(1)}{(2)...(1)} B_0 (n-1)^2 + \frac{(2)}{(1)} B_1 (n-1)}{2n^2}= \frac{B_0 (n^2-2n+1) + 2 B_1 (n-1)}{2n^2} =\frac{n^2-n}{2n^2}$
So for n = 2, this is (4 -2)/8 or 25%
for n = 20 this is (400-20)/(800) = 47.5%

For M = 2 we have $\frac{\sum_{i=0}^2 \frac{(3)...(i+1)}{(3-i)...(1)} B_i (n-1)^{3-i}}{3n^3} = \frac{ \frac{(3)...(1)}{(3)...(1)} B_0 (n-1)^{3}+ \frac{(3)...(2)}{(2)...(1)} B_1 (n-1)^{2}+ \frac{(3)}{(1)} B_2 (n-1)}{3n^3} = \frac{ 2(n-1)^3+ 3 (n-1)^2 + (n-1)}{6n^3} = \frac{2 n^3 - 3n^2 + n}{6 n^3}$

So for n = 2 this is (16 - 12 + 2)/(48) = 6/48 = 12.5%
and for n = 20 this is (16000 - 1200 + 20)/(48000) = 30.875%

For M = 10 we have $\frac{\sum_{i=0}^{10} \frac{(11)...(i+1)}{(11-i)...(1)} B_i (n-1)^{11-i}}{11n^{11}} = \frac{ B_0(n-1)^{11} + 11 B_1(n-1)^{10} + 55 B_2(n-1)^9 + 330 B_4(n-1)^7 + 462 B_6(n-1)^5 + 165 B_8(n-1)^3 + 11 B_{10}(n-1)^1 }{11n^{11}} \\ = \frac{ 6(n-1)^{11} + 33(n-1)^{10} + 55 (n-1)^9 - 66(n-1)^7 + 66 (n-1)^5 - 33(n-1)^3 + 5 (n-1) }{66n^{11}} = \frac{6n^{11} - 33n^{10} + 55n^9-66n^7+66n^5-33n^3+5n}{66n^{11}}$

So for n = 2 this is 66/135168 = 1/2048 = 0.048828125%
and for n = 20 this is 918955730936100/13516800000000000 = 6.7986189847900390625%

Last edited: Oct 18, 2009
14. rpennerFully WiredValued Senior Member

Messages:
4,833
So for a n-sided die beating the best of M n-sided dice, the odds are smaller than 1/(M+1) because of ties. If n > 10M, the odds are close to $\frac{1}{M+1} - 1/{2n}$ , but that is not the case with n = 20 and M = 50.

But what does it even mean for K n-sided dice to "beat" M n-sided dice?
I would read that as the best of the K has to beat the best of the M, and then the best of the K-1 must beat the best of the M-1 and so on, with no ties. So M >= K would seem to be a requirement.

15. phytiRegistered Senior Member

Messages:
625
Two group values a and b.
probability of a = 1/20.
probability of b = 1/20.
probability of a>b is (1-b/20), b range (1 to 20).
s = sum of probabilities = [20-(20*21/2*20)] = 9.5
average of the range is s/20 = .475

Geometrically draw a square 20x20 representing all ab combinations.
Label one side a, the other b.
Excluding the diagonal when a = b, 400-20 = 380 events.
Half on either side of the diagonal represent a>b or b>a, i.e. 190/400 = .475

rpenner beat me to it!

Last edited: Oct 16, 2009
16. rpennerFully WiredValued Senior Member

Messages:
4,833
For K=2, and M=2 and n>=9, I get a depressing complex fit to numerical experiments:
$\frac{4(5545)*r^4 +8(1109o-6066)r^3+4(333o^2-3642o+26300)r^2 +2(42o^3-708o^2+10534o-58731)r + (2o^4-40o^3+973o^2-11591o+51840)}{6n^4}$
Where n = 10r + o.

Tested from 9...84

Somewhat simpler as:
$\frac{\left. { 1109n^4 \\ \, - 24264n^3 \\ \, \,+ 2(3(n \textrm{mod} 10)^2-24(n \textrm{mod} 10)+263000)n^2 \\ \, \, \,- 4(62(n \textrm{mod} 10)^3-522(n \textrm{mod} 10)^2-350(n \textrm{mod} 10)+1468275)n \\ \, \, \, \,+ 133(n \textrm{mod} 10)^4+2224(n \textrm{mod} 10)^3-40900(n \textrm{mod} 10)^2+77600(n \textrm{mod} 10)+25920000 } \right. }{3000n^4}$

Anyone want to tackle writing this simpler and extending for lower n?

17. rpennerFully WiredValued Senior Member

Messages:
4,833
But those numerical experiments had a coding error which introduced artifacts above n=9. So what I discovered has nothing to do with the actual problem.

For K=2, M=2 and n>2 we have the probability as $\frac{n(n-1)(2n^2-2n-1)}{6n^4}$ Whew... that makes more sense for a number of reasons.

18. Betrayer0fHopeMY COHERENCE! IT'S GOING AWAYYRegistered Senior Member

Messages:
2,311
This is orgasmic.

19. rpennerFully WiredValued Senior Member

Messages:
4,833
But nearly correct (I left some B's in the next to last equation even though I already evaluated them), as opposed to post #13 - which is also entirely correct but describes a program error, not what was asked for.

20. CptBorkValued Senior Member

Messages:
6,411
The important point here is that it's far from a divine miracle if one should roll a 20-sided die and find that value beats the next 10 rolls. I would guess that was the original idea of the thread, to establish whether something fishy might be going on and God could be found just by rolling dice.