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2024-09-14 17:34:41
2 votes, rating 6
2 votes, rating 6
Calculating Probability of double skull-double skull (Lesson 2)
A few days ago there was a nice discussion on discord between some of us about what was the probability of rolling a double skull followed by another double skull (presumably therefore using a reroll) during a game in which let's say about seventy rolls of two six-faces dice.
Based on brute-force algorithm and/or an article found on the web the result should be a little over 5% (5.05562242% rounded to the eighth decimal place to be precise) but several methods have been proposed in the chat giving values some times only little bit different.
I did not find an easy way to bring the discussion with some figures back to this blog (the column on the side truncates some figures if you go in the comments mode)so I wrote a public note which link is given below, which seems quite readable to me. This note also proposes two programmes that perform the calculation (one in C# and another in MATLAB)
Perhaps if there is any mathematician in the community who would like to give us a slightly simpler explanation than the one proposed by Prof. Villarino's article (or a confirmation of the correctness of the result) he would be welcome!
Happy reading!
Probability of Double-Double-Skull
Based on brute-force algorithm and/or an article found on the web the result should be a little over 5% (5.05562242% rounded to the eighth decimal place to be precise) but several methods have been proposed in the chat giving values some times only little bit different.
I did not find an easy way to bring the discussion with some figures back to this blog (the column on the side truncates some figures if you go in the comments mode)so I wrote a public note which link is given below, which seems quite readable to me. This note also proposes two programmes that perform the calculation (one in C# and another in MATLAB)
Perhaps if there is any mathematician in the community who would like to give us a slightly simpler explanation than the one proposed by Prof. Villarino's article (or a confirmation of the correctness of the result) he would be welcome!
Happy reading!
Probability of Double-Double-Skull
Comments
Posted by Joost on 2024-09-14 19:38:59
Using my phone’s calculator I get to 5,4%. Every time you roll there is a chance of (1/36)*(1/36) that a quad skull happens. That is a chance of 99,92284% of a quad skull NOT happening. The chance of at least one quad skull happening in 70 tries, assuming unlimited rerolls, is 70 times (100-99,92284) = roughly 5,4%
The assumption of 70 rerolls available is of course wrong since the max is 36 (including leaders) and an ax amount of brilliant coaching rerolls but that should be negligible.
The assumption of 70 rerolls available is of course wrong since the max is 36 (including leaders) and an ax amount of brilliant coaching rerolls but that should be negligible.
Posted by PatKnez on 2024-09-14 20:31:15
Your odds of double skulls is 1 in 36 or about 2.97%. Your odds of doing this 2 times in a row is 1 in 1296, the result of 36x36 or about .07 percent.
Posted by PatKnez on 2024-09-14 21:04:24
But that whole paper is a lot of work to see that happen. What's interesting to me though is that with dice rolling the probability resets after every roll. I know this won't relate to a game algorithm through a computer but with real dice every time you roll your probability actually stays the same each time. The way I look at double skulls is that if I have just rolled them I have the same 1 in 36 chance of doing it again I don't ever feel that the likelihood is less than 1 percent but is actually still just another 2.97 precent chance of reoccurring... Then again I think Nuffle just doesn't like some people :)
I also found the numbers to be slightly off as they were referencing the number of 2d6 rolls at 70. That number is very variable as the 2d6 rolls don't include block dice rolls as they are calculated separately I believe in fumble in the dice stats. Are you including both teams dice rolls in your formula or a singular team and does the algorithm account for that? I guess that is why it is such a variable game and why rolling less dice is usually safer.
I also found the numbers to be slightly off as they were referencing the number of 2d6 rolls at 70. That number is very variable as the 2d6 rolls don't include block dice rolls as they are calculated separately I believe in fumble in the dice stats. Are you including both teams dice rolls in your formula or a singular team and does the algorithm account for that? I guess that is why it is such a variable game and why rolling less dice is usually safer.
Posted by sirvejlance on 2024-09-14 22:27:08
Always 50%…it happens or it doesnt. 😎
Posted by RDaneel on 2024-09-14 23:47:03
I too was initially misled and tried to calculate this probability with a quad skull event. But as I wrote in the notes the quad skull experiment is different from the experiment of rolling TWO dice twice in a row. If you use the quad skull probability you are implicitly thinking of rolling four six dice all at once for a large number of times and then you want to see when four skulls come up in a row. So 5,4% for me is result of another experiment not this one.
Concerning reroll i was not clear. Forget the reroll , and let's think only to an experiment where you roll 70 times 2D6 and you want to see how frequent happen a sequence of double skull followed by double skull. The notes state that this probability is 5.05562242%. Which is close but not 5,4%.
Concerning reroll i was not clear. Forget the reroll , and let's think only to an experiment where you roll 70 times 2D6 and you want to see how frequent happen a sequence of double skull followed by double skull. The notes state that this probability is 5.05562242%. Which is close but not 5,4%.
Posted by JackassRampant on 2024-09-15 03:00:14
Didn't read the article, but the math is not that crazy if you don't consider RR economy or alternative block conditions.
70 x 2d block with Block, unlimited TRR that only work on skullouts, you can expect 70/1296 quads, which my calculator says is 5.4012345%. Your chance of having to eat any skulls is (1295/1296)^70, or almost exactly 5.26%.
70 x 2d block with Block, unlimited TRR that only work on skullouts, you can expect 70/1296 quads, which my calculator says is 5.4012345%. Your chance of having to eat any skulls is (1295/1296)^70, or almost exactly 5.26%.
Posted by Joost on 2024-09-15 09:32:32
"I too was initially misled and tried to calculate this probability with a quad skull event. But as I wrote in the notes the quad skull experiment is different from the experiment of rolling TWO dice twice in a row. If you use the quad skull probability you are implicitly thinking of rolling four six dice all at once for a large number of times and then you want to see when four skulls come up in a row. So 5,4% for me is result of another experiment not this one."
I must be missing something, because I don't understand how the probability can be different depending on whether 4 dice are rolled at once or in two separate instances. The only condition for a block to achieve 'Quad Skull Status' is to role a skull on 4 dice, regardless of order in which they are rolled, or even if 2 of the 4 dice are only rolled if the first two were both a skull.
I must be missing something, because I don't understand how the probability can be different depending on whether 4 dice are rolled at once or in two separate instances. The only condition for a block to achieve 'Quad Skull Status' is to role a skull on 4 dice, regardless of order in which they are rolled, or even if 2 of the 4 dice are only rolled if the first two were both a skull.
Posted by Joost on 2024-09-15 09:37:10
"70 x 2d block with Block, unlimited TRR that only work on skullouts, you can expect 70/1296 quads, which my calculator says is 5.4012345%. Your chance of having to eat any skulls is (1295/1296)^70, or almost exactly 5.26%."
@JackassRampant I was unsure of this myself. I believe the chance of running into at least one Quad Skull is 5,4%, but it does include scenarios where you run into more than 1 Quad Skull.
Does the 5.26% odds represent the chance that you run into 1 Quad Skull exactly, out of 70 tries?
@JackassRampant I was unsure of this myself. I believe the chance of running into at least one Quad Skull is 5,4%, but it does include scenarios where you run into more than 1 Quad Skull.
Does the 5.26% odds represent the chance that you run into 1 Quad Skull exactly, out of 70 tries?
Posted by RDaneel on 2024-09-15 10:19:28
One has to be very careful with the definition of the experiment. Here we are looking for the probability of a double skull followed by a double skull on a series of 70 throws of two six-face dice (without reroll). Every time you roll 2 dices. NOT four.
If we denote by S the skull result of the dice and X any other non-skull result this sequence "XX SS SS XX" is a ‘favorable’ sequence .
If, on the other hand, you use the quad skull probability you are implicitly implying that you are rolling FOUR dice at the same time: the sequence above can NEVER happen to you because happen in turn launch 2 and 3 where your 4 dices can only generate something like XXXX XXXX SSSS XXXX followed or preceded by a number of trials multiple of four.
As I wrote in the notes I tried calculating the probability using the QUAD SKULL prob (1/36)*(1/36) on 69 trials, so n=69, (not 70) and it comes out very close to the calculation in the article (5,052%) but not 5.05562242% so from mathematical point of view is not exactly the same. Therein lies the curiosity that we have not been able to explain.
If we denote by S the skull result of the dice and X any other non-skull result this sequence "XX SS SS XX" is a ‘favorable’ sequence .
If, on the other hand, you use the quad skull probability you are implicitly implying that you are rolling FOUR dice at the same time: the sequence above can NEVER happen to you because happen in turn launch 2 and 3 where your 4 dices can only generate something like XXXX XXXX SSSS XXXX followed or preceded by a number of trials multiple of four.
As I wrote in the notes I tried calculating the probability using the QUAD SKULL prob (1/36)*(1/36) on 69 trials, so n=69, (not 70) and it comes out very close to the calculation in the article (5,052%) but not 5.05562242% so from mathematical point of view is not exactly the same. Therein lies the curiosity that we have not been able to explain.
Posted by RDaneel on 2024-09-15 10:23:01
The chance to have at least one quad skull launching 70 times 4 dices is 5,121%
You have to use Eq.2 in the note, with n=70, k=1, p=(1/36)*(1/36), q=(1-p).
You have to use Eq.2 in the note, with n=70, k=1, p=(1/36)*(1/36), q=(1-p).
Posted by Joost on 2024-09-15 11:47:45
I still don’t see how your example is fundamentally different, unless you lay the sequence out in a non-random way. Assuming every next roll is randomly determined you need 4 skulls right after each other, each at a 1/6 opportunity to occur. I don’t see how it matters whether you roll them in twos or fours, even if your condition is that you only roll another two dice after the first two are skulls.
Posted by RDaneel on 2024-09-15 13:07:27
The reason is that we are talking about two different experiments. We must all agree that the definition of classical probability is Eq. (1) in the note.
Now in the quad skull case that gives you 5.4% I think you are omitting a favorable case compared to the case of the double skull followed by the double skull.
For example this sequence XX SS SS is "a success" or "favorable case" in the experiment "roll 2d6 in sequence". This event is "impossible" in the case of quad skull because XX SS followed by SS XX is not a favorable case.
In sum, for me, the values of NA and N referring to definition in eq .(1) in the two cases are slightly different. There is One probability over 1296 possibilities of making ONE quad skull by rolling four dice.
The probability of seeing this event over 70 trials is definitely 70/1296= 5.4%.
But is not what we are speaking about in this blog in my opinion ( i had exactly the same thinking at the beginning but then the brute force algorithm told me that i was wrong).
Now in the quad skull case that gives you 5.4% I think you are omitting a favorable case compared to the case of the double skull followed by the double skull.
For example this sequence XX SS SS is "a success" or "favorable case" in the experiment "roll 2d6 in sequence". This event is "impossible" in the case of quad skull because XX SS followed by SS XX is not a favorable case.
In sum, for me, the values of NA and N referring to definition in eq .(1) in the two cases are slightly different. There is One probability over 1296 possibilities of making ONE quad skull by rolling four dice.
The probability of seeing this event over 70 trials is definitely 70/1296= 5.4%.
But is not what we are speaking about in this blog in my opinion ( i had exactly the same thinking at the beginning but then the brute force algorithm told me that i was wrong).
Posted by Beltroniko on 2024-09-15 18:50:52
This part of the discussion is quite interesting to me, in fact my initial thoughts on discord were that it could be calculated with 69 4 dice rolls because it seemed logical to me that 70 2-dice rolls could be represented as 69 4-dice rolls. But Christer wrote down all possibilities for rolling heads twice in a row with 4 coin flips and it happens 50% of the times, while using the above method (which would mean 3 double coin throws) gives out a different result. I still don't understand why though. Maybe drop by the Math Thread in the #general discord channel Joost, maybe you can help me understanding why this is different
Posted by RDaneel on 2024-09-15 19:15:53
On the other hand, I still do not understand why calculating the quad skull on 69 throws (of four dice) is very similar to calculating the double skull on 70 throws (but as you say, it is not the same thing). Yes I need help
I think Joost is now trying to calculate a different exercise:
"what is a probability to roll doubleskull-doubleskull in a row, then doubleskull-doublebothdown in a row and finally a double skull in , let's say , 45 2d6 blocks ( so during a drive)? "
Seems too low? Oh no... look 1st drive here
https://fumbbl.com/p/match?id=4572283
8-0 ! ! !
welcome to the real world...
I think Joost is now trying to calculate a different exercise:
"what is a probability to roll doubleskull-doubleskull in a row, then doubleskull-doublebothdown in a row and finally a double skull in , let's say , 45 2d6 blocks ( so during a drive)? "
Seems too low? Oh no... look 1st drive here
https://fumbbl.com/p/match?id=4572283
8-0 ! ! !
welcome to the real world...
Posted by JackassRampant on 2024-09-15 20:55:01
@RDaneel, the .054 is the expected mean number of quads, while the slightly lesser number is the chance of any quads.
Posted by Joost on 2024-09-16 08:33:11
Daneel is referring to the game we played yesterday, where I (oh irony) rolled a quad skull AND a double skull / double both down. In my first 7 blocks.