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Made in in
[MOD]
Otiose in a Niche






Hyderabad, India

OK so I have a math question, doesn't affect anything but it's bugging me.

Let's say I put down 6 objective markers and each time the players open on there's a 1 in 6 chance it will have the prize.

Well this leaves the chance that none will have the prize if no one rolls a 6 (or whatever). So I decide that if 1-5 don't have the prize then #6 must.

So what are the chances that the last box will be the one with the prize?

My first thought was it would be 1-(chance of 5 failed rolls)

1 failed roll 5/6
5 failed rolls in a row 25/36
So chance of #6 having the prize is 11/36

But then the chance of the first 5 having the prize is not 1/6, more like 1/7 (25/36 divided by 5)... which doesn't make sense either.

I feel like there's an obvious answer but I can get it.

If I did playing cards it would be easier since you'd eliminate a card each time.

Box 1 - 1 in 6
Box 2 - 1 in 5
etc
Box 6 - 1 in 1

 
   
Made in gb
Leader of the Sept







With combined probabilities should it not be an And combination? You are

5/6 x 5/6 x 5/6 x 5/6 x 5/6 = 3125/7776

Then the final roll needs to be 1/6 leading to

3125/46656

On my phone so not sure what that collapses to.

Also I haven’t done probabilities for a long time so this might be tiredly nonsense

This message was edited 1 time. Last update was at 2022/05/03 20:51:51


Please excuse any spelling errors. I use a tablet frequently and software keyboards are a pain!

Terranwing - w3;d1;l1
51st Dunedinw2;d0;l0
Cadre Coronal Afterglow w1;d0;l0 
   
Made in gb
Decrepit Dakkanaut







Aye, Kid, you done goofed - right idea on 1-(likelihood of failing 5 rolls), but you didn't work out the bit in brackets correctly.

5 failed rows in a row isn't a 25/36 chance. You start with the initial 5/6, then for each additional failed roll you're multiplying by 5/6 again.

5/6
25/36 (5/6 x 5/6)
125/216 (5/6 x 5/6 x 5/6)
625/1296 (5/6 x 5/6 x 5/6 x 5/6)
3125/7776 (5/6 x 5/6 x 5/6 x 5/6 x 5/6)

1-3125/7776 gives you 4651/7776, or roughly 59.8% chance, if we assume that the sixth box must be right if the other five fail.

2021-4 Plog - Here we go again... - my fifth attempt at a Dakka PLOG

My Pile of Potential - updates ongoing...

Gamgee on Tau Players wrote:we all kill cats and sell our own families to the devil and eat live puppies.


 Kanluwen wrote:
This is, emphatically, why I will continue suggesting nuking Guard and starting over again. It's a legacy army that needs to be rebooted with a new focal point.

Confirmation of why no-one should listen to Kanluwen when it comes to the IG - he doesn't want the IG, he want's Kan's New Model Army...

tneva82 wrote:
You aren't even trying ty pretend for honest arqument. Open bad faith trolling.
- No reason to keep this here, unless people want to use it for something... 
   
Made in in
[MOD]
Otiose in a Niche






Hyderabad, India

 Dysartes wrote:
Aye, Kid, you done goofed - right idea on 1-(likelihood of failing 5 rolls), but you didn't work out the bit in brackets correctly.

5 failed rows in a row isn't a 25/36 chance. You start with the initial 5/6, then for each additional failed roll you're multiplying by 5/6 again.

5/6
25/36 (5/6 x 5/6)
125/216 (5/6 x 5/6 x 5/6)
625/1296 (5/6 x 5/6 x 5/6 x 5/6)
3125/7776 (5/6 x 5/6 x 5/6 x 5/6 x 5/6)

1-3125/7776 gives you 4651/7776, or roughly 59.8% chance, if we assume that the sixth box must be right if the other five fail.


Thanks, I was wondering if the answer wasn't 5/6^5 rather than 5/6 * 5.

But going back, is the chance of any individual box having the treasure still 1/6 ~17% or is 59.8%/5 = ~12%?


 
   
Made in gb
Leader of the Sept







I don’t think so. If you keep going until you find the thing, then the first is 1/6

The second is 5/6 x 1/6
Third 5/6 x 5/6 x 1/6
Etc


Please excuse any spelling errors. I use a tablet frequently and software keyboards are a pain!

Terranwing - w3;d1;l1
51st Dunedinw2;d0;l0
Cadre Coronal Afterglow w1;d0;l0 
   
Made in gb
Decrepit Dakkanaut







 Kid_Kyoto wrote:
 Dysartes wrote:
Aye, Kid, you done goofed - right idea on 1-(likelihood of failing 5 rolls), but you didn't work out the bit in brackets correctly.

5 failed rows in a row isn't a 25/36 chance. You start with the initial 5/6, then for each additional failed roll you're multiplying by 5/6 again.

5/6
25/36 (5/6 x 5/6)
125/216 (5/6 x 5/6 x 5/6)
625/1296 (5/6 x 5/6 x 5/6 x 5/6)
3125/7776 (5/6 x 5/6 x 5/6 x 5/6 x 5/6)

1-3125/7776 gives you 4651/7776, or roughly 59.8% chance, if we assume that the sixth box must be right if the other five fail.


Thanks, I was wondering if the answer wasn't 5/6^5 rather than 5/6 * 5.

But going back, is the chance of any individual box having the treasure still 1/6 ~17% or is 59.8%/5 = ~12%?

The chance of any individual box having the prize is 1/6 - if it is the first box looked in.

As Flinty points out, the probability then changes for subsequent pulls - it isn't a flat distribution.

Box 1 - 1/6 = 1/6 = 16.66%
Box 2 - 5/36 = 1/6 x 5/6 = 13.89%
Box 3 - 25/216 = 1/6 x 5/6 x 5/6 = 11.57%
Box 4 - 125/1296 = 1/6 x 5/6 x 5/6 x 5/6 = 9.64%
Box 5 - 625/7776 = 1/6 x 5/6 x 5/6 x 5/6 x 5/6 = 8.03%
Box 6 - 4651/7776 = 1-(5/6 x 5/6 x 5/6 x 5/6 x 5/6) = 59.8%

I do note that these don't add up to 100%, but I think that's OK in this case, given what we're stating about each box.

If I had a way of knowing which the last box would be, that's where I'm camping for the scenario

2021-4 Plog - Here we go again... - my fifth attempt at a Dakka PLOG

My Pile of Potential - updates ongoing...

Gamgee on Tau Players wrote:we all kill cats and sell our own families to the devil and eat live puppies.


 Kanluwen wrote:
This is, emphatically, why I will continue suggesting nuking Guard and starting over again. It's a legacy army that needs to be rebooted with a new focal point.

Confirmation of why no-one should listen to Kanluwen when it comes to the IG - he doesn't want the IG, he want's Kan's New Model Army...

tneva82 wrote:
You aren't even trying ty pretend for honest arqument. Open bad faith trolling.
- No reason to keep this here, unless people want to use it for something... 
   
Made in au
Owns Whole Set of Skullz Techpriests






Versteckt in den Schatten deines Geistes.

Is this a Monty Hall problem?

Industrial Insanity - My Terrain Blog
"GW really needs to understand 'Less is more' when it comes to AoS." - Wha-Mu-077

 
   
Made in us
Stone Bonkers Fabricator General






A garden grove on Citadel Station

 Kid_Kyoto wrote:
OK so I have a math question, doesn't affect anything but it's bugging me.

Let's say I put down 6 objective markers and each time the players open on there's a 1 in 6 chance it will have the prize.

Well this leaves the chance that none will have the prize if no one rolls a 6 (or whatever). So I decide that if 1-5 don't have the prize then #6 must.

So what are the chances that the last box will be the one with the prize?

My first thought was it would be 1-(chance of 5 failed rolls)

1 failed roll 5/6
5 failed rolls in a row 25/36
So chance of #6 having the prize is 11/36

But then the chance of the first 5 having the prize is not 1/6, more like 1/7 (25/36 divided by 5)... which doesn't make sense either.

I feel like there's an obvious answer but I can get it.

If I did playing cards it would be easier since you'd eliminate a card each time.

Box 1 - 1 in 6
Box 2 - 1 in 5
etc
Box 6 - 1 in 1


If you have 6 chances to roll a 6 on 1d6, and you get it guaranteed on the last chance, really what is being wondered is, what's the chance that I roll a 1-5, 5 times in a row? The chance to roll 1-5 on a d6 is 5/6.

5/6 * 5/6 * 5/6 * 5/6 * 5/6 = 3125 / 7776, or a 40.2% chance that you don't get a 6 on your first 5 rolls, which is a 59.8% chance that you get a 6 on any of the first 5 rolls.

EDIT: And on each individual roll, the chance to get a 6 is always 17%. First roll? 1 in 6 chance it will be a 6. Failed the first roll, on to the second roll? Still a 17% chance this 1 in 6 roll is a 6. Every roll is individually a 17% chance to get that 6 when you are on that roll.

This message was edited 1 time. Last update was at 2022/05/03 22:57:26


ph34r's Forgeworld Phobos blog, current WIP: Iron Warriors and Skaven Tau
+From Iron Cometh Strength+ +From Strength Cometh Will+ +From Will Cometh Faith+ +From Faith Cometh Honor+ +From Honor Cometh Iron+
The Polito form is dead, insect. Are you afraid? What is it you fear? The end of your trivial existence?
When the history of my glory is written, your species shall only be a footnote to my magnificence.
 
   
Made in ch
Irked Necron Immortal




Switzerland

 H.B.M.C. wrote:
Is this a Monty Hall problem?

no, this one is actually much simpler then it looks at first




Automatically Appended Next Post:
if I understand the question correctly

: "6th" iff "5 failures"
ergo
0.402... = (5/6)^5


Automatically Appended Next Post:
 Flinty wrote:
I don’t think so. If you keep going until you find the thing, then the first is 1/6

The second is 5/6 x 1/6
Third 5/6 x 5/6 x 1/6
Etc

ye, this is simple enough

This message was edited 2 times. Last update was at 2022/05/03 23:07:06


 
   
Made in gb
Leader of the Sept







@Dysartes - I think your last line has an error. It’s not just 1-(the last one). If the last box is guaranteed to have the prize then it’s 1-(16+13+11+9+8) because there is a cumulative aspect as well I think.might be wrong though.

If the last box is also a straight roll then It does t have to add up to 1 as it’s not a zero sum condition. There is a chance of just not getting the prize in 6 rolls.

Please excuse any spelling errors. I use a tablet frequently and software keyboards are a pain!

Terranwing - w3;d1;l1
51st Dunedinw2;d0;l0
Cadre Coronal Afterglow w1;d0;l0 
   
Made in au
FOW Player




 Kid_Kyoto wrote:
OK so I have a math question, doesn't affect anything but it's bugging me.

Let's say I put down 6 objective markers and each time the players open on there's a 1 in 6 chance it will have the prize.

Well this leaves the chance that none will have the prize if no one rolls a 6 (or whatever). So I decide that if 1-5 don't have the prize then #6 must.

So what are the chances that the last box will be the one with the prize?


I don't want to spoil this very interesting maths conversation, but if you were to do this in real life, wouldn't you just mark the underside of one objective? (e.g. six cardboard chits that all look the same on one side, but on the other side there's one marked 'PRIZE'?)
   
Made in fi
Locked in the Tower of Amareo





Zenithfleet wrote:
 Kid_Kyoto wrote:
OK so I have a math question, doesn't affect anything but it's bugging me.

Let's say I put down 6 objective markers and each time the players open on there's a 1 in 6 chance it will have the prize.

Well this leaves the chance that none will have the prize if no one rolls a 6 (or whatever). So I decide that if 1-5 don't have the prize then #6 must.

So what are the chances that the last box will be the one with the prize?


I don't want to spoil this very interesting maths conversation, but if you were to do this in real life, wouldn't you just mark the underside of one objective? (e.g. six cardboard chits that all look the same on one side, but on the other side there's one marked 'PRIZE'?)


Sure you could.

That however does alter the odds and probababilities IN GAME. Which might or might not be what you want.

2024 painted/bought: 109/109 
   
Made in gb
Decrepit Dakkanaut







 Flinty wrote:
@Dysartes - I think your last line has an error. It’s not just 1-(the last one). If the last box is guaranteed to have the prize then it’s 1-(16+13+11+9+8) because there is a cumulative aspect as well I think.might be wrong though.

If the last box is also a straight roll then It does t have to add up to 1 as it’s not a zero sum condition. There is a chance of just not getting the prize in 6 rolls.

Hang on - I actually think the 1- element on my calculation is the problem.

The only way we're getting to box 6 is if the first five are all wrong, hence the 5/6^5 - but if I do 1- that, I'm not then saying that figure. The chance of box 6 being right is just the 3125/776.

Gives this distribution instead:

Box 1 - 1/6 = 1/6 = 16.66%
Box 2 - 5/36 = 1/6 x 5/6 = 13.89%
Box 3 - 25/216 = 1/6 x 5/6 x 5/6 = 11.57%
Box 4 - 125/1296 = 1/6 x 5/6 x 5/6 x 5/6 = 9.64%
Box 5 - 625/7776 = 1/6 x 5/6 x 5/6 x 5/6 x 5/6 = 8.03%
Box 6 - 3125/7776 = 5/6 x 5/6 x 5/6 x 5/6 x 5/6 = 40.2%

2021-4 Plog - Here we go again... - my fifth attempt at a Dakka PLOG

My Pile of Potential - updates ongoing...

Gamgee on Tau Players wrote:we all kill cats and sell our own families to the devil and eat live puppies.


 Kanluwen wrote:
This is, emphatically, why I will continue suggesting nuking Guard and starting over again. It's a legacy army that needs to be rebooted with a new focal point.

Confirmation of why no-one should listen to Kanluwen when it comes to the IG - he doesn't want the IG, he want's Kan's New Model Army...

tneva82 wrote:
You aren't even trying ty pretend for honest arqument. Open bad faith trolling.
- No reason to keep this here, unless people want to use it for something... 
   
Made in gb
Dakka Veteran




Lincoln, UK

You have it right @Dysartes. It really is (5/6)^5, or 40.18% There's a 59.82% chance that there will be one or more successful finds before you reach the 6th objective.

The best way to draw it out is as a tree, where each branch "level" splits again between the 5/6 "fail" and 1/6 "success" branches.

The number of branches doubles (since there are 2 outcomes for each branch). Each endpoint on the final branch tips has a cumulative probability found by multiplying the five numbers at each level together (although the individual number on any given branch is either 5/6 or 1/6, because any single dice roll has a 1/6 chance of succeeding). There will be a different number of successes, from 0 to 5 at each tip point too.

To get to the condition where it must be in the 6th box, you're following the branching path where it fails five times - there can only be one tip where all five fail, out of 2^5 or 32 tips. That's 5/6 x 5/6 x 5/6 x 5/6 x 5/6, or 40.18%

(Binomial theorem also works - again, you're looking for the probability of five occurrences of a 5/6 probability event on 5 tests. You can also add up probabilities for the counts of "number of successes" if the Binomial Theorem feels like too much of a faff).

This message was edited 4 times. Last update was at 2022/05/05 16:46:20


 
   
Made in gb
Longtime Dakkanaut





Oxfordshire

 Dysartes wrote:

Box 1 - 1/6 = 1/6 = 16.66%
Box 2 - 5/36 = 1/6 x 5/6 = 13.89%
Box 3 - 25/216 = 1/6 x 5/6 x 5/6 = 11.57%
Box 4 - 125/1296 = 1/6 x 5/6 x 5/6 x 5/6 = 9.64%
Box 5 - 625/7776 = 1/6 x 5/6 x 5/6 x 5/6 x 5/6 = 8.03%
Box 6 - 3125/7776 = 5/6 x 5/6 x 5/6 x 5/6 x 5/6 = 40.2%


This solves Kid_Kyoto's original question but another way of representing this which makes intuitive sense (and would help understand the likely outcome of the game) is to ask:
"Before I start turning boxes, what are the chances I will have found the prize after the turn of a certain box?"
Then you get (we will have different rounding errors):

Box 1 - 16.67%
Box 2 - 30.56%
Box 3 - 42.13%
Box 4 - 51.77%
Box 5 - 59.81%
Box 6 - 100%
   
Made in in
[MOD]
Otiose in a Niche






Hyderabad, India

Zenithfleet wrote:


I don't want to spoil this very interesting maths conversation, but if you were to do this in real life, wouldn't you just mark the underside of one objective? (e.g. six cardboard chits that all look the same on one side, but on the other side there's one marked 'PRIZE'?)


There's lots of ways to do it of course. I like the idea of rolling because it gives players the feeling that they did something, they rolled their lucky dice and did their lucky dance and wore their lucky underwear so the player made it happen, it wasn't just there.

It's the same reason I like GW's armor save mechanism over a straight "weapon vs armor" type chart. A player feels like he saved the unit by saying "I need sixes" rather than the cold equations determining everything.

So you could use chits or secretly note where the treasure is but this way no one knows. Drawing cards would also work but I think dice are more fun, they add a kinetic element to the game. Which raised the possibility of no one rolling a six which led to the question.

For what it's worth, the scenario I was thinking of would require finding the treasure and then getting off the board with it so it would not be the end of the game.

 
   
Made in au
FOW Player




 Kid_Kyoto wrote:
Zenithfleet wrote:


I don't want to spoil this very interesting maths conversation, but if you were to do this in real life, wouldn't you just mark the underside of one objective? (e.g. six cardboard chits that all look the same on one side, but on the other side there's one marked 'PRIZE'?)


There's lots of ways to do it of course. I like the idea of rolling because it gives players the feeling that they did something, they rolled their lucky dice and did their lucky dance and wore their lucky underwear so the player made it happen, it wasn't just there.

It's the same reason I like GW's armor save mechanism over a straight "weapon vs armor" type chart. A player feels like he saved the unit by saying "I need sixes" rather than the cold equations determining everything.

So you could use chits or secretly note where the treasure is but this way no one knows. Drawing cards would also work but I think dice are more fun, they add a kinetic element to the game. Which raised the possibility of no one rolling a six which led to the question.

For what it's worth, the scenario I was thinking of would require finding the treasure and then getting off the board with it so it would not be the end of the game.


Now now Kid, you're not allowed to have fun in a way I disapprove of!

Fair enough. I was just reminded of all those threads on Boardgamegeek or similar where someone will say, "I don't like shuffling and drawing cards, so I've come up with a very simple mathematical method of randomising it," and then describe a homebrew dice and chart system that takes 4x as long to do and 10x as long to explain ...

Your point about the psychological enjoyment of certain methods (even if they might be inefficient) is a really good one that could be its own thread. I'm fairly sure GW designers themselves have said that the save mechanic was important to keep each player engaged in an otherwise 'Igo Ugo' system where they might tune out. As well as allowing some extra finely granulated probabilities on six-sided dice by having three steps instead of two.

One of my favourite mechanics in GW games is the 'Brace for Impact' rule in Battlefleet Gothic, where (as you know Bob) you have to decide whether or not you want to sacrifice firepower now in order to roll saves against damage on your opponent's turn.

Conversely, as a kid I used to get frustrated by Milton Bradley's Space Crusade (even though it was my first approximation of a wargame). Not only did it lack a save mechanic, it combined 'to hit' and 'to wound' into a single roll on specialty dice. Most of the faces of those dice said '0'. It might have been mathematically fine, but it felt as if you were constantly missing at point-blank range because you were constantly rolling zeroes.

BTW, did you know that many animals, including pigeons, will develop superstitions about 'lucky dances' and so on if you randomise their feeding times?
   
Made in fi
Locked in the Tower of Amareo





Prize on 1 card and not rolled is def valid idea. But does alter odds and as such strategy notably. Especially as possible locations are reduced. When you look at first odds are 1/6. When 2 left your odds off finding treasure is 50% with your method. Still 1/6 with dice.

2024 painted/bought: 109/109 
   
Made in gb
Leader of the Sept







Roll the dice. If you don’t find it in the 6th one, then the lucky prize pig is unleashed and whatever remains of both forces must chase and subdue it in order to find the treasure!


Automatically Appended Next Post:
Sorry TRESHUUUUUUURE!

This message was edited 1 time. Last update was at 2022/05/06 21:36:09


Please excuse any spelling errors. I use a tablet frequently and software keyboards are a pain!

Terranwing - w3;d1;l1
51st Dunedinw2;d0;l0
Cadre Coronal Afterglow w1;d0;l0 
   
 
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