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Strategy Math Question 2


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#1 astros

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Posted 04 May 2018 - 06:26 AM

Assume that your opponent's flag is randomly placed on his back row. The remainder of his pieces are randomly placed.

You have a special piece, the Zero, that kills any piece except bombs. Let the Zero start at A4, your left hand-side front row, moves vertically to A10 and then(if needed) horizontally to J10. What is the probability that you capture your opponents flag before hitting a bomb?
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#2 Napoleon 1er

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Posted 04 May 2018 - 07:00 AM

It is equal to the average of the probability to hit a bomb within 4 lottos and the probability to hit a bomb within 13 lottos, assuming the game starts with all 40 pieces on the board.
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#3 TemplateRex

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Posted 04 May 2018 - 08:18 AM

If you want to compute this accurately, you have to do some conditional probability. It's like Russian roulette, after every empty "click", the probability of a bullet coming up increases. So I did the following.

 

Standing at A6, the chances of A7 being a bomb are 6/39. It's not 6/40 because the flag is on the back row, and that leaves 39 squares for bombs. Surviving a bomb on A7, the chances of A8 being a bomb are 6/38, then on A9 it's 6/37. After hitting on A9, the Zero piece is alive in 59.7% of the cases. Then hitting on A10, it's a flag in 1/10 of the cases, and a bomb in 6/36 of the cases. Hitting on B10, the flag is now 1/9 of the cases, given that you still have to hit. Doing all the multiplications in Excel, I get a 28.8% chance of hitting the flag before losing the Zero piece.

 

Another interesting tidbit: the Zero piece wins on average only 1.89 pieces of the opponent before it dies.

 

Below a graph showing how the cumulative probabilities of being alive, having hit a bomb or the flag change as you move from A6 to A10 and then J10.

 

TIYLcGx.jpg


Edited by TemplateRex, 04 May 2018 - 08:39 AM.

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#4 TemplateRex

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Posted 04 May 2018 - 09:12 AM

BTW, the title suggests there was a Strategy math question 1, but I can't find it, do you have a link? (or was it the Q about the number of initial setups?)


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#5 Major Nelson

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Posted 04 May 2018 - 09:42 AM

Are all 6 bombs (in rows 7-10) still in the game?
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#6 TemplateRex

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Posted 04 May 2018 - 09:45 AM

How many bombs are there at the back row?

Posted 3 hours ago

Assume that your opponent's flag is randomly placed on his back row. The remainder of his pieces are randomly placed. 

Edited by TemplateRex, 04 May 2018 - 09:46 AM.

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#7 Morx

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Posted 04 May 2018 - 09:46 AM

Based on the puzzle input, they are randomly distributed on the field:  "The remainder of his pieces are randomly placed."

 

Any bombs on A7,A8 and A9 or on the back row before the flag is found make this lottoing with the ZERO piece a bad plan.



#8 Major Nelson

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Posted 04 May 2018 - 09:47 AM


Posted 3 hours ago

I changed my question.
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#9 TemplateRex

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Posted 04 May 2018 - 09:55 AM

I changed my question.

 

OK, the OP wasn't 100% clear, but I assume a full setup, and red having a Zero piece on A4, starting to lotto right out of the gate.


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#10 TemplateRex

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Posted 04 May 2018 - 10:33 AM

Based on the puzzle input, they are randomly distributed on the field:  "The remainder of his pieces are randomly placed."

 

Any bombs on A7,A8 and A9 or on the back row before the flag is found make this lottoing with the ZERO piece a bad plan.

 

I think I could write a program to simulate lotto-ing out of the gate with any piece rank using the Gravon database, and see what it would give on average. I doubt there is a 28.8% chance of finding a flag unopposed though. 


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#11 astros

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Posted 04 May 2018 - 04:14 PM

If you want to compute this accurately, you have to do some conditional probability. It's like Russian roulette, after every empty "click", the probability of a bullet coming up increases. So I did the following.

 

Standing at A6, the chances of A7 being a bomb are 6/39. It's not 6/40 because the flag is on the back row, and that leaves 39 squares for bombs. Surviving a bomb on A7, the chances of A8 being a bomb are 6/38, then on A9 it's 6/37. After hitting on A9, the Zero piece is alive in 59.7% of the cases. Then hitting on A10, it's a flag in 1/10 of the cases, and a bomb in 6/36 of the cases. Hitting on B10, the flag is now 1/9 of the cases, given that you still have to hit. Doing all the multiplications in Excel, I get a 28.8% chance of hitting the flag before losing the Zero piece.

 

Another interesting tidbit: the Zero piece wins on average only 1.89 pieces of the opponent before it dies.

 

Below a graph showing how the cumulative probabilities of being alive, having hit a bomb or the flag change as you move from A6 to A10 and then J10.

 

TIYLcGx.jpg

Not quite what I got. To make the calculations easier, you can assume that there is a 1/10 chance that the flag is any given column of the back row, then just calculate the probability of reaching that column.

 

Also reconsider your treatment of A10.


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#12 TemplateRex

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Posted 04 May 2018 - 04:41 PM

Not quite what I got. To make the calculations easier, you can assume that there is a 1/10 chance that the flag is any given column of the back row, then just calculate the probability of reaching that column.
 
Also reconsider your treatment of A10.


I did compute chances of reaching different squares, conditional on not reaching flag/bomb earlier, see my post. I assume flag is randomly assigned to the 10 back row squares. What is your number?

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#13 astros

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Posted 04 May 2018 - 04:44 PM

I did compute chances of reaching different squares, conditional on not reaching flag/bomb earlier, see my post. I assume flag is randomly assigned to the 10 back row squares. What is your number?

I will message you privately as I do not want to publicly post my answer yet.


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#14 Napoleon 1er

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Posted 04 May 2018 - 05:30 PM

If you want to compute this accurately, you have to do some conditional probability. It's like Russian roulette, after every empty "click", the probability of a bullet coming up increases. So I did the following.

 

Standing at A6, the chances of A7 being a bomb are 6/39. It's not 6/40 because the flag is on the back row, and that leaves 39 squares for bombs. Surviving a bomb on A7, the chances of A8 being a bomb are 6/38, then on A9 it's 6/37. After hitting on A9, the Zero piece is alive in 59.7% of the cases. Then hitting on A10, it's a flag in 1/10 of the cases, and a bomb in 6/36 of the cases. Hitting on B10, the flag is now 1/9 of the cases, given that you still have to hit. Doing all the multiplications in Excel, I get a 28.8% chance of hitting the flag before losing the Zero piece.

 

Another interesting tidbit: the Zero piece wins on average only 1.89 pieces of the opponent before it dies.

 

Below a graph showing how the cumulative probabilities of being alive, having hit a bomb or the flag change as you move from A6 to A10 and then J10.

 

TIYLcGx.jpg

I'm not an expert in statistics but it seems to me that you shall not have a 0% chance to be alive in J10. At a guess I would say it should be 3-5%


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#15 astros

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Posted 04 May 2018 - 05:45 PM

Guess I did mess it up a bit
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#16 The Prof

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Posted 04 May 2018 - 06:31 PM

I can confirm TemplateRex's answer.  I get 28.834%.


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#17 TemplateRex

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Posted 04 May 2018 - 06:46 PM

I'm not an expert in statistics but it seems to me that you shall not have a 0% chance to be alive in J10. At a guess I would say it should be 3-5%


With alive I mean not having hit a bomb or flag.

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#18 Napoleon 1er

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Posted 04 May 2018 - 06:48 PM

yes I agree but in J10 your probability to be alive is higher than 0, right? you have a 0 probability to be alive only after 33 lottos, right?


Edited by Napoleon 1er, 04 May 2018 - 06:52 PM.

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#19 astros

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Posted 04 May 2018 - 06:55 PM

yes I agree but in J10 your probability to be alive is higher than 0, right? you have a 0 probability to be alive only after 33 lottos, right?


Yes you can be alive after J10. After 34 lottos you can be alive too because you can get the flag on lotto 34.

Edited by astros, 04 May 2018 - 06:57 PM.

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#20 Napoleon 1er

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Posted 04 May 2018 - 06:58 PM

agree, please read 34 lottos instead of 33 lottos


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