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Strategy Math Question 2


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#21 The Prof

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Posted 04 May 2018 - 07:31 PM

For average number of pieces captured I get 3.8873.

 

This is a bit tricky because capturing 4 pieces, for example, could happen like M,M,M,M,B or M,M,M,F.

 

I would add this somewhat counter-intuitive  tidbit.  The probability of capturing a certain number of pieces, in order from most likely to least likely, is:  0, 1, 4, 2, 5, 3, 6, 7, 8, 9, 10, 11, 12, 13.



#22 TemplateRex

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Posted 04 May 2018 - 08:01 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?

No, after hitting J10, you have hit the flag, guaranteed, since it is on the last row. With “alive" I meant not having hit a bomb or flag, so you have to continue hitting. After hitting a flag earlier you are no longer alive since the game is over and you cannot continue lottoing (but you are in heaven :)). Sorry if this is confusing terminology.

Edited by TemplateRex, 04 May 2018 - 08:14 PM.

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#23 TemplateRex

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Posted 04 May 2018 - 08:09 PM

For average number of pieces captured I get 3.8873.

This is a bit tricky because capturing 4 pieces, for example, could happen like M,M,M,M,B or M,M,M,F.

I would add this somewhat counter-intuitive tidbit. The probability of capturing a certain number of pieces, in order from most likely to least likely, is: 0, 1, 4, 2, 5, 3, 6, 7, 8, 9, 10, 11, 12, 13.

I computed 1.89 as the number of pieces the Zero captures before hitting a bomb (in 71.2% of the cases), but I disregarded captured pieces if the flag gets captured. I also would not count the flag as a captured piece.

Starting at A6, the trade off is getting 1.89 pieces on average 71.2% of the time for your Zero, or capturing the flag outright in the remaining 28.8% of the games. Of course, this disregards stopping earlier, eg after capturing Marsh on A7 you could savely stop lottoing.

Edited by TemplateRex, 04 May 2018 - 08:17 PM.

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#24 Unladen Swallow

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Posted 04 May 2018 - 08:22 PM

Of course, this disregards stopping earlier, eg after capturing Marsh on A7 you could savely stop lottoing.

 

It depends what piece the "ZERO" replaces. 

 

If the Zero was replacing a scout, then surely, it makes sense to lotto it until it loses to a bomb. 


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#25 TemplateRex

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Posted 04 May 2018 - 08:49 PM

It depends what piece the "ZERO" replaces. 
 
If the Zero was replacing a scout, then surely, it makes sense to lotto it until it loses to a bomb. 


Why not keep it hidden like a super marsh?

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#26 Napoleon 1er

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Posted 04 May 2018 - 10:08 PM

No, after hitting J10, you have hit the flag, guaranteed, since it is on the last row. With “alive" I meant not having hit a bomb or flag, so you have to continue hitting. After hitting a flag earlier you are no longer alive since the game is over and you cannot continue lottoing (but you are in heaven :)). Sorry if this is confusing terminology.

... look ... i guess that in stratego "remaining alive" means "not have died" ... so who captures the flag does not die but is still alive then. .... but you are right Astros was not really precise enough in the description of his maths .... :P  ... still how do  you calculate the probability to have hit a bomb after X lotto? ... for me the right answer is the average between the probability to hit a bomb after 4 lotto and the probability to hit a bomb after 12 lotto (cause if you make 12  successful lotto in a row then the 13th is obviously the flag) ... but which average ... mathematical average or geometrical average or another average?


If you don't know where you go ... you have a lot of chance to arrive elsewhere ...

#27 TemplateRex

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Posted 04 May 2018 - 10:22 PM

... look ... i guess that in stratego "remaining alive" means "not have died" ... so who captures the flag does not die but is still alive then. .... but you are right Astros was not really precise enough in the description of his maths .... :P ... still how do you calculate the probability to have hit a bomb after X lotto? ... for me the right answer is the average between the probability to hit a bomb after 4 lotto and the probability to hit a bomb after 12 lotto (cause if you make 12 successful lotto in a row then the 13th is obviously the flag) ... but which average ... mathematical average or geometrical average or another average?


There are several ways to compute it. I have a spreadsheet that I will post in a few days. I use a simple decision tree and multiply probabilities along paths. You can also use an averaging strategy, but you need complicated weights per back row square.

Edited by TemplateRex, 04 May 2018 - 10:26 PM.

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#28 Napoleon 1er

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Posted 04 May 2018 - 10:23 PM

but what is the formula?


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#29 TemplateRex

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Posted 04 May 2018 - 10:28 PM

but what is the formula?


Google for binomial distribution. I am not near my pc and tools to write out the formulas. Will update in a few days.

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#30 The Prof

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Posted 04 May 2018 - 11:43 PM

Napoleon, you can compute the answer like this.

 

(0.1)(33/39)(32/38)(31/37) +

(0.1)(33/39)(32/38)(31/37)(30/36) +

(0.1)(33/39)(32/38)(31/37)(30/36)(29/35) + 

(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34) +

(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34)(27/33) +

(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34)(27/33)(26/32) +

(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34)(27/33)(26/32)(25/31) + 

(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34)(27/33)(26/32)(25/31)(24/30) + 

(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34)(27/33)(26/32)(25/31)(24/30)(23/29) +

(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34)(27/33)(26/32)(25/31)(24/30)(23/29)(22/28) = 0.28834


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#31 astros

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Posted 05 May 2018 - 12:20 AM

Napoleon, you can compute the answer like this.
 
(0.1)(33/39)(32/38)(31/37) +
(0.1)(33/39)(32/38)(31/37)(30/36) +
(0.1)(33/39)(32/38)(31/37)(30/36)(29/35) + 
(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34) +
(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34)(27/33) +
(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34)(27/33)(26/32) +
(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34)(27/33)(26/32)(25/31) + 
(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34)(27/33)(26/32)(25/31)(24/30) + 
(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34)(27/33)(26/32)(25/31)(24/30)(23/29) +
(0.1)(33/39)(32/38)(31/37)(30/36)(29/35)(28/34)(27/33)(26/32)(25/31)(24/30)(23/29)(22/28) = 0.28834


That was my solution, I just incorporated an extra term into each sequence.
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#32 GaryLShelton

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Posted 05 May 2018 - 12:49 AM

There needs to be a statistical corner for The Prof and TemplateRex to bide their time.  :)

 

Has there been any kind of practical analysis of how the numbers might change with low ranked opponents to higher?



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#33 Morx

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Posted 05 May 2018 - 08:23 AM

This is just a simplified abstract probliem. What Rex is suggesting is using a large dataset of real games to compute lotto chances on setups.

There the spy and marshall trading need to be modelled.

#34 Guldin

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Posted 05 May 2018 - 02:16 PM

is there a newer math i don't know about or forgot? there are 6 bombs and 1 flag so the odds would be 6 to 1 against capturing your opponents flag before hitting a bomb (14.2857143 %). this would also be the same probability of rolling a 2 or rolling a 12 before rolling a 7 with a pair of dice. i guess this is why it is so hard for many of you to calculate why a tourney in vegas would popularize the game and make great pot odds for any of us! don't any of you guys ever try to make a living at a craps table...


Edited by Guldin, 05 May 2018 - 02:41 PM.


#35 Guldin

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Posted 05 May 2018 - 02:50 PM

lets try an easier question; how many months have 30 days?



#36 Morx

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Posted 05 May 2018 - 03:05 PM

Only blackjack is plus EV and only if you count well.

#37 TemplateRex

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Posted 05 May 2018 - 03:44 PM

is there a newer math i don't know about or forgot? there are 6 bombs and 1 flag so the odds would be 6 to 1 against capturing your opponents flag before hitting a bomb (14.2857143 %). this would also be the same probability of rolling a 2 or rolling a 12 before rolling a 7 with a pair of dice. i guess this is why it is so hard for many of you to calculate why a tourney in vegas would popularize the game and make great pot odds for any of us! don't any of you guys ever try to make a living at a craps table...


The dice analogy doesn’t work. The lotto example is more like a deck of 40 cards (ace through 10 for all 4 suits), where the bombs are the ace through 6 of spades, the flag is the ace of hearts. Now you take the flag apart, randomly shuffle the remaining deck, and place back the flag randomly between positions 4 through 13. Then you keep drawing from the top of the deck until you hit a bomb or the flag.

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#38 Guldin

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Posted 05 May 2018 - 05:10 PM

The dice analogy doesn’t work. The lotto example is more like a deck of 40 cards (ace through 10 for all 4 suits), where the bombs are the ace through 6 of spades, the flag is the ace of hearts. Now you take the flag apart, randomly shuffle the remaining deck, and place back the flag randomly between positions 4 through 13. Then you keep drawing from the top of the deck until you hit a bomb or the flag.

well since you could not draw a flag on the first 3 cards, you're odds of catching a bomb on the first try would be 33 to 6, then 32 to 6, then 31 to 6, after that the odds would be 14.2857143 % each time to capture the flag until you got down to 6 pieces. then it would be 5-1, 4-1, 3-1, 2-1, 1-1, then 100%.



#39 The Prof

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Posted 05 May 2018 - 05:12 PM

is there a newer math i don't know about or forgot? there are 6 bombs and 1 flag so the odds would be 6 to 1 against capturing your opponents flag before hitting a bomb (14.2857143 %). 

 

This would be correct if both the bombs and flag were randomly placed.  However, in Astros example the flag is known to be on the back row, so your chance of finding it before a bomb is higher than 1/7.



#40 Guldin

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Posted 05 May 2018 - 08:23 PM

This would be correct if both the bombs and flag were randomly placed.  However, in Astros example the flag is known to be on the back row, so your chance of finding it before a bomb is higher than 1/7.

then the odds of hitting a bomb before the back row are about 37%. 






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