Position # 50

[PersianLord]

Position information:

Pip-counts : Red 61-125 White
Score:      : Red   4-4 White
Match:                5-pointer

Red to play 5-1

Spoiler

This is for beginners. Rudimentary math:  After  11/10-11/6, White's hitting roll is 55 with a probability of 1/36, after other moves, White's hitting roll is 65 with a probability of 2/36.

[close]

[spielberg]

Spoiler

Duping the 5s is obvious - why give twice the chances of a miracle roll to white?

[close]

Easy but keep them coming Ali , no matter how easy - the reason the move is correct is far more valuable to learn than which it is.


Steve

[PersianLord]

Spoiler

Duping the 5s is obvious - why give twice the chances of a miracle roll to white?

[close]

Easy but keep them coming Ali , no matter how easy - the reason the move is correct is far more valuable to learn than which it is.


Steve

Thanks Steve.

As you correctly pointed out, the reasoning behind the best move is always of more educational value. And yes, I will keep them coming 

Regards

[dorbel]

As you say, the reason for making the play is more interesting than the answer itself. 11/6, 11/10 is indeed the correct play, but not because it duplicates fives! We can easily see that being hit with a 6-5 is twice as likely as with 5-5, but the hit with 5-5 is much more damaging because it enables White to hit and make her 4pt. After being hit with 5-5 Red's winning chances drop to around 13% (Snowie 3-ply), whereas after being hit with 5-6 they are about 46% (ditto). Red is better off being hit twice than he is being hit once.
The true reason for playing 11/10 is that it gets Red closer to home. If not hit, then only 2-1, 5-5 and the anti-joker 6-5 (!) leave a blot next turn. If he stays back on the 11pt, He has 6-6, 4-2, 3-3, 3-1, 2-2, 2-1 and 1-1, ten numbers that still leave a blot, although none leave a direct shot of course. Alll numbers that leave a blot also have their own little thread of numbers that give White second helpings if he misses next turn.

[PersianLord]

As you say, the reason for making the play is more interesting than the answer itself. 11/6, 11/10 is indeed the correct play, but not because it duplicates fives! We can easily see that being hit with a 6-5 is twice as likely as with 5-5, but the hit with 5-5 is much more damaging because it enables White to hit and make her 4pt. After being hit with 5-5 Red's winning chances drop to around 13% (Snowie 3-ply), whereas after being hit with 5-6 they are about 46% (ditto). Red is better off being hit twice than he is being hit once.
The true reason for playing 11/10 is that it gets Red closer to home. If not hit, then only 2-1, 5-5 and the anti-joker 6-5 (!) leave a blot next turn. If he stays back on the 11pt, He has 6-6, 4-2, 3-3, 3-1, 2-2, 2-1 and 1-1, ten numbers that still leave a blot, although none leave a direct shot of course. Alll numbers that leave a blot also have their own little thread of numbers that give White second helpings if he misses next turn.

I never claimed it's because of duplication. I just said after 11/10, red's chance of being hit is 1/36 (55) and after other moves, it's 2/36 (65). And I do think this is a more-than-enough reason for 11/10, without even considering what a god damn bot later would say about the probabilities of winning/losing.

[inim]

red's chance of being hit is 1/36 (55) and after other moves, it's 2/36 (65). And I do think this is a more-than-enough reason for 11/10, without even considering what a god damn bot later would say about the probabilities of winning/losing.

That's exactly the difference between a 0-ply (PL) and 2-ply (dorbel) analysis

[donzaemon]

The other part of the problem that nobody has considered is what happens if YOU roll a 66 after not moving up , vs a 55 after moving up after the respective plays.   a 55 after you move up to 5 away crunches the 6 point while leaving the blot in the outfield ,  a very bad scene ,  the 66 after not moving up keeps the board intact and is much better all together.  This gives some weight to not moving up , but not enough weight to justify leaving the extra shot.

dorbel was going on about this in shouts , claiming that not moving up 11/10 and leaving the extra shot was correct (and even calling people dense in the process).  His reason was that 55 was such a joker that you don't win hardly any games after it , and the return wins after 65 justify it in the end.  (seems like he overlooked the crunch factor mentioned above which argues for his position) He defended his idea by offering a prop where he pays 2 to 1 for the relative positions after the two plays .... but ,  I think he is either overthinking of just trolling for a prop (that doesn't show everything about the position). 

it's true his prop would probably win up to 2 to 1 after giving opp 2 to 1 on the position but it doesn't tell the whole story.
The rest of the story is " you're already a clear favorite , that's not the time to be increasing your risk just because there might be better returns after your opponent makes a game out of it"

Think about it this way.... 
if you get hit with a 65 ,  your chances or winning are around 50%
if you get hit with a 55 , your chances of winning are around 10 - 15% 
if you don't get hit you're at around 95% in both cases
(just estimates but should be in the ballpark)

so :
( 95 - 50 ) * 2 = 90  ->  how much you lose on the 65's
95 - 10 = 85  or  95 - 15 = 80  ->  how much you lose on the 55

of course we'd want accurate numbers when it starts getting close, rough estimates might skew it a bit.

You could just think simply and say 2 shots that lose around 50%  would be worth about the same as 1 shot that loses 100% for your opponent. That should tell you that if you reduce it to 1 shot and get ANY wins out of the position that you're probably in a better place than leaving the 2 shots. So that 10-15% you're getting when you win after getting hit with the super joker is all extra gravy and makes it well worth leaving only the one shot.

Sometimes we get caught up and want to be creative but leaving extra immidiate shots when you're ahead is just that : leaving extra shots.  It's like shooting yourself just because you are wearing a bullet proof vest, there's no real need to do it.

[dorbel]

Donz, you didn't listen to what I said in shouts and you can't have read my post above. Or perhaps you are deliberately misrepresenting me. I'll say it again. 11/10 is the correct play, but having to leave a 5-5 to hit is a downside of the play, not a benefit.
Giving two shots after each of which you are 50% is the same as leaving one shot where you lose 100%? I think not. The queue to play that one as a prop would stretch around the block. In 100 games where you lose every time you lose 100 points. In 200 games where you win 50% you don't lose anything!
In this position, 100 games at 13% loses 74 points. 200 games at 46% loses 16 points. It's your choice.
Going back to the original reason for posting on this position, the reasoning behind the play is more important than knowing the right answer.

[donzaemon]

That is clearly NOT what you were saying in shouts , I have the dialog , you were going on about how the 2 shot prop would hugely win and saying it's right etc.
aside from that , you can argue that the estimation shortcut for losing 2/36 * .5 doesn't compare to 1/36 * 1   

you said : In 100 games where you lose every time you lose 100 points. In 200 games where you win 50% you don't lose anything!

This is not the case.

if you lose every time in 100 games , you lose exactly 100 games
if you lose 50% in 200 games , you lose 100 games  ( and win 100 games )
voila ,  you lose 100 games in each scenario.

this shows a similar mistake that you're making in the original problem ,  we are only tallying up the losses for both cases, the wins go in the default pool because they come in a various ways.   

but before you go on and on about it , ask me if I really care about tutoring you further.

[dorbel]

Yes, but I win 100 games to counterbalance my 100 losses, you dont win any!
Get NIHo to explain it to you.

[socksey]

Gee, I'd just be happy if I knew what either of you were talking about!   

socksey



"The extent of the catastrophe that threatens gives us the measure of the transformation that will be necessary in order to master it." - Lewis Mumford

[PersianLord]

That's exactly the difference between a 0-ply (PL) and 2-ply (dorbel) analysis

Thank you for introducing noise into GNUBG's analysis.      

[donzaemon]

again , dorbel is omitting significant parts of the problem in order to try to sound like he has a point.
Either he has no understanding of probability , or he's just looking for that shread of something to hold on to in order to sound he's not completely stupid.

here's the idea
say for example ,  I have a roulette wheel with 36 numbers (no zeroes) it is numbered from 1 to 36

I give you 2 choices  you can either pick the number 1 only ,
or you can pick both numbers 2 and 3 .

if you choose 1 and it hits ,  you lose  $1 

if you choose 2 and 3 ,  when the 2 hits you win $1  ,  and when the 3 hits you lose $1 

If anything besides 1 , 2 , or 3 hits , you win $1 .

OK,  so obviously ,you've been given an extermely good deal here

now ,  which one do you choose ?   

most logical people will say ,  " it doesn't matter ,  I win huge either way and have 1 chance in 36 of losing and 35 chances in 36 of winning no matter which one I choose"

but dorbel would say   "  you win much more with the 2 and 3 because they're balanced out and equal zero  and the　 1 only loses "

duh ,  this is the kind of mistake we can make if we only isolate one aspect of a problem and/or make the mistake of looking at probability as a result.


in our backgammon problem it comes to a similar problem but it's more like :

number 1 loses $0.75

number  2 wins $1

number 3 loses $1

every other number wins $1


in this case we can clearly see that choosing  number 1 loses LESS than choosing both 2 and 3 .... 

so we choose 1, minimize our losses ,  and win even MORE  than in the other case.


now,  even with this type of remedial explanation ,  I'd be willing to bet that dorbel STILL doesn't get it.

[PersianLord]

now,  even with this type of remedial explanation ,  I'd be willing to bet that dorbel STILL doesn't get it.

I would like to wager a bet of 500 NACCI points that, if you provide your foolproof explanation to an anthropoid, even a retarded one, it would grasp the idea in about 10 mins. Unfortunately, I am not sure of the gentleman you mentioned his name, though. Thus, I kindly decline your invitation.

Regards

[spielberg]

Fun Fun Fun -- please prolong this argument. I await dorbel's riposte.

[dorbel]

"number 1 loses $0.75

number  2 wins $1

number 3 loses $1 "

Now donz, you play 100 games choosing number one. You lose 100 x $0.75 = $75 lost.
I play 100 games of each of two and three. I win 100 x $1 and lose 100 x $1 = Nothing lost. I look forward to your explanation of how this is good for you.

However, all my posts have been about position 50. Consider this proposition.
You play 100 games as Red after being hit from the barwith 5-5. I play 200 games as Red after being hit from the bar with 6-5. I dont fancy flying to Japan to play this but I have no doubt you can find a European to play it for you!

[donzaemon]

my my aren't we dense.

Look again at the roulette explanation

[dorbel]

I've wasted enough time on this. Everybody make your own mind up.

[PersianLord]

Well, my math skills are not good enough, but I think the main part of the problem is that dorbel ignores the fact that right now red is a huge favorite to win. In other words, he ignores the scenario of match after non-hitting rolls. He extracts a tiny part of the puzzle (the probable story of the game after white's hitting rolls, aka jokers) and bases his analysis on it, and guess, he gets puzzled even more.

Regards

[dorbel]

Well I guess the moral of that story is never go into print without checking your ar
ithmetic! My theory would be right if the double play won a little more and the single play won a little less, but here it isn't. Donz is correct. I am retiring to a monastery and taking a vow of silence.