Base stealing

[Katz's Killers]

Can someone please give me the full explanation of stolen base probability? For example, if Pete Pitcher has a -1 hold, and Steve Stealer has stealing (A) *2-6, 11/- (17-13), and Carl Catcher has a -3 arm, what are the odds that Steve will steal 2nd base. Thanks.

[geekor]

From the FAQ:

[quote:07e8eeba3a]Detailed steal rating (following the A-E rating) -- The first number is the dice-roll (2-12) probability that the runner will get a good lead; the second number is the roll he would be picked off. Numbers not listed are the numbers he doesn't get a good lead. A wide range of getting a good lead usually means the player attempts many steals. A wide range of the second number means he gets picked too often.

Detailed steal safe chances (in parentheses at the end of the steal rating) -- If the runner gets a good lead (see number description to the left), the first number is the possibility he will be safe stealing, between 1-20. The second number is the possibility if he doesn't get a good lead. Nineteen is the highest number possible. A high first number indicates an accurate basestealer - if he gets a good lead, he usually makes it. A high second number means he is a true speed demon, able to steal even if he doesn't get a good lead. [/quote:07e8eeba3a]

All the hold does is minus (or plus_ from the numbers in the parenthesis. The arm of the C is for when he actually tried to steal, it minus (or plus) from his Speed (I think).

[Katz's Killers]

I don't think that's the complete story. I'm hoping someone can give a richer explanation. What I'm looking for is the complete probabilty formula so that one would get some "answer" like the base stealer has a "70.4%" chance of getting into 2nd...taking into account how often the pitcher holds the runner.

[Mean Dean]

[quote:94784bcbed]if Pete Pitcher has a -1 hold, and Steve Stealer has stealing (A) *2-6, 11/- (17-13), and Carl Catcher has a -3 arm, what are the odds that Steve will steal 2nd base.[/quote:94784bcbed]You can assume that any baserunner with a star is being held. Each roll on a d20, as you probably know, is 5%. So assuming Steve gets the lead, his chance of stealing the base is 85% (his first steal number, 17, * 5%), minus 10% for having the lead and being held on base, minus 20% for the hold/arm (-4 * 5%) = 55%.

His chance to get the lead is [url=http://www.tauonline.org/Article/121/SixSidedDiceProbabilities/]the chance of rolling a 2, 3, 4, 5, 6 or 11 on 2d6[/url], which is 47.2%.

If he [i:94784bcbed]doesn't[/i:94784bcbed] get the lead, then his chance to steal will be 65% (his second steal number, 13, * 5%), minus 20% this time for not having the lead and being held on base, minus the same 20% for the hold/arm = 25%.

The chance of his not getting the lead, obviously, is (1 - 47.2%) = 52.8%.

So, what we've got so far is that the chance that he will end up on 2nd after a steal attempt is (55% * 47.2%) + (25% * 52.8%) = 39.2%.

(If we knew the catcher's T-rating, we could factor it in at this point. There is a 15% chance on a steal attempt that the catcher's T-rating will be tested. If he rolls his T-rating or lower on that test, then it's a throwing error and the basestealer goes to 3rd base. For instance, if t = 4, the chance of an error is 15% * 20% = 3%, which would be added to the stealer's success chance. Since we don't know it in this example, we'll assume t = 0, so it's still a 39.2% total chance for Steve.)

Finally, we come to pickoffs and balks. Every time a runner goes for the lead, there is a 5% chance of a possible pickoff. This chance is based on his second stealing number; he has to roll that number or less to avoid the pickoff. So in his case, the pickoff will occur 30% of the times it has the possibility to occur. 5% * 30% = 1.5%. This brings the total chance of the steal attempt placing him on 2nd to 37.7%.

A balk also has a 5% chance to happen on each lead attempt, and happens if the pitcher rolls his balk number or less (so for instance, if bk = 5, the balk occurs 25% of the time that the possibility arises, and the total chance is 5% * 25% = 1.25%). That would, naturally, be added to the baserunner's safe chance. We don't know the balk rating here, so we'll leave it out.

That is as complete as the calculation gets. Of course, in reality, you have the option to bail out on the steal attempt if you don't get the lead. That means that, unless it's the rare situation involving terrific basestealers and/or terrible defensive pitchers/catchers where stealing without a lead becomes plausible, the only things you generally need to worry about are a) the chance to succeed given the lead, and b) the chance of getting the lead or a balk, as compared to the chance of an automatic CS or pickoff. (Automatic CS are the numbers right after the slash in the stealing info; Steve Stealer as you described him doesn't have any, which made this a little easier ;)

[durantjerry]

How did you come up with the hold rating?

[Katz's Killers]

Your work is greatly appreciated; but I want to make sure I have it down completely;

[b:a236709946]You wrote;[/b:a236709946]
You can assume that any baserunner with a star is being held. Each roll on a d20, as you probably know, is 5%. So assuming Steve gets the lead, his chance of stealing the base is 85% (his first steal number, 17, * 5%), [u:a236709946]minus 10% for having the lead and being held on base[/u:a236709946], minus 20% for the hold/arm (-4 * 5%) = 55%.

OK. Because the runner has the * the pitcher will always try to hold him. Steve Stealer has (17-13), which means when he gets a good lead he gets in 85% of the time, and when he does not get a good lead he gets in 65% of the time (that 5% for each digit (1/20th)). So far so good. The hold/arm of -4 is the combined -1 hold from the pitcher and the -3 arm from the catcher, and that translates into minus 20% (the same 1/20th times 4). [u:a236709946]Where is the minus 10% coming from???[/u:a236709946]


[b:a236709946]You wrote[/b:a236709946]
His chance to get the lead is the chance of rolling a 2, 3, 4, 5, 6 or 11 on 2d6, which is 47.2%.

Got it.


[b:a236709946]You wrote[/b:a236709946]

If he doesn't get the lead, then his chance to steal will be 65% (his second steal number, 13, * 5%), [u:a236709946]minus 20% this time for not having the lead and being held on base[/u:a236709946], minus the same 20% for the hold/arm = 25%.

[u:a236709946]Where is the minus 20% for not having the lead coming from? [/u:a236709946] I get the same -20% for the pitcher's hold plus the catcher's arm rating.


[b:a236709946]You wrote[/b:a236709946]
The chance of his not getting the lead, obviously, is (1 - 47.2%) = 52.8%.

So, what we've got so far is that the chance that he will end up on 2nd after a steal attempt is (55% * 47.2%) + (25% * 52.8%) = 39.2%.

Got that...and I get the rest.

Question; are the odds different for stealing 3rd base and home, and what is the math for that?

Many, many thanks.

[Mean Dean]

I think you're both asking the same question; if not, let me know :) The -10%/-20% is just the penalty that you get for being held on base. It's -10% if you have the lead and are attempting to steal 2nd; -20% if you don't have the lead and are attempting to steal 2nd, or if you are attempting to steal 3rd or home. It's a separate thing from the hold rating of the pitcher; anytime any pitcher holds, this same amount is deducted.

(The one thing I think we didn't mention is that the total of the arm and hold cannot exceed +25% or -25%. For instance, a -3 pitcher and a -3 catcher still only adds up to a -5, or a 25% worse chance. The -10%/-20% is figured out after that, so if you are facing that -3 pitcher/-3 catcher [-25%] and are attempting to steal 2nd without a lead [-20%], your chance is going to be your second steal number minus 45%.)

You [i:2fcdd153d7]must[/i:2fcdd153d7] get the lead in order to even attempt a steal of 3rd or home. If you do, then stealing 3rd will be exactly the same success chance as stealing 2nd without a lead would have been (second steal number, minus 20% if held, plus/minus arm/hold rating). Stealing home is usually a straight -45% from the second steal number, minus 20% if held (which is almost never). Neither the catcher's arm nor the pitcher's hold rating count on steals of home.

With runners on 1st and 3rd, there's a delayed steal play. No lead-taking is involved. The chance to steal 2nd is the second steal number of the runner on 1st, plus/minus catcher's arm only. The defensive manager can then either choose to hold the ball, or throw. If he throws, the runner on 3rd can attempt the steal of home, at a success chance of his second steal number, minus 25%. The defensive manager can then decide whether he wants to try to throw out the runner attempting to steal 2nd, or the runner attempting to steal home. The other runner will be safe (unless his teammate gets caught for the third out.)

[Katz's Killers]

Outstanding! So let me give this a try (just stealing not factoring in balks). Imagine Omar Moreno is on first. His stealing reads (AA) *2-6, 12/- (19-15), Ira Thomas is the catcher with an arm of -4, and Dick Tidrow is the pitcher with a hold of +3.

Adding up 2-6 + 12 (on 2d) you get 44.45%, which is Moreno’s chances of getting a good lead. Conversely, 55.55% is Moreno’s odds of not getting a good lead. Adding Tidrow’s hold and Thomas’s arm you get -1, or -5%.

Now the 19 good lead gets converted to 95%, minus the 10% standard penalty for stealing, and then the -5% reduction from hold/arm, yielding 80%. This is now multiplied with 44.45 to yield .3556. The 15 not getting a good lead gets converted to 75%, minus the 20% standard penalty for stealing, then the -5% reduction for hold/arm, yielding 50%. This is multiplied by 55.55 to yield .27775. Finally .3556 + .27775 = 63.33% chance of being safe at 2B.

Is this correct? This seems to give me a percentage which intuitive seems lower than I would expect.

Lastly, you mentioned that the hold/arm can’t total over +5 or else is it reduced down to a +5. Is it true that this works the other way? Imagine a catcher with -3 arm and a pitcher with a -3 hold. Does that -6 get converted into a -5?

[Mean Dean]

[quote:fd17c70bf0]Lastly, you mentioned that the hold/arm can’t total over +5 or else is it reduced down to a +5. Is it true that this works the other way? Imagine a catcher with -3 arm and a pitcher with a -3 hold. Does that -6 get converted into a -5?[/quote:fd17c70bf0]Yeah, it works both ways.

Your math looks right. Think about it: If he were on 1st base 100 times with no one on ahead of him, against this (pretty normal) catcher/pitcher combo, he'd steal 63 bases. That's quite a good rate!

That said, basestealers as good as Moreno are pretty rare. The system is set up such that most of the time, if you don't get the lead, your chance to successfully steal the base is not going to be high enough to justify the attempt, and you'll stay put. So usually, you're not factoring in that second part of the calculation without the lead. That means your success chance when you actually do try is going to be much higher, although the chance of your staying put and not going anywhere will also be much higher.

[Katz's Killers]

[quote="DeanTSC"][quote:33e79de8b2] Think about it: If he were on 1st base 100 times with no one on ahead of him, against this (pretty normal) catcher/pitcher combo, he'd steal 63 bases. That's quite a good rate!

That said, basestealers as good as Moreno are pretty rare. The system is set up such that most of the time, if you don't get the lead, your chance to successfully steal the base is not going to be high enough to justify the attempt, and you'll stay put. So usually, you're not factoring in that second part of the calculation without the lead. That means your success chance when you actually do try is going to be much higher, although the chance of your staying put and not going anywhere will also be much higher.[/quote:33e79de8b2]

.....Obviously I have some control over the decision to steal by setting the team's and the player's aggressive to steal, but are you suggesting that if this baserunner [u:33e79de8b2]attempts[/u:33e79de8b2] to steal 100 times the baserunner would be safe more than 63% because the computer will "restrain" the runner when he does not get a good lead?? His decison to try to steal will actually increase his overall stolen base percentage because he is being more selective. So instead of being safe 63%, maybe he's safe 70% of the time.

Another related question; when the runner is held, groundballs (GB) that have the "++" are automatically turned into a hit (if you click on the explanation of the pitcher's card it says that). I can not find any instance of this ++ symbol used on any pitcher's card. Is this a mistake? This should be part of the game in some format. Am I not reading this correctly?

Again, many thanks.