From today’s *Times*, an article by **David Waldstein** called “Mets’ Stretch Without a Slam? Gone. Gone“:

The Mets had gone 299 games and 280 plate appearances with the bases loaded since their last grand slam, while their opponents had hit 18 during that span. So when the opportunity arose in the fourth inning Tuesday night — with

Jason Bayat the plate, no less — the chance of a Mets grand slam was slim.

Was the chance of a grand slam really so slim?

If you flip a coin 10 times in a row and come up with 10 heads, is the 11th flip any more likely to be heads (or, for that matter, tails)?

Of course, it could be that such a long streak without a grand slam feeds into the Mets’ collective psyche and discourages them from hitting one.

It could also be that the Mets are simply worse than everyone else at hitting home runs in general (which they’re not, even though they’re pretty bad).

But more likely the streak was an essentially random run of bad luck.

Which finally ended with Jason Bay hitting a grand slam.

And then, seven batters later, **Carlos Beltran** hit *another* grand slam for the Mets.

Random is as random does. Which is why predicting the future is such a fool’s errand.

Question: how many at-bats until the Mets hit their *next* grand slam?

No fewer than 4.

There’s a distinction between at-bats and plate appearances. A plate appearance resulting in a walk does not count as an at-bat, so it’s entirely possible to have grand slams in consecutive at-bats — Grand Slam, Walk, Walk, Walk, Grand Slam.

Correction: No fewer than 1.

As always, there’s a Venn Diagram for predicting the future:

http://www.vennding.com/2010/11/prognostication-is-fools-game.html

Although it’s true that the chance of a grand slam wasn’t lessened because of the streak preceding it, it was still a pretty slim chance, particularly with Jason Bay at the plate.

Prior to yesterday’s game, Bay had 3 homeruns in 192 at bats, giving him — with some disclaimers — about a 1/64 chance of hitting a homerun. Looking at his numbers with runners in scoring position going into the game, it’s an even slimmer chance: Bay was 8 for 51 with RISP with no homers this year.

In his career prior to last night’s game, Bay had 2 grand slams in 97 chances, but he’s also fallen off considerably since his halcyon days, so it is more reasonable to look at a more recent sampling as opposed to career numbers.

The Mets will hit their next grand slam in 178 at bats. Mark my words.

I would look at:

MLB GS Rate (since new park)

Mets GS Rate (since new park)

Individuals Mets players HR rates and likelihood of encountering bases loaded

And make some sort of weighted average. You might also want to include the time of year and quality of upcoming opponents pitching/defense.

I am probably forgetting a lot of stuff, but this really isn’t a hard question like most of the ones posted here. It has an approximation of a right answer.

I think grand slams are not subject to random distribution; I would think particular hitter/pitcher combinations would affect the outcome greatly. Jose Bautista is much more likely to hit a grand slam off, say, me, than Tim Lincecum.

You have to consider that baseball players aren’t coins, and their chances of hitting a grand slam are neither 50% each at bat nor equal for all players (or teams). Going nearly 300 games without a grand slam sort of implies that the team lacks the ability to do so–or rather that the probability of any at bat with bases loaded resulting in a grand slam for the team is very slim.

Important statistical point: if you get heads 10 times in a row, chances of an 11th head are slightly better than 50%. If you get heads 100 times in a row, the 101st is a virtual certainty. And in sport you don’t have to look hard for self-fulfilling prophecies.

Important statistical point: you need to enroll in a Statistics class.

No, really. Ask Levitt. It’s one of those things that are almost too obvious to make sense.

fraac is right. Note that he did NOT presuppose a fair coin. He’s taking a coin someone hands him and flipping it. If you know nothing a priori about the coin flip’s probability distribution, then a basic Bayesian analysis suggests that the conditional probability of flipping a head if all 100 previous observed tosses were heads is high, as is the probability that you have a crooked coin.

Sorry fraac.

The 101st coin flip is the same except if someone made a bet before the 1st coin flip saying they could flip 101 heads in a row.

Once the first 100 happen, they are history, The next flip is an independent occurence.

Sorry, John B, but no. Your problem is you’re operating in a world of abstract math, implicitly assuming an honest coin. In the real world, if you’ve gotten 10 heads in a row, it’s a good bet that someone slipped in a loaded coin, is practicing sleight-of-hand, or something of the sort. If it’s 100 in a row, this is a virtual certainty.

Let’s not overlook that the probable outcomes in Mr. Bay’s at bat. It’s not simply HOME RUN/NOT HOME RUN. It’s HOME RUN vs. TRIPLE vs. DOUBLE vs. SINGLE vs. FLY OUT vs. GROUND OUT vs. STRIKE OUT vs. WALK vs. HBP, and each of those possibilities have nearly infinite sub-possibilities. There were a few pitches to Mr. Bay before his home run. What if one of those hit him? What if one of those was a wild pitch, thus turning a grand slam into a three-run home run? The thing about baseball is that each pitch (not each at-bat with the bases loaded, which is a loaded situation to begin with) carries a very low probability for the scenarios listed above(outside of contact/no contact, which is 50/50.)

Now, even if you take the simplistic approach that the home run was simply a fly ball that went beyond the wall and was not foul; in this case, were the ball hit 50 feet or so towards center field, it would not have been a home run.