Baseball Question

[Xenod]

I'm a year out of college and I've forgotten way too much probability...

Someone related an anecdote from a book they read about a man who made a lot of money in college by offering the following wager.

You can pick any 3 Major League hitters. If they get 6 or more hits, you get paid at 10:1 odds. If they don't you lose.

Assuming that the three hitters you pick are each hitting .350 and they get a combined 12 at bats:

a) is this a good bet to take?
b) what would the odds need to be for this to be a good bet?


Thanks

[T50_Omaha8]

My spreadsheet shows this bet losing 0.212735383
of the time, which way worse than the .0909090909... breakeven treshold. You're going to need 4-1 odds to show a healthy profit on this one.

The chance they get EXACTLY 6 hits is even larger than 1/11:

.35^6*.65^6*12!/(6!6!) = 0.128103318

If you were to do MORE than 6 hits, then the bet loses 0.084632065 of the time for a slight profit.

[jay_shark]

All three players would need to average 0.500 in hitting which is pretty much impossible .

Lets reduce the 3 hitters to 1 hitter and that this player gets at bat 12 times with an average of 0.350 .


The probability you hit any particular pitch is 0.350 . The number of ways you can hit 6 times out of 12 at bats is simply 12c6 .

12c6*0.350^6*0.650^6 =12.8%
12c7*0.350^7*0.650^5= 5.91%
12c8*0.350^8*0.650^4= 1.98%
12c9*0.350^9*0.650^3=0.00476
The other cases are negligible

Which means there is about a 20.69% that this player goes at least 6 for 12 . You're getting paid 10:1 which means it's break even if you win once and lose 10 times .

So of course this is a good bet since (100-20.69)/20.69 = 3.83 :1 which is also the break even point .

[T50_Omaha8]

I see you interpreted the OP as WE get to choose the 3 batters, while I did it the other way. It was pretty unclear to me.

[Xenod]

Thanks for the responses.

Just to confirm, you're saying that there will be 6 or more hits about 21.2% of the time?

[jay_shark]

Yes , just continue summing the way I did and you should arrive at the number t50_omaha8 has quoted .

sigma [12ck*0.35^k*.65^(12-k) { sum from k=6 to k=12}]=0.212735383

[DarkMagus]

Jay_shark, all of those calculations can be done easily with the cumulative binomial distribution function.

In excel,

=1-binomdist(5,12,0.35,TRUE)

That is, 1 minus the probability they get 5 or less hits in twelve trials, with a 0.35 probability of success.

This gives 21.27%.

You need at least 3.7:1 odds to break even on this bet, so you should definitely be taking 10:1 odds.

[jay_shark]

I know Doug , you are right but I feel like it's cheating . People are becoming reliant on calculators and programs like excel to work out all the calculations for them . Of course if you're confident in your mathematical abilities and you want to save yourself some time then you could use excel and plug away . For those who want to know how excel comes up with that answer then it is good to work it out the rigorous way .

If you ask me to compute sqrt2 , I will probably use a pencil and paper and approximate the roots to the equation y=x^2-2 using calculus .

Ok , you can call me crazy now :P

[T50_Omaha8]

Don't worry man, for my excel spreadsheet I typed the numbers 1 through 12 in rows, then typed in the binomial theorem a copied and pasted it to come up with a value for each number, then summed it all to make sure it was 1, then was able to do partial sums or whatever. It's a lot easier to interpret the data when you've got everything laid out right in front of you.

And if you ask me to approximate sqrt(2), I'll start multiplying it by other perfect squares in order to find a suitable fraction approximation.

I always did like ghetto math. [img]/images/graemlins/wink.gif[/img]

[AaronBrown]

Shame on you guys, you're blinded by models. You have to look at the data. First of all, the average full-time player gets only 3.5 at-bats per game (walks and sacrifices don't count as at-bats). Second, the guys with the best batting average often get fewer at bats, since it's less costly to walk them and they keep their average high by not swinging at bad pitches.

You wouldn't pick the guys with the highest batting averages, you'd pick the guys with the most hits per game, among the players you were sure would play. If you had picked the top three players yesterday, based on hits per game of healthy everyday players with games, you would have had two guys with one hit and one with zero.

Among the top 20 guys you could have picked, two had 3 hits, five had 2 hits, ten had 1 hit and three had 0 hits. If you had picked at random from among these, your chance of getting 6 or more hits was 148/1,140 or 13%.

This is close enough to 10% that it might be a good bet, especially when you figure most people probably don't pick optimally and you might pick a guy who doesn't play that day.

To get the real answer, you'd want to look at more data, of course. My point is only that models without data are dangerous.