Probablities and Odds

[tommy]

I know there was a thread at some point in the past but I find the search function impossible to use on these boards, so I can't find it. But anyway, for those of you who like probabilities and odds types of questions/puzzles:

If a team has three consecutive batters coming to the plate, and each of them has a .333 batting average (that we are assuming continues), what are the chances that at least one of the three gets a hit?

I think it's 19 of 27 but I'm not sure of it, and I did that by writing out possible combinations instead of using a proper formula. Anyone have that formula or at least tell me if I got the right answer? My credibility with my high school-aged son depends on it.

[darthur]

I think that's right.

Probability that at least one person gets a hit
= 1 - Probability that all three don't get a hit
= 1 - (Probably that a specific player doesn't get a hit)^3 (*)
= 1 - (2/3)^3
= 1 - 8/27
= 19/27

The formula for (*) is based on the fact that if two events are independent, then the probability they both happen is equal to the product of the probabilities that each one happens.

Hope that helps!

[PackMan97]

The answer you are looking for is 9f.

[tommy]

I think that's right.

Probability that at least one person gets a hit
= 1 - Probability that all three don't get a hit
= 1 - (Probably that a specific player doesn't get a hit)^3 (*)
= 1 - (2/3)^3
= 1 - 8/27
= 19/27

The formula for (*) is based on the fact that if two events are independent, then the probability they both happen is equal to the product of the probabilities that each one happens.

Hope that helps!

Makes perfect sense. Thanks !

I was wondering why it wouldn’t be the sorta inverse of that — 1/3 to the 3rd power - but that couldn’t be right and with your explanation then of course I see what I was missing. On further reflection, duh. I’m usually not a dum guy. 😊

[Skydog]

Makes perfect sense. Thanks !

I was wondering why it wouldn’t be the sorta inverse of that — 1/3 to the 3rd power - but that couldn’t be right and with your explanation then of course I see what I was missing. On further reflection, duh. I’m usually not a dum guy. 😊

1/3 to the third power is the probability all three get hits.

[acdevil]

Assuming the chances are truly independent and the pitcher isn’t tipping his pitches in the stretch like in the Yankees/Tigers game the other day.

[camion]

I don't like hard problems so I would solve an easier one and take what's left.

The easier problem: What is the probability that none of the the batters get a hit?

It should be 2/3 * 2/3 * 2/3 = 8/27.

Whatever is left is the probability that at least one batter gets a hit. 1 - 8/27 = 19/27.

[TruBlu]

The probability of me figuring this out = 0

[JasonEvans]

Speaking of probabilities, on consecutive seasons of Survivor, they have presented contestants with the Monty Hall Problem. In each case the competitor chose to keep the box they had chosen, essentially taking a 33% chance over a 67% chance... and in both season the smaller chance proved to be the correct box.

I hate it when stupidity is rewarded like that.

[camion]

Speaking of probabilities, on consecutive seasons of Survivor, they have presented contestants with the Monty Hall Problem. In each case the competitor chose to keep the box they had chosen, essentially taking a 33% chance over a 67% chance... and in both season the smaller chance proved to be the correct box.

I hate it when stupidity is rewarded like that.

I've learned during my years in baseball (also true in parenting) that you can do the right things and have a bad outcome. You can do the wrong things and have a good outcome.


The odds are better if you do the right things, but there are no guarantees.

[Indoor66]

Isn't that because odds on outcomes are based on large numbers not singular events (even though the odds still apply). I can flip heads 100 times in a row and the odds on flip 101 are still 50-50.

[rsvman]

I don't like hard problems so I would solve an easier one and take what's left.

The easier problem: What is the probability that none of the the batters get a hit?

It should be 2/3 * 2/3 * 2/3 = 8/27.

Whatever is left is the probability that at least one batter gets a hit. 1 - 8/27 = 19/27.

This is how I would do it, too. It just seems a lot easier this way.

[CDu]

I know there was a thread at some point in the past but I find the search function impossible to use on these boards, so I can't find it. But anyway, for those of you who like probabilities and odds types of questions/puzzles:

If a team has three consecutive batters coming to the plate, and each of them has a .333 batting average (that we are assuming continues), what are the chances that at least one of the three gets a hit?

I think it's 19 of 27 but I'm not sure of it, and I did that by writing out possible combinations instead of using a proper formula. Anyone have that formula or at least tell me if I got the right answer? My credibility with my high school-aged son depends on it.

It's one minus the probability that none of the 3 gets a hit. So 1 - 0.667^3, which is indeed 19/27.

ETA: should have read the rest of the thread and realized that I didn't need to respond as someone had already done so.

[CDu]

Isn't that because odds on outcomes are based on large numbers not singular events (even though the odds still apply). I can flip heads 100 times in a row and the odds on flip 101 are still 50-50.

Minor edit for statistical specifics: the odds of a coin flip being heads are 1. The probability of it being heads is 50%. Odds of an event are calculated as probability of the event divided by (1 - probability of the event). So a 50/50 bet has a probability of 50% and an odds of 1 (50%/50%).

But yes, the overall sentiment of your statement is correct. The idea behind playing the probability (or the odds) is that in the long run the higher-probability event should win out. But in any single event, the less likely outcome is possible and sometimes you get the wrong outcome despite the right process (or in Jason's example the "right" outcome despite the wrong process).

[Wander]

Here's my annoying "well, actually" moment of the day. If you flip heads on a coin 100 times in a row, then the probability (not odds, thanks CDu ) is that flip 101 is absolutely, definitely, no question, essentially 100%, going to be heads.

[LasVegas]

Assuming the chances are truly independent and the pitcher isn’t tipping his pitches in the stretch like in the Yankees/Tigers game the other day.

Throw a trash can into the equation and probability is 100%!

[Skydog]

Here's my annoying "well, actually" moment of the day. If you flip heads on a coin 100 times in a row, then the probability (not odds, thanks CDu ) is that flip 101 is absolutely, definitely, no question, essentially 100%, going to be heads.

Well not 100%. It could be a fair coin and you just had a 1 in 1,267,650,600,228,229,401,496,703,205,376 occurrence!

[snowdenscold]

Well not 100%. It could be a fair coin and you just had a 1 in 1,267,650,600,228,229,401,496,703,205,376 occurrence!

But even if offered 1000:1 odds on tails, I’m still shoving a ton of money in on heads, even with reciprocal odds.

[snowdenscold]

I think that's right.

Probability that at least one person gets a hit
= 1 - Probability that all three don't get a hit
= 1 - (Probably that a specific player doesn't get a hit)^3 (*)
= 1 - (2/3)^3
= 1 - 8/27
= 19/27

The formula for (*) is based on the fact that if two events are independent, then the probability they both happen is equal to the product of the probabilities that each one happens.

Hope that helps!

And to make it more general, if you’re looking at an event with probability A occurring and looking at it over N times:

The event happens all N times: prob = A^N
The event happens at least 1 time: prob = 1 - (1-A)^N