Calculating the odds of matching 3 things out of 9

[Kelemvor]

So a local grocery store has a lame promotion right now where you get a card with 9 things to scratch off. You can only scratch off 3 and if you happen to get the right 3 that match, you win something.

First scratch you can get any of the winning 3 from the 9.
Second scratch you can get either of the remaining 2 from the 8.
Third scratch you have to get the last one from the 7

I haven't done statistics for years but to calculate that would you do:
3/9 * 2/8 * 1/7 which his about 1%?

Or is that wrong?

 

[911paramedic]

I always forget the stats multiplication orders. I'm a moran.

 

[Zorander]

I haven't done statistics for years but to calculate that would you do:
3/9 * 2/8 * 1/7 which his about 1%?

Or is that wrong?

Sounds about right to me. You have 1.19% chance of winning.

 

[Greenman]

Is it really worth the effort for a bag of cheatos?

 

[Matthiasa]

That makes the assumption that each is a potential winning card with only 3 of the things that need to be matched. If it was more random then that some cards might have more then 3 of those and some might have less. Skewed heavily to having less typically.

Though if this is for stats hw then yes.

 

[Spikesoldier]

So a local grocery store has a lame promotion right now where you get a card with 9 things to scratch off. You can only scratch off 3 and if you happen to get the right 3 that match, you win something.

First scratch you can get any of the winning 3 from the 9.
Second scratch you can get either of the remaining 2 from the 8.
Third scratch you have to get the last one from the 7

I haven't done statistics for years but to calculate that would you do:
3/9 * 2/8 * 1/7 which his about 1%?

Or is that wrong?

formula is correct.

 

[Kelemvor]

Is it really worth the effort for a bag of cheatos?

Exactly the point. 99% of the cards have the "Prize" of $1. All I do is scratch off the prize area and then throw it out when I see a $1. So worthless.

 

[Fenixgoon]

That makes the assumption that each is a potential winning card with only 3 of the things that need to be matched. If it was more random then that some cards might have more then 3 of those and some might have less. Skewed heavily to having less typically.

Though if this is for stats hw then yes.

this.

 

[DCal430]

So a local grocery store has a lame promotion right now where you get a card with 9 things to scratch off. You can only scratch off 3 and if you happen to get the right 3 that match, you win something.

First scratch you can get any of the winning 3 from the 9.
Second scratch you can get either of the remaining 2 from the 8.
Third scratch you have to get the last one from the 7

I haven't done statistics for years but to calculate that would you do:
3/9 * 2/8 * 1/7 which his about 1%?

Or is that wrong?

Is it a game with something like 6 X and 3 Diamonds, and you need to uncover the 3 Diamonds to win, or it 3 X, 3 Circles, and 3 Diamonds and you need scratch any 3 of the same thing.

 

[yhelothar]

The 3/9th should not be in the equation. It seems that there are three sets of three. So in the first scratch, it doesn't matter which of the three types you uncover. What matters is just your second and third scratches have to match the first.

So make that 3.6%

 

[Nebbers]

That makes the assumption that each is a potential winning card with only 3 of the things that need to be matched. If it was more random then that some cards might have more then 3 of those and some might have less. Skewed heavily to having less typically.

Though if this is for stats hw then yes.

this.

Yes. This.

And how many "things" are there in total? Without that information any attempt to figure this out is a waste of time. If there are 500 different "things" and they're all randomly scattered about and you have to have three of them, one in each column, on one card... you may just as well burn your money.

 

[shortylickens]

Damn, I'm in statistics right now and this is stumping me.

 

[Kelemvor]

The 3/9th should not be in the equation. It seems that there are three sets of three. So in the first scratch, it doesn't matter which of the three types you uncover. What matters is just your second and third scratches have to match the first.

So make that 3.6%

No, there are 3 that match and 6 other random choices that do not match. There are not 3 set of 3. You have to get exactly the right 3 in order to win.

 

[shortylickens]

No, there are 3 that match and 6 other random choices that do not match. There are not 3 set of 3. You have to get exactly the right 3 in order to win.

So you need to get 1/9, then 1/8 of whats left, then 1/7 of whats left.
9x8x7 = 504, so 1/504.
Less than .2 percent.

 

[Kelemvor]

So you need to get 1/9, then 1/8 of whats left, then 1/7 of whats left.
9x8x7 = 504, so 1/504.
Less than .2 percent.

No. as I stated in the OP, the first pick can be for any of the correct 3 of the 9 so 3/9. Then there are 2 left of the 8 choices so 2/8. Then there is 1 left of the 7 choices so 1/7.

 

[gorcorps]

OP is correct. Others seem to be overthinking it for some reason.

The way the card is set up is that there's 9 spots, and 3 specific ones need to be chosen. Think of the wrong spots as X's and good spots as O's. The order those O's are uncovered doesn't matter as long as only the O's are found when finished. Therefore the first scratch is a 3/9 chance of finding an O. The 2nd has a 2/8 chance, and the 3rd has a 1/7 chance.

 

[shortylickens]

OP is correct. Others seem to be overthinking it for some reason.

I just bombed my last two statistics tests and have been paranoid for weeks.

 

[superccs]

I always forget the stats multiplication orders. I'm a moran.

Dial "M" for Moran, and hope that a moron doesn't answer the phone.

 

[gorcorps]

I just bombed my last two statistics tests and have been paranoid for weeks.

Oh I won't say it's easy... lord knows I didn't pass stats with flying colors. It really did usually come down to over complicating things though. My brain goes, "no... it couldn't be THAT, could it? let's try it this way"

 

[Rubycon]

Use a bright light source/laser to see what's under the pads. :biggrin:

 

[Kelemvor]

Use a bright light source/laser to see what's under the pads. :biggrin:

I tried holding them up to a light but couldn't see anything. If I find one where the prize is actually decent I might try my 500w halogen work lamp or something.

 

[Rubycon]

I tried holding them up to a light but couldn't see anything. If I find one where the prize is actually decent I might try my 500w halogen work lamp or something.

You need a pretty concentrated source for it to work. The work light will probably blind you!

While not exactly something everyone has in their drawer a 5W 808nm pump diode and IR filter equipped video camera can see through most stuff. There is a real danger of ignition however if the beam is allowed to play over paper too close to the stripe (ouch!) or fiber coupled head (which one SHOULD be using hehe!)

Bright LEDs such as LumiLEDs, Cree, etc. can work too. Again what's important is how intense the light is in a small area. A small 10x10 Fresnel in the sun can be used but is extremely dangerous if you're looking UP at its focused rays!

 

[Kelemvor]

You need a pretty concentrated source for it to work. The work light will probably blind you!

While not exactly something everyone has in their drawer a 5W 808nm pump diode and IR filter equipped video camera can see through most stuff. There is a real danger of ignition however if the beam is allowed to play over paper too close to the stripe (ouch!) or fiber coupled head (which one SHOULD be using hehe!)

Bright LEDs such as LumiLEDs, Cree, etc. can work too. Again what's important is how intense the light is in a small area. A small 10x10 Fresnel in the sun can be used but is extremely dangerous if you're looking UP at its focused rays!

Way too much work for $1. Maybe if I find one that I can win a TV or something.

 

[Rubycon]

Way too much work for $1. Maybe if I find one that I can win a TV or something.

Definitely but get a mechanism down to crack the lottery ones.

You would be surprised at what loops people will go through to get free money. :biggrin:

 

[DrPizza]

Damn, I'm in statistics right now and this is stumping me.

Don't feel too bad. In NY, it's a high school level problem. About 20% of the students are stumped by it, at least the first time they encounter such a problem.

If it helps, there's an alternative method for finding the solution. Since the probability is the number of desired outcomes over the total number of possible outcomes, it's 1 (there's only one winning combination of 3 spots) out of 9C3. (The number of combinations of 3 items taken from a set of 9 items.) So, one out of 9*8*7/(3*2*1). Since 1/ something is the reciprocal of something, it's clear to see that this works out to (3*2*1)/(9*8*7) i.e. the same thing as in the OP.