Fun with math -- TPiR stat question

[lobster]

This comes up every time I order TPiR tickets, and I wanted to ask the pros if my math is correct on this --

Barring the producer interview table, and just purely based on the statistics before entering the studio, if you were to belong to a group of 20, I have calculated that you have better than a 1:2 chance (1 in 1. that someone in your group will be called upon --- here is my math..

330 seats ÷ 9 contestants = overall 1:36 chance of any one person getting picked

then, of that 1:36, if you have 20 people in your group, that ratio becomes 20:36, which gives us 1:1.8.   Is this correct?

cheers
LObs

[DoorNumberFour]

[quote name=\'lobster\' post=\'177774\' date=\'Feb 13 2008, 11:14 AM\']
This comes up every time I order TPiR tickets, and I wanted to ask the pros if my math is correct on this --

Barring the producer interview table, and just purely based on the statistics before entering the studio, if you were to belong to a group of 20, I have calculated that you have better than a 1:2 chance (1 in 1. that someone in your group will be called upon --- here is my math..

330 seats ÷ 9 contestants = overall 1:36 chance of any one person getting picked

then, of that 1:36, if you have 20 people in your group, that ratio becomes 20:36, which gives us 1:1.8.   Is this correct?

cheers
LObs
[/quote]
Seems correct to me, but your personal chance of getting called stays at about 1:36, no matter what size group you're in.

Then, of course, there's the chance that nobody from your group gets picked.

[Fedya]

[quote name=\'lobster\' post=\'177774\' date=\'Feb 13 2008, 12:14 PM\']
This comes up every time I order TPiR tickets, and I wanted to ask the pros if my math is correct on this --

Barring the producer interview table, and just purely based on the statistics before entering the studio, if you were to belong to a group of 20, I have calculated that you have better than a 1:2 chance (1 in 1. that someone in your group will be called upon --- here is my math..

330 seats ÷ 9 contestants = overall 1:36 chance of any one person getting picked

then, of that 1:36, if you have 20 people in your group, that ratio becomes 20:36, which gives us 1:1.8.   Is this correct?

cheers
LObs
[/quote]
So if you're in a group of 36, then there's a 100% chance that somebody in your group will be chosen?

If the nine contestants are picked randomly from the 330 seats, the chance that none of them will be in your group of 20 is:

(310 * 309 * 308 * 307 * 306 * 305 * 304 * 303 * 302) divided by
(330 * 329 * 328 * 327 * 326 * 325 * 324 * 323 * 322)

This comes out to 56.56%.

[Jay Temple]

Not quite. Consider instead the probability that the nine contestants do NOT include anyone from your group.

The probability of the first contestant not being in your group is 310/330. If so, the prob. of the next contestant not being in your group is 309/321, and so on through 302/322. This works out to (310x309x...x302)/(330x329x...x322), which is approximately .566. That means there's only a .434 chance that someone from your group will be chosen.

[clemon79]

Simulmath! (And well done. Probability was my favorite part of pre-calc.)

[lobster]

[quote name=\'Fedya\' post=\'177780\' date=\'Feb 13 2008, 12:04 PM\']
So if you're in a group of 36, then there's a 100% chance that somebody in your group will be chosen?
[/quote]

Right, I thought about that too, which obviously presents the flaw in my simplified math ... but when I put that theory on a smaller scale (keeping with the TPiR theme), you have 20 spaces on the wheel, so there's a 1:20 chance you'd hit the dollar in one spin (forget about the 2nd spin stuff for this example)... but of course it's not 100% guaranteed that someone will hit the $1 on the first spin in 20 attempts... but if you write it out, wouldn't we say:

1:20 chance of hitting the dollar spot
20 spins
20:20 = 100% .. ?

which obviously isn't reality.. so which step in that formula is incorrect?

[clemon79]

[quote name=\'lobster\' post=\'177798\' date=\'Feb 13 2008, 12:18 PM\']
1:20 chance of hitting the dollar spot
20 spins
20:20 = 100% .. ?

which obviously isn't reality.. so which step in that formula is incorrect?
[/quote]
What you are describing is the likelihood of an event happening. Yes, 20 trials of a 1 in 20 event does work out to a probability of 1. Which means nothing more than it SHOULD happen, and that if it does not it's a statistical anomaly.

(The other problem is that you are trying to apply multiple trials to what is a single-trial activity. You could spin the wheel 400 times, and on the 401st time the odds of hitting the buck are still 1 in 20. Which means that you *should* have gotten 20 $1.00 spins, but doesn't mean you're going to.)

[Fedya]

[quote name=\'lobster\' post=\'177798\' date=\'Feb 13 2008, 04:18 PM\']
[quote name=\'Fedya\' post=\'177780\' date=\'Feb 13 2008, 12:04 PM\']
So if you're in a group of 36, then there's a 100% chance that somebody in your group will be chosen?
[/quote]

Right, I thought about that too, which obviously presents the flaw in my simplified math ... but when I put that theory on a smaller scale (keeping with the TPiR theme), you have 20 spaces on the wheel, so there's a 1:20 chance you'd hit the dollar in one spin (forget about the 2nd spin stuff for this example)... but of course it's not 100% guaranteed that someone will hit the $1 on the first spin in 20 attempts... but if you write it out, wouldn't we say:

1:20 chance of hitting the dollar spot
20 spins
20:20 = 100% .. ?

which obviously isn't reality.. so which step in that formula is incorrect?
[/quote]
There's a 95% chance (p=0.95) that you won't hit the $1 space.  Each spin is independent of each other (for the purposes of this question), so the probability that you won't hit the $1 space in 20 spins is (0.95)^20, which is just under 36%.

Probability question for the calculus students: New York used to have a lotto that was pick 6 out of 54 numbers, meaning there were 54C6, or 25,827,165 possibile combinations.  They also had a "quick pick" that would randomly pick the combination for your ticket.  Assuming that the "quick picks" are fully random (and not pseudorandom) in that on any given quick pick, each and every combination is exactly as likely to come up as any other combination, if you play 25,827,165 quick picks, what's the probability that you won't have the winning combination?

[lobster]

[quote name=\'clemon79\' post=\'177800\' date=\'Feb 13 2008, 02:30 PM\']
[quote name=\'lobster\' post=\'177798\' date=\'Feb 13 2008, 12:18 PM\']
1:20 chance of hitting the dollar spot
20 spins
20:20 = 100% .. ?

which obviously isn't reality.. so which step in that formula is incorrect?
[/quote]
What you are describing is the likelihood of an event happening. Yes, 20 trials of a 1 in 20 event does work out to a probability of 1. Which means nothing more than it SHOULD happen, and that if it does not it's a statistical anomaly.

(The other problem is that you are trying to apply multiple trials to what is a single-trial activity. You could spin the wheel 400 times, and on the 401st time the odds of hitting the buck are still 1 in 20. Which means that you *should* have gotten 20 $1.00 spins, but doesn't mean you're going to.)
[/quote]

I gotcha.. it's kind of like how every spin of the roulette wheel has its own odds of hitting red/black but if you realize that the last 8 spins were red, even though the next spin has its own same near-even odds, you can still look at the overall picture in hindsight and say "the chances that 8 spins coming up red were this...."  So!  

Having said all of that, what ARE the true odds of saying "What are the odds of hitting the $1 spot at least once in twenty tries?", and how do you come to that figure?

[lobster]

[quote name=\'Fedya\' post=\'177801\' date=\'Feb 13 2008, 02:32 PM\']
 Assuming that the "quick picks" are fully random (and not pseudorandom) in that on any given quick pick, each and every combination is exactly as likely to come up as any other combination, if you play 25,827,165 quick picks, what's the probability that you won't have the winning combination?
[/quote]

Could you not use your TPiR wheel formula for this, if only on a larger scale?
That is, calculate the odds of NOT winning once and send that figure to the 25,827,165th power?

[clemon79]

[quote name=\'lobster\' post=\'177811\' date=\'Feb 13 2008, 01:12 PM\']
Having said all of that, what ARE the true odds of saying "What are the odds of hitting the $1 spot at least once in twenty tries?", and how do you come to that figure?
[/quote]
You're not following me. The probability is 1. It SHOULD happen. If it does not, they're bucked the odds. That was my entire point.

Probability math assumes a perfect distrubution of results. If the wheel gives that distribution, yes, by definition the dollar MUST come up. The wheel does not give that distrubution.

(and, again, you are trying to apply probability math to multiple trials, when one trial affects the next one not at all.)

EDIT: Fedya might have covered this more correctly than I; it's been a few years. But my point holds.

[tvrandywest]

Boy did we need you guys on Sunday at a taping of Jonathan Goodson's BIG SPIN for the California lottery. Those with the theoretical knowledge couldn't do the math in their heads, and those with calculators didn't understand the theory. A producer in the booth had the answer, but it took an AUDIENCE MEMBER to do the math before any of us on the floor had it.

Simply, round 2 of the ACES HIGH game has 2 players head-to-head trying to create a better 4 CARD poker hand using only Aces, Kings and Queens. Player 1 has already drawn cards and ends up with 4 Kings. Player 2 has 4 Queens and has to draw cards. The only winning hand would be 4 Aces.

Player 2 discards what looked like a great hand, 4 Queens, hoping that the 4 cards he draws will be all Aces. And he makes it!

Chance of each of the 4 cards being an Ace is 1 in 3. So  all four cards being Aces is 3 x 3 x 3 x 3 = 1 in 81.

Back to TPiR:

What are the odds of getting exactly a dollar in one or a combination of two spins on the big wheel?

My uneducated guess: 1 in 20 PLUS 1 in 20? 1 in 40???

Randy
tvrandywest.com

[mcsittel]

[quote name=\'tvrandywest\' post=\'177829\' date=\'Feb 13 2008, 04:55 PM\']
Back to TPiR:

What are the odds of getting exactly a dollar in one or a combination of two spins on the big wheel?

My uneducated guess: 1 in 20 PLUS 1 in 20? 1 in 40???
[/quote]

Spin 1:

1/20 chance of $1
19/20 of something else

Spin 2:

If not $1 on the first spin, 1/20 chance of getting the number that, added to your non-dollar first spin, will equal $1 total.

So...

1/20 + ((19/20)*(1/20)) = 1/20 + 19/400 = 20/400 + 19/400 = 39/400 = 0.0975.

And 1/20 + 1/20 is 2/20, not 1/40.  I've had many a student tell me otherwise!

[clemon79]

[quote name=\'tvrandywest\' post=\'177829\' date=\'Feb 13 2008, 02:55 PM\']
What are the odds of getting exactly a dollar in one or a combination of two spins on the big wheel?

My uneducated guess: 1 in 20 PLUS 1 in 20? 1 in 40???

Randy
tvrandywest.com
[/quote]
Well, no, it's gonna be slightly better than 1 in 20, because you can do that in a single spin.

I'm pretty sure it's 1 in 10, because you're effectively getting two shots at hitting a 1 in 20; if you miss it the first time, you have another 1 in 20 shot to hit what you need to make a buck. Which works out to 2 in 20, or 10%.

[TLEberle]

[quote name=\'tvrandywest\' post=\'177829\' date=\'Feb 13 2008, 02:55 PM\']Back to TPiR:

What are the odds of getting exactly a dollar in one or a combination of two spins on the big wheel?

My uneducated guess: 1 in 20 PLUS 1 in 20? 1 in 40??? [/quote]Out of the 380 ways that you can spin the wheel twice, nineteen of them result in a total of $1, plus the 1/20 of getting $1 off the bat and not needing that second spin enters a form of probability that I can't do in my fifteen minutes worth of break time without drawing a probability tree. But I think it's still close to P = 0.05.