Topps 582 Montgomery Club

[tribefan26]

Originally Posted by JoshMN

So the odds are worse than 1 in a quadrillion, but it's already happened 3 times in 7400 boxes (0.04% of the time)?

I have to believe those people aren't lying- who would admit to just setting 900 bucks on fire if it didn't actually happen. So seems like either the calculation may be a bit off or it's just one of the craziest coincidences ever.

The calculation doesn't apply - the cards are not packed randomly.

[theThrill22]

Topps with a picture of the contents showing on my box, thanks Topps.


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[LittleJimmies]

Originally Posted by brauncj1

Assumption- 1000 cards total
150 rookies

850/1000 is the probability the first card drawn is a non rookie
849/999 is the probability the second card drawn is a non rookie
Etc…

The .85^x would be correct if the card drawn was replaced with the same type of card, rookie or non-rookie.

Ahhhhhhhh, you were taking it to a level of preciseness that is fully correct, but doesn't affect anything in practical terms. Odds for first card would be around 1,517,400/1,782,000 (85.1515%) and 1,517,185/1,781,785 (85.1497%) for the last card in the run. So more impossible than we thought haha!

[tribefan26]

Originally Posted by Dbacksbaseball

Fricken Math guys taking over the thread

I can't help it- I'm an actuary!

[brauncj1]

Originally Posted by SmokeyJ

Both ideas are valid given our lack of knowledge of how Topps does it.

In your scenario, a box can have no doubles, but that clearly happens.

Nope you’re wrong

[SmokeyJ]

Originally Posted by brauncj1

Nope you’re wrong

Can you elaborate? If a card that was selected earlier can be selected again, why would the underlying probabilities change?


In your scenario, Topps lays out 330 cards (1 of each) and pulls 216 of them. I agree that the probability changes with each pull.

In my scenario, Topps has a list of the 330 cards and has a random number generator pick 216 numbers with doubles allowed. In that case, the probability of picking a base card with each random number generator is always .85.

[JoshMN]

Originally Posted by brauncj1

Assumption- 1000 cards total
150 rookies

850/1000 is the probability the first card drawn is a non rookie
849/999 is the probability the second card drawn is a non rookie
Etc…

The .85^x would be correct if the card drawn was replaced with the same type of card, rookie or non-rookie.

Thanks. I was thinking along those lines, but am no expert. So the delta between that calc and 0.0004 (or whatever reality is) = Topps deficiency with respect to random packing. Quite a few orders of magnitude in this case.

[slogue2]

Originally Posted by theThrill22

Topps with a picture of the contents showing on my box, thanks Topps.


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I noticed that when my first box came in yesterday. Smaller, more appropriate sized shipping box, but then the packing slip showing on the outside. Wonder if they are changing up how they fulfill orders.

[CapeTownCards]

Where them #BowmanHeritage at? Come on Topps

[CapeTownCards]

Originally Posted by theThrill22

Topps with a picture of the contents showing on my box, thanks Topps.


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Topps just Toppsing along… doing what Topps do best.

[eastbayak]

Originally Posted by theThrill22

Topps with a picture of the contents showing on my box, thanks Topps.


Sent from my iPhone using Tapatalk

Haha yeah, basically reads "STEAL ME"

[brauncj1]

Originally Posted by SmokeyJ

Can you elaborate? If a card that was selected earlier can be selected again, why would the underlying probabilities change?


In your scenario, Topps lays out 330 cards (1 of each) and pulls 216 of them. I agree that the probability changes with each pull.

In my scenario, Topps has a list of the 330 cards and has a random number generator pick 216 numbers with doubles allowed. In that case, the probability of picking a base card with each random number generator is always .85.

Because every time a card is selected the number of cards to choose the next card from is reduced by 1.

3 card example with 1 rookie

Chance of first card not being a rookie is 2/3
—Given that the first card wasn’t a rookie
Chance of second card not being a rookie is 1/2
This is because there are only 2 cards left.

[zworykin]

Originally Posted by brauncj1

Because every time a card is selected the number of cards to choose the next card from is reduced by 1.

3 card example with 1 rookie

Chance of first card not being a rookie is 2/3
—Given that the first card wasn’t a rookie
Chance of second card not being a rookie is 1/2
This is because there are only 2 cards left.

True - though worth noting that we're talking about a total print run of hundreds of thousands of cards (at least), so (as I think someone already pointed out) the actual effect of this clarification is miniscule.

[jduds]

Originally Posted by brauncj1

Because every time a card is selected the number of cards to choose the next card from is reduced by 1.

3 card example with 1 rookie

Chance of first card not being a rookie is 2/3
—Given that the first card wasn’t a rookie
Chance of second card not being a rookie is 1/2
This is because there are only 2 cards left.

That's all correct and means the last card will have roughly a .43 chance at a rookie instead of .85 making the odds substantially higher if I am right in counting the rookies at 49. They are astronomical anyways though.

[brauncj1]

Originally Posted by zworykin

True - though worth noting that we're talking about a total print run of hundreds of thousands of cards (at least), so (as I think someone already pointed out) the actual effect of this clarification is miniscule.

I pointed it out multiple times that the other method was close enough in this instance but it was incorrectly stated that because we don’t know topps collation that the are both valid and technically that statement is incorrect. It doesn’t matter whether we know topps collation one method is correct and the other is a quick estimation.

[brauncj1]

Originally Posted by jduds

That's all correct and means the last card will have roughly a .43 chance at a rookie instead of .85 making the odds substantially higher if I am right in counting the rookies at 49. They are astronomical anyways though.

The .43 is incorrect. I’m not 100% sure where you got that number but I’m guessing you were just assuming that a box wouldn’t have doubles and only had the 330 cards to pick from which, in reality, isn’t the case.

Just a side note: I don’t want it to get lost that none of this math really means anything in this instance because the collation isn’t random.

[brauncj1]

On a side note- I just opened my box and pulled 48 different base rookies and 4 rookie colors. If there are only 49 rookies I find that highly unlikely as well. Someone early suggested that there had to be some kind of break down in the collation process with the rookie sheets and I lean towards that theory.

[jusdukky]

Originally Posted by brauncj1

On a side note- I just opened my box and pulled 48 different base rookies and 4 rookie colors. If there are only 49 rookies I find that highly unlikely as well. Someone early suggested that there had to be some kind of break down in the collation process with the rookie sheets and I lean towards that theory.

Consider yourself one of the fortunate ones for this collation.

[Jamesebee]

Originally Posted by brauncj1

On a side note- I just opened my box and pulled 48 different base rookies and 4 rookie colors. If there are only 49 rookies I find that highly unlikely as well. Someone early suggested that there had to be some kind of break down in the collation process with the rookie sheets and I lean towards that theory.

I concur. My box from Blowout included a RC or (RC) as the top card in every single pack. There were no other base RC's except for parallels.

[brauncj1]

I just finished putting my box in numerical order and didn’t have a single base duplicate.

[brauncj1]

Originally Posted by Jamesebee

I concur. My box from Blowout included a RC or (RC) as the top card in every single pack. There were no other base RC's except for parallels.

Most of my packs had RC as the 2nd and 3rd card. I found another RC when putting my box in numerical order and if there are only 49 as someone stated then I got them all.

[yoadrian49]

I would jump in here and say one box I opened had 0 RC's, 0 combo cards and 0 league leader cards. All base with no duplicates.

[RobDerhak]

Woof the last two pages

[SmokeyJ]

Originally Posted by brauncj1

Because every time a card is selected the number of cards to choose the next card from is reduced by 1.

3 card example with 1 rookie

Chance of first card not being a rookie is 2/3
—Given that the first card wasn’t a rookie
Chance of second card not being a rookie is 1/2
This is because there are only 2 cards left.

I completely understand that. I agree that it's correct if a card can't be chosen twice. In fact, I stated that specifically in the post your replied to. It's like you aren't reading my posts before responding.

You're doing choosing without replacement and I've suggested choosing with replacement. I lean toward the latter becuase boxes *could* include duplicates.

It's two different ways of trying to estimate the probaility. Both are valid ways as Topps doesn't use either of those methods specifically to pack boxes, but they're both reasonable models. So, the "you're wrong" to me earlier was...well...wrong.

Got my box delivered today, but not planning on any experiments with it. Keepin' it sealed.

[cj828282]

Originally Posted by brauncj1

I just finished putting my box in numerical order and didn’t have a single base duplicate.

Yeah, same with my box. I thought that was a positive.