Matthew Oberholtzer
United States Snellville Georgia

A friend and I had a discussion over the stocks in Life. If I remember correctly, when you buy a stock, you get a card with a number 19, and whenever it is spun, you get money. She selected 9 because it was her lucky number. I selected 1 because I thought I stood to gain the most with that number. My argument was that if nothing but 1s are spun the rest of the game, I will make more than if nothing but 9s are spun the rest of the game. She countered with the argument every number is just as likely to be spun.
Now that she and I are long broken up, I ask you, BGG, does my argument make any sense? Is stock 1 a better investment than stock 9?

Ben Lott
United States Mason Michigan
Being a Lions fan is a gift...
...and a curse.

I think your logic is flawed. The number of spins in a game does not increase the odds of a 1 being spun. I understand your reasoning, I just don't think it's accurate.

Rusty McFisticuffs
United States Arcata California

I think Moberho is right. If there are 20 spaces between you and the end of the track, and you invest in 9 and spin three 9s in a row, the game will be over, and your stock will have paid off three times. However, if you invest in 1 and spin three 1s in a row (which is just as likely, right?), your stock will have paid off three times, and you'll still have an average of ~3 more opportunities for it to pay off again. Interesting!

Tim Mossman
United States Gaithersburg Maryland

If a low number is spun . . . potentially there are more future spins for that player before they reach retirement. So, if you are on a terrific hot streak and your low number is hitting, you will potentially have more opportunities to spin more low numbers.
However, if you have that same low number, and only big numbers are appearing, your opportunities for future "1" are diminished without any corresponding benefit, while a high number stockowning player will benefit.
My intuition tells me there may be a very, very slight advantage over a very large number of plays (i.e., more times than most of us would care to play Life) for having the low number (because game #312 will feature a bunch of consecutive 1s), but not so much to go against a "gut feeling" on a lucky number.
Probably could write a Monte Carlo simulation to find out . . .
Moberho wrote: Now that she and I are long broken up . . .
For the next girlfriend, I'd probably try Lost Cities, Schotten Totten or Balloon Cup.

Rusty McFisticuffs
United States Arcata California

IronMoss wrote: Probably could write a Monte Carlo simulation to find out . . . Dammit! Either my code is wrong, or Moberho is wrong, ha ha. The results I'm seeing show roughly the same payout for each number. (And varying the length of the track from 5 spaces to 100 appears to make no difference; I would have thought that 9 would be worth less, for example, if there were only 5 spaces remaining.)

Nick Reed
United Kingdom Southampton Hampshire

kuhrusty wrote: IronMoss wrote: Probably could write a Monte Carlo simulation to find out . . . Dammit! Either my code is wrong, or Moberho is wrong, ha ha. The results I'm seeing show roughly the same payout for each number. (And varying the length of the track from 5 spaces to 100 appears to make no difference; I would have thought that 9 would be worth less, for example, if there were only 5 spaces remaining.)
Not worthless, no  even though there might only be 5 squares left, there's still an equal chance of getting 9 on the spinner, even though it'll make your piece overshoot on the board.
And it shouldn't make any statistical difference what number you pick. Whether there's 1 square left or the entire board, each time the spinner spins there's an equal chance for any of the numbers to appear. It doesn't matter whether the additional result is that the piece moves 1 square or 9, when it comes round to the next spin, all the numbers still have equal chance again, and that's the important thing  the spinner will spin X times throughout the game, and for each of them you've an equal chance of any number appearing.

Rusty McFisticuffs
United States Arcata California

Erkk, it's time for bed, because my brain isn't working.
Out of 100,000 runs down tracks of random length (between 5 and 99 spaces), the number of times each number pays out is about the same, but the number of games in which the chosen number was the best choice (i.e. it paid out the most times that game, or tied for most) is significantly biased toward the larger numbers! (Like, 16552 games in which 1 was the best choice, up to 19968 games in which 10 was the best choice!) Does that make sense to anyone?? (The alternative is that there's something wrong with my code, but... let's try to stay within the realm of possibility here, ha ha ha.)
So, Moberho, the good news is that you don't have to call her up and apologize, because it looks like she was wrong too.
Ncik wrote: Not worthless, no  even though there might only be 5 squares left, there's still an equal chance of getting 9 on the spinner, even though it'll make your piece overshoot on the board. (Well, I meant "worth less," as in, I thought 1 would pay off more than 9. I believe I was wrong about that, though.)

Scott Muldoon (silentdibs)
United States Astoria New York

This is the most counterintuitive thread of all time.
Cool.

Paul Schulzetenberg
United States Los Angeles CA

kuhrusty wrote: Erkk, it's time for bed, because my brain isn't working.
Out of 100,000 runs down tracks of random length (between 5 and 99 spaces), the number of times each number pays out is about the same, but the number of games in which the chosen number was the best choice (i.e. it paid out the most times that game, or tied for most) is significantly biased toward the larger numbers! (Like, 16552 games in which 1 was the best choice, up to 19968 games in which 10 was the best choice!) Does that make sense to anyone?? (The alternative is that there's something wrong with my code, but... let's try to stay within the realm of possibility here, ha ha ha.)
So, Moberho, the good news is that you don't have to call her up and apologize, because it looks like she was wrong too.
This example isn't a definitive answer, but I thought it might be interesting to see.
S  stock # chosen x  spin x1  first spin x2  second spin etc.. l  length of board B  best stock m  money paid out (in thousands) m1  money paid out for stock #1 m2  money paid out for stock #2 ms  money paid out for stock #s etc.
Trivial example: l = 1 x = (1, 2, ... 9) are all equivalent and will pay out at the same rate.
Second example: l = 2 if x1 = (2, 3, ..., 9, 10) then B = x mx = 1 if x1 = 1 then if x2 = 1 then B = 1 m1 = 2 if x2 = (2, 3, ..., 9, 10) then B = (1, x2) m1 = 1 mx2 = 1
In plain english, if the length of the board is two spaces: 90% of the time 1 will not be your best option 1% of the time it will be your (exclusive) best option and pay out $2000 9% of the time it will be tied for your best option with one other number and pay out $1000 average payout: $2000 * 1/100 + $1000 * 9/100 = $110
89% of the time, 9 will not be your best option 10% of the time, 9 will be your (exclusive) best option and pay out $1000 1% of the time, 9 will be tied with 1 for your best option and pay out $1000 average payout: $1000 * 10/100 + $1000 * 1/100 = $110
I don't have the energy to find a formal mathematical proof, but I bet that there's one that could be found inductively here. My hypothesis is that all of the average payouts will be equal for any board length. The number of times that each number is "best" will skew to the large numbers, but it will be counterbalanced by the fact that when the small numbers are best, they'll pay out more money.
So, the takeaway message is that my gut instinct is that all numbers will average the same payout but that 1 is "riskier"  it won't be the best number as often, but when it is, it will pay out more than when 9 is the best number.
Is there a way to put subscripts in BGG? It would make this explanation a lot clearer.

Ben Vincent
United States Ridgefield Washington

Quote: Out of 100,000 runs down tracks of random length (between 5 and 99 spaces), the number of times each number pays out is about the same, but the number of games in which the chosen number was the best choice (i.e. it paid out the most times that game, or tied for most) is significantly biased toward the larger numbers! (Like, 16552 games in which 1 was the best choice, up to 19968 games in which 10 was the best choice!) Does that make sense to anyone?? (The alternative is that there's something wrong with my code, but... let's try to stay within the realm of possibility here, ha ha ha.)
I think this is reasonable, within the constraints you've established. Consider an extreme example  a track 1 space long. On the next spin, each number from 110 has a 10% chance of showing up. Whatever the result, the game will end after the next spin. Therefore the number you choose doesn't matter.
How about a track with 10 spaces. The odds say the game will end after 2 spins (the average spin being a 5.5). There is a 10% chance the game will end after 1 spin, and only the 10 pays out. These 10% of games are clear wins for the 10. On any result other than 10, there will be a second spin. There is a 90% chance that the second spin does not match the first spin  meaning it's very likely that 2 different numbers will pay out, and no one gains an advantage. Possibly the game will continue to a third or further round, to the extreme of 10x 1s, but the odds of getting even 2x 1 are much less than the odds of getting 1x 10.
I think if you ran your simulation for fixed track lengths (of say 10, 100, and 1000), you'd find that as the track gets longer, the results results tend to equalize.

Rusty McFisticuffs
United States Arcata California

Unitoch wrote: My hypothesis is that all of the average payouts will be equal for any board length. The number of times that each number is "best" will skew to the large numbers, but it will be counterbalanced by the fact that when the small numbers are best, they'll pay out more money. I wasn't seeing the last part, which was why this was mindbending to me"if they all pay out the same number of times, then how can the larger numbers win more games?" But you're right; on a 25space track, 1 will win only ~74% as many games as 10, but in the games where it does win, it will pay out ~1.58 times, while 10 will pay out ~1.26 times in the games where 10 wins.
SabreRedleg wrote: I think if you ran your simulation for fixed track lengths (of say 10, 100, and 1000), you'd find that as the track gets longer, the results results tend to equalize. Yes, mostly: on 10, 100, and 1000space tracks, 1 wins ~79%, ~87%, and ~97% as many games as 10, respectively. But it seems to bottom out around 73% in the 2025 area, and then head back up as you approach length 1.
sdiberar wrote: This is the most counterintuitive thread of all time.
Cool. Yeah! But now I think we're in agreement that, regardless of board length, it's better to bet on 10 than 1, right?
(Now, Moberho, I bet she knew higher numbers were better, and she was playing you with that "oh, 9 is my favorite number" business. It's a good thing you guys broke up!)


