Pepsi Play For a Billion: statistical analysis

[Blaq]

Let me preface by suggesting that we're entering into a territory filled with landmines. The Monty Hall Paradox has been known to throw friends and family members into screaming fits, and it's a seemingly simpler problem than PFaB!

Even though I usually have a great grasp of probability analysis, while emailing a friend to describe Sunday night's special, I found myself repeatedly changing my analysis. I hope we can crunch the numbers together and deal with the game's puzzling twists without turning this into a flamefest.

I'll now let some of you pipe in with your analysis. Let's make this a thread to remember, and in a good way...

[clemon79]

I'm not sure what there is to crunch: Upon entering the building, each of the 992 players (8 \"had a wedding\", according to Drew, and couldn't make it or send a proxy in their place) had an equal chance (so, 1 in 992) of winning a million, and a 992 in 1,000,000 chance of winning the billion. (More on that later.)

Once the knockout round started, any player who hit their buzzer to take the bribe was bucking odds against them, since only one player per round is eliminated and the chances oif their name never showing up in an envelope (because they had the right number in the first place) are just as good as anyone else's.

The whole \"inheritance\" of numbers was smoke and mirrors. All that did was guarantee that the closest (and therefore most interesting) number, whether or not it was initially selected by the million-dollar winner, was the one that would be compared against the number that Holly and the chimp came up with. You might as well have given the winner the 982 numbers of the non-finalists, because we know those weren't as close to correct as the ten finalists, and you might as well have given the winner the numbers of those players knocked out with no money, because those weren't as close as his either (or else HE would have been knocked out instead). And since they DID give the numbers of the players who took the guaranteed cash to him, ultimately, they gave him ALL of the numbers. For the purposes of TV, they only \"gave\" him the numbers he might have a vested interest in, but mathematically it's the same thing.

Since 992 people picked numbers, that effectively means there were 992 chances to win the billion, and that up until the final minute, 991 of those had been eliminated as being too far away.

So the chances of their BEING a billion-dollar payout were about 1 in 1,000. (I'm rounding up the 992 for simplicity.)

The chances of any of the individual players winning the million, and therefore qualifying for the billion dollar payout were also about 1 in 1000.

The chances of both happening to a single specified player would be found by multiplying the two together, which brings us back to 1 in 1,000,000. Which, since they selected a six-digit number between 000,000 and 999,999 which they had to match exactly to win, makes perfect sense.

[Matt Ottinger]

Unlike the Monty Hall Paradox, here the choices that are made affect the outcome.  In other words, despite appearances, this isn't a totally random game.  As I think Davies himself suggested a while ago, it amounted to a high-stakes game of chicken.  I don't think it's the sort of thing that lends itself to a *purely* mathematical analysis.  Every individual is going to have a different, personal POV about how much a \"bird in the hand\" is worth to them.

I'll give Davies credit for making it somewhat engaging, but as Chris said, the vast majority of it was smoke and mirrors.

[Jay Temple]

By one measure, this is pretty simple:  At no stage of the game was the bribe enough to be worth taking.  (And this is true even if we ignore the possibility of the billion dollars.)

At the first stage of the game, ten players, if you do not take the bribe, there is a .1 chance of winning the million, for an expected value of $100,000, which is more than the $20,000 bribe.  At every stage after that, it is more than $100,000.

In the last stage of the game, two players, the bribe climbs to $100,000, but the expected payoff is .5 x $1,000,000, or $500,000.

So, at any stage between those, bribe < $100,000 < expected payout.

I said at the top, \"By one measure, ... at no stage of the game was the bribe enough to be worth taking.\"  The measure in question is the expected value.  That is what your average winnings would be per game if you played the scenario that way lots and lots of times.

However, the phrase \"worth taking\" is subjective.  One could just as easily argue that the strategy with the highest guaranteed payout is the best.  That would be taking the first bribe, because the least (and most) you can end up with is $20,000, while not taking the bribe could mean that you end up with nothing.

If I have $21,000 in student loans, I might well conclude that it is better to refuse the $20,000 bribe and try for the $30,000 bribe:
> If I take the $20,000 bribe, the probability of paying off my loans is 0.
> If I pass on the $20K and take the $30,000 bribe, the probability of paying off my loans is about 0.9  (I'm out with nothing if I have the 10th best guess and no one takes the $20,000 bribe.  I also have to ring my buzzer first to take the $30,000, but this is offset by the expected value going up if I'm not.)
> If I pass on the $20K and $30K, the probability of paying off my loans is about 0.8.  This represents the probability that I have one of the eight best guesses, or two people before me quit, etc., as above.
In other words, this strategy maximizes the probability of winning at least $21,000.

To sum up, the average payout is maximized by not taking a bribe at all, but different people will have different, equally valid opinions on what thing they want to maximize.

[clemon79]

[quote name=\'Jay Temple\' date=\'Sep 16 2003, 01:47 PM\'] To sum up, the average payout is maximized by not taking a bribe at all, but different people will have different, equally valid opinions on what thing they want to maximize. [/quote]
 That's an excellent way of addressing the bribe portion of the game, Jay. I particularly like the application of \"expected payout\". Very well put.

[Speedy G]

This seems much like the argument about Russian Roulette's bonus round: do you play for the $100K or take the $10K, given that you have a better expectation if you go for the $100K?

Sure, you should mathematically never take a bribe to maximize your winnings, but you run the risk.  If you play the odds, there's a chance every round that you have, effectively, won a trip to wherever they taped the show.  Does the bribe sound like enough money for you?

To win a million (ignoring the rounds before the final, since I take it that there wasn't much personal choice involved), you have to:
A) be standing on stage with 9 people who all find a need for X ten thousand dollars,
B) be lucky enough to not have your name pulled for elimination 9 times, or
C) a combination of the two.

If everyone randomly chose friend or foe, then only a quarter of the teams should ever leave with nothing to show on either side of the desk, but we all know that's not nearly the case.  It's not just a math thing, it's a personality thing too.

I'm sure Pepsi would love for everyone to be math whizzes, though.  Then they don't have to pay out any money on top of the grand prize.  =D

Not to drag it off topic with what everyone would personally do, but if it were me up there, I'd wait out 90% of the clock to see if anyone else feels they need the money before taking a walk myself.  Improve your guaranteed money if possible.

[cyberjoek]

Ok, here is the extra problem, \"what are the odds of winning with only one bottle cap?  (There were 20 million caps entered)\"

So we figure the odds of getting on the shows are 1,000 in 20,000,000.  Simplified 1 in 20,000.
Multiply that by 1 in 1000 for the million to get 1 in 20,000,000 (duh!) and for the billion, 1 in 20,000,000,000 (one in twenty billion).

A. Can anyone see errors in the analysis?
B. Those odds don't look so good do they?

-Joe Kavanagh

[clemon79]

[quote name=\'cyberjoek\' date=\'Sep 16 2003, 06:01 PM\'] B. Those odds don't look so good do they?
 [/quote]
 Do you think anyone would have insured them if they were?

[Peter Sarrett]

The thing that baffled me the most were the people (all women) who hit the buzzer immediately or almost immediately.  In each round, there was no reason to buzz in until there was only 1 second left on the clock.  Almost nobody buzzed in that late.

If you'd already decided you wouldn't stick it out to the end, your best strategy was to wait until the last possible moment in each round before bailing.  The longer you wait, the greater the chance someone else will buzz in before you, allowing you to last another round.  The more rounds you last, the more money you take home.  The people who buzzed in early obviously hadn't thought it through.  They cracked.

I loved the fact that the \"right\" guy won the million.  Can you imagine if you'd accepted $30,000 only to find out you'd given up a million bucks?

  - Peter

[Ian Wallis]

I loved the fact that the \"right\" guy won the million. Can you imagine if you'd accepted $30,000 only to find out you'd given up a million bucks?

I think it comes down to what you'd be willing to accept.  As a lottery player for 19 years, I've unfortunatly never come close to winning the big prize.  When you're on that stage you've probably got better odds than ever of winning a big payout - but still have the possibility of leaving with nothing.  I'd be disappointed if I made it that far and left with nothing.

I certainly wouldn't accept a $20,000 payout - not when there's a chance at a million.  When I was watching the show my original thought was I'd be in it to the end - but if your number wasn't the closest you still might not have won anything.  I guess it depends on how confident you are that you're closest - but at that point it's just a toss of a coin because nobody in the studio had any idea what the number actually was.

If I was going to take the buyout, I would wait until it was something that would make a significant difference.  If you're down to the last two and hit the button, you're handing the other guy a million dollars.  That would be hard to live with if I eventually found out it was my number that was the closest.  I think if I was in the last 3, I would have considered hitting it at $90,000.  That's a big purse to take that would be a great financial help - and certainly more than I had ever won before.  By doing that you're not handing the other guy a million - they still have to \"play chicken\" in the final two.

I guess the better question is - when you have that kind of chance, do you gamble and stay in it to the end with the possibility of leaving with nothing, or do you take the biggest payout possible and at least leave with a nice check?  It's a tough choice...

[clemon79]

[quote name=\'Ian Wallis\' date=\'Sep 17 2003, 07:05 AM\'] If you're down to the last two and hit the button, you're handing the other guy a million dollars.  That would be hard to live with if I eventually found out it was my number that was the closest. [/quote]
 I wasn't at all surprised when they got down to the last two and both of them stuck it out. At that point the odds are 50/50 that you're the big winner. If you've bucked the odds up until that point (or watched other people do it for you), you're gonna stay in until the end.

At the same time, I can understand why a lot of the players weren't clear-headed and buzzed in quicker than they should have. Consider: Out of 2 million entrants, you're one of the final 1,000 to be flown to Florida, and now you're one of the final 10. For ONE MEEEELION DOLLARS. Show me the player who _isn't_ a bundle of nerves at that point. Everything's gonna be a whirlwind.

[Jay Temple]

[quote name=\'cyberjoek\' date=\'Sep 16 2003, 08:01 PM\'] Ok, here is the extra problem, "what are the odds of winning with only one bottle cap?  (There were 20 million caps entered)"

So we figure the odds of getting on the shows are 1,000 in 20,000,000.  Simplified 1 in 20,000.
Multiply that by 1 in 1000 for the million to get 1 in 20,000,000 (duh!) and for the billion, 1 in 20,000,000,000 (one in twenty billion).

A. Can anyone see errors in the analysis?
B. Those odds don't look so good do they?

-Joe Kavanagh [/quote]
 The odds you cite are correct:  The chances that one bottle cap gets you a million are 1 in 20,000;  a billion, 1 in 20,000,000.  However, I'm not sure what you mean by B.  The chances of Pepsi having to pay the billion to someone are 1,000 (number of different guesses)/1,000,000 (number of possible outcomes) = 1 in 1,000.

Incidentally, I think there is a strategy that increases your chances of winning the million, but it has no effect on your chances of winning the billion.

Recall how the \"best\" guesses are determined:
1) number of digits that match exactly
2) number of remaining digits that are correct but in the wrong place
3) \"mathematically\" closest (So, if the correct number is 444451, then 444449 is closer than 444460)
4) lower number

123456 and 777777 have the same chance (.000001) of being exactly right, and the expected number of digits that are in the exact right place is 0.6.

However, if the last four digits are 7777 and neither the first nor the second is 7, the expected number of digits in 777777 that are right-but-in-the-wrong-place is 0.0.  If the last four digits are 3456 and the first is not 1 and the second is not 2, the expected number of RBIWP's is 0.2.

So, you're better off choosing a number with six different digits.

The 3rd and 4th tie-breakers are \"mathematically closest,\" and lower number.  The expected value for the six-digit number generated is 499,999.5, so you should choose a number that gets you close to that number, but less than the number.  The six-digit number that best meets my criteria is ... 498765.  (For all this, of course, I probably wasn't even in the top 100, let alone top 10.)