## Posts tagged ‘probability’

### Two the Hard Way (and an Easy Way)

During our trip to Arizona for winter break, two problems surfaced organically while we were on holiday. (Sorry. Two *math* problems. There were lots of non-math problems, too, but we don’t have time for all that.)

On the plane, the interactive in-flight map showed the outside temp, toggling between Fahrenheit and Celsius. That led to the following MJ4MF original problem, which I thought — and still think — is pretty good:

The conversion between Fahrenheit and Celsius temperatures follows the rule

F= 9/5C+ 32. Sometimes, the temperature is positive for both Celsius and Fahrenheit; sometimes, the temperature is negative for both Celsius and Fahrenheit; and other times, Fahrenheit is positive while Celsius is negative. What is the least possible product of the Fahrenheit temperature and its corresponding Celsius temperature?

You can pause here if you’d like to solve this before I present a spoiler.

Before I reveal two different solutions, allow me to digress. It could be that the statement about the temps sometimes being positive and sometimes being negative is denying a teachable moment. The graph below shows the linear relationship between the two temperature scales. Perhaps a good classroom question is:

When will the product

CFbe positive and when will it be negative?

Or maybe a better question is:

When are

CandFboth positive, when are they both negative, and when do they have different signs?

So, to the problem that I posed. As I thought about it on the plane, I concluded that if *F* = 1.8*C* + 32, then the product *CF* = 1.8*C*^{2} + 32*C*. I then used calculus, found the derivative (*CF*)’ = 3.6*C* + 32, set that equal to 0, and concluded that *C* = ‑8.89, approximately. The corresponding Fahrenheit temperature is *F* = 16, so the minimum product is roughly ‑142.22.

Using calculus was like rolling a pie crust with a steamroller, though. I could have just as easily graphed the parabola and noted its vertex:

If I had used the other form of the rule, namely *C* = 5/9 (*F* ‑ 32), things might have been a little easier. Maybe. In that case, setting the derivative equal to 0 yields *F* = 16, which is arguably a nicer number. But then you still have to find the corresponding Celsius temperature, which is *C* = ‑8.89, and the product is still roughly ‑142.22. So, not much easier, if at all, and again graphing the parabola and noting its vertex would have done the trick:

The only real benefit to using this alternate version of the rule is that it provides a reasonable check. Since both methods — and both graphs — yield an answer of ‑142.22, we can feel confident in the result.

But there’s an easier way to solve this one.

Thinking this was a good problem — and because I like when my sons make me feel stupid — I gave it to Eli and Alex. Within seconds, Eli said, “Well, *F* is positive and *C* is negative between 0°F and 32°F, so the minimum will occur halfway between them at *F* = 16. That means *C* = ‑80/9, so it’s whatever ‑1280/9 reduces to.” (Turns out, -1280/9 = ‑142 2/9 ≈ ‑142.22.)

Eli hasn’t taken calculus, so he doesn’t *know* — or, at least, he hasn’t *learned* — that the minimum product should occur halfway between the *x*– and *y*‑intercepts of the linear graph. Yet, he had an intuitive insight that just happens to be true. As a result, what took me about five minutes of deriving and manipulating took him about five seconds.

The second problem arose at the grocery store. Among our purchases was a box of sugar cubes, which contained, surprisingly, 126 cubes.This number is surprising in the sense that it’s not a number you’ll see very often, except for an occasional appearance in the ninth row of Pascal’s Triangle, or maybe if you’re a chemist searching for stable atoms.

A question that could have arisen from this situation involves surface area and volume:

A rectangular prism with integer dimensions has a volume of 126 cubic units. What is the least possible surface area?

That’s not the question that was shared with Alex and Eli, though. (The answer, if you care, is 162 square units, which results from a 3 × 6 × 7 arrangement — which, in fact, is the exact arrangement of cubes in the box above. I suspect this is not a coincidence.)

The problem that I shared with my sons involved probability:

Imagine that the arrangement of cubes is removed from the box intact, and all six faces of the prism are painted red. Then one of the sugar cubes is selected at random and rolled. What is the probability that the top face of the rolled cube will be red?

The boys made an organized list, as follows:

Painted Faces |
Number of Cubes |

3 | 8 |

2 | 40 |

1 | 58 |

0 | 20 |

Further, the boys reasoned:

- P(cube with 3 red faces, red face lands on top) = 8/126 x 1/2 = 8/252
- P(cube with 2 red faces, red face lands on top) = 40/126 x 1/3 = 40/378
- P(cube with 1 red face, red face lands on top) = 58/126 x 1/6 = 58/756

Therefore,

- (P of getting a red face) = 8/252 + 40/378 + 58/576 = (24 + 80 + 58) / 576 = 162/576 = 9/42

Wow! That seems like a lot of work to get to the answer. Surely there’s an easier way, right?

Indeed, there is.

Notice that the penultimate step yielded the fraction 162/576. The numerator, 162, may look familiar. It’s the answer to the question that wasn’t asked above, the one about the least possible surface area of the prism. That’s no coincidence. In total, there will be 162 faces painted red. And there are 6 × 126 = 576 total faces on all of the sugar cubes (that is, six faces on each cube). This again suggests that the probability of rolling a red face is 162/576.

Did you happen to notice that the volume and surface area use the same digits in a different order? Cool.

So there you have it, two problems, each with two solutions, one easy and one hard. Or as mathematicians might say, one elegant and one common.

It’s typical for problems, especially problems worth solving, to have more than one solution strategy. What’s the trick to finding the elegant solution? Sadly, no such trick exists. Becoming a better problem solver is just like everything else in life; your skills improve with practice and experience. It’s akin to Peter Sagal’s advice in *The Incomplete Book of Running*, where he says, “You want to be a writer? […] Just sit down and write. The more you write, the better a writer you will become. You want to be a runner? Run when you can and where you can. Increase your mileage gradually, and your body will respond and you’ll find yourself running farther and faster than you ever thought possible.” You want to be a problem solver? Then spend your time solving problems. That’s the only way to increase the likelihood that you’ll occasionally stumble on an easy, elegant solution.

And every once in a while, you may even solve a problem faster than your kids.

### Mr. Consistency, Khris Davis

If you flipped four coins, the probability of getting exactly one head would be 0.25.

But the probability of doing that four times in a row is much lower, somewhere closer to 0.0039, or about 1 in 250.

Now, imagine flipping 100 coins four times, and getting the same number of heads each time. The odds of that happening are only slightly better than impossible. In fact, if every person *in the entire world* were to flip 100 coins four times, it would still be highly unlikely that this would ever happen.

That’s how rare it is, and it gives you some idea of what Major League Baseball player Khris Davis just pulled off. The Oakland Athletics outfielder just finished his fourth consecutive season with a batting average of .247. That’s right — the same average four seasons in a row.

Davis had some advantage over our coins, though. For starters, he wasn’t required to have the same number of at-bats every year. Moreover, batting averages are rounded to three decimal places, so his average wasn’t *exactly* the same during those four years; it was just really, really close:

**2015**: .24745 (97 hits in 392 at-bats)**2016**: .24685 (137 in 555)**2017**: .24735 (140 in 566)**2018**: .24653 (142 in 576)

How could something like this happen? According to Davis, “I guess it was meant to be.“

Perhaps it *was* predestination, but I prefer to put my faith in numbers.

Empirically, we can look at the data. From 1876 to present, there have been 19,103 players in the major leagues. The average length of an MLB career is about 5.6 years, which means that an average player would have about three chances to record the same batting average four seasons in a row. It’s then reasonable to say that there have been approximately 3 × 19,103 = 57,309 opportunities for this to happen, yet Khris Davis is the only one to accomplish this feat. So experimentally, the probability is about 1 in 60,000.

Theoretically, we can look at the number of ways a player could finish a season with a .247 batting average. In 2007, the Phillies’ Jimmy Rollins recorded an astounding 716 at-bats. That’s the most ever by a Major League Baseball player. So using a sample space from 1 to 716 at-bats, I determined the number of ways to achieve a .247 batting average:

- 18 hits, 73 at-bats
- 19 hits, 77 at-bats
- 20 hits, 81 at-bats
- 21 hits, 85 at-bats
- 22 hits, 89 at-bats
- 36 hits, 146 at-bats
- …
- 161 hits, 652 at-bats
- 161 hits, 653 at-bats
- …
- 177 hits, 716 at-bats

And, of course, there are the examples above from Davis’s last four seasons.

It’s interesting that it’s not possible to obtain a batting average of .247 if the number of at-bats is anywhere from 90 to 145; yet it’s possible to hit .247 with 161 hits for either 652 or 653 at-bats. I guess it’s like Ernie said: “That’s how the numbers go.“

All told, **there are 245 different ways to hit .247** if the number of at-bats is 716 or fewer.

That may sound like a lot, but consider the alternative: there are 256,441 ways to **not** hit .247 with 716 or fewer at-bats.

So, yeah. No matter how you look at it, what Davis did is pretty ridiculous. Almost as ridiculous as what happened to Saul…

Saul is working in his store when he hears a voice from above. “Saul, sell your business,” the voice says. He ignores it. His business is doing well, and he’s happy. “Saul, sell your business,” the voice repeats. The voice goes on like this for days, then weeks. “Saul, sell your business.” Finally, Saul can’t take it any more. He finds a buyer and sells his business for a nice profit.

“Saul, take your money, and go to Las Vegas,” the voice says.

“But why?” asks Saul. “I have enough to retire!”

“Saul, take your money to Las Vegas,” the voice repeats. It is incessant. Finally, Saul relents and heads to Vegas.

“Saul, go to the blackjack table and bet all your money on one hand.”

He hesitates for a moment, but he knows the voice won’t stop. So, he places his bet. He’s dealt 18, while the dealer has a 6 showing. “Saul, take a card.”

“What? The dealer has…”

“Saul, take a card!” the voice booms.

Saul hits. He gets an ace, 19. He sighs in relief.

“Saul, take another card.”

“You’ve got to be kidding me!” he pleads.

“Saul, take another card.”

He asks for another card. Another ace, 20.

“Saul, take another card,” the voice demands.

But I have 20!” Saul shouts.

“TAKE ANOTHER CARD, SAUL!”

“Hit me,” Saul says meekly. He gets another ace, 21.

And the voice says, “Un-fucking-believable!”

### Nationals Win Probability, and Other Meaningless Statistics

The first pitch of last night’s Nationals-Phillies game was 8:08 p.m. That’s pretty late for me on a school night, and when a 38-minute rain delay interrupted the 4th inning, well, that made a late night even later.

The Phillies scored 4 runs in the top of the 5th to take a 6‑2 lead. When the Nationals failed to score in the bottom of the 5th, I asked my friends, “What are the chances that the Nationals come back?” With only grunts in response and 10:43 glowing from the scoreboard, we decided to leave.

On the drive home, we listened as the Nationals scored 3 runs to bring it to 6‑5. That’s where the score stood in the middle of the 8th inning when I arrived home, and with the Nats only down by 1, I thought it might be worth tuning in.

The Nats then scored 3 runs in the bottom of the 8th to take an 8-6 lead. And that’s when an awesome stat flashed on the television screen:

Nats Win Probability

- Down 6-2 in the 6th: 6%
- Up 8-6 in the 8th: 93%

Seeing that statistic reminded me of a Dilbert cartoon from a quarter-century ago:

I often share Dogbert’s reaction to statistics that I read in the newspaper or hear on TV or — *egad!* — are sent to me via email.

I had this kind of reaction to the stat about the Nationals win probability.

For a weather forecast, a 20% chance of rain means it will rain on 20% of the days with exactly the same atmospheric conditions. Does the Nats 6% win probability mean that *any team* has a 6% chance of winning when they trail 6-2 in the 6th inning?

Or does it more specifically mean that the Nationals trailing 6-2 in the 6th inning to the Phillies would only win 1 out of 17 times?

Or is it far more specific still, meaning that this particular lineup of Nationals players playing against this particular lineup of Phillies players, late on a Sunday night at Nationals Stadium, during the last week of June, with 29,314 fans in attendance, with a 38-minute rain delay in the 4th inning during which I consumed a soft pretzel and a beer… are **those** the right “atmospheric conditions” such that the Nats have a 6% chance of winning?

As it turns out, the win probability actually includes lots of factors: whether a team is home or away, inning, number of outs, which bases are occupied, and the score difference. It does not, however, take into account the cost or caloric content of my mid-game snack.

A few other stupid statistics I’ve heard:

- Fifty percent of all people are below average.
- Everyone who has ever died has breathed oxygen.
- Of all car accidents in Canada, 0.3% involve a moose.
- Any time Detroit scores more than 100 points and holds the other team below 100 points, they almost always win.

**Have you heard a dumb stat recently?** Let us know in the comments.

### Morelli, Coleman, and Statistical Outliers

You won’t see Pete Morelli and crew officiating tonight’s Monday Night Football game — and Philadelphia Eagles’ fans couldn’t be happier.

At kick-off, more than 74,000 fans had signed a petition to have Morelli banned from serving as the referee for any Eagles’ game. That’s because last Thursday night, Morelli and his crew called 10 penalties for 126 yards against the Eagles, whereas they only called 1 penalty for 1 yard against their opponents, the Carolina Panthers.

But Philadelphia sports reporter Dave Zangaro pointed out that Morelli has a history of lopsided officiating against the Eagles. In the last four Eagles’ games that Morelli has covered, his crew has called 40 penalties for 396 yards against the Eagles, but only eight penalties for 74 yards against the opponents.

No doubt, that’s quite a disparity.

But I’m curious if any of the petition signers have actually checked the numbers. Statistical anomalies happen, and I suspect that the imbalance they’ve identified is likely one of many. I didn’t run the numbers to determine if Morellli’s stats constitute an outlier; that would be too much work. But, I did take a quick peek at the other referees in the league to see what I can see.

And what I found leads me to wonder, **Why hasn’t anyone started a petition to get Walt Coleman banned from officiating Atlanta Falcons games?** Maybe it’s because Coleman officiates *in favor* of the Falcons.

Check it. In the last six Falcons’ games that Coleman has officiated, the Falcons have been penalized only 29 times for 216 yards. Their opponents, by comparison, have been penalized 53 times for 463 yards. That’s an average of four fewer penalties and half as many penalty yards per game.

And it’s even worse if you consider only home games. In those four games, the advantage is just 16 penalties for 111 yards against the Falcons to 37 penalties for 320 yards against their opponents.

Don’t believe me? Take a look…

Date | Game | Opponent’s Penalties | Opponent’s Penalty Yds | Falcons’ Penalties | Falcons’ Penalty Yds |

9/30/12 | Panthers @ Falcons | 9 | 64 | 2 | 15 |

1/13/13 | Seahawks @ Falcons | 6 | 35 | 3 | 11 |

9/29/13 | Patriots @ Falcons | 9 | 93 | 6 | 55 |

12/23/13 | Falcons @ 49ers | 7 | 45 | 5 | 37 |

12/4/16 | Chiefs @ Falcons | 13 | 128 | 5 | 30 |

9/24/17 | Falcons @ Lions | 9 | 98 | 8 | 68 |

Totals |
53 |
463 |
29 |
216 |

Admittedly, those numbers aren’t quite as stark as Morelli’s, but they don’t exactly paint a picture of Coleman as an impartial ref, either.

In 2012, replacement official Brian Stropolo was banned from working a New Orleans Saints’ game when pictures of him donning Saints’ attire were found on his Facebook page. So there is precedence if the NFL wants to use my analysis to ban Coleman from Falcons’ games, or if they want to accept the petition and ban Morelli from Eagles’ games.

But let’s keep this in perspective and remember one thing: **It is Philadelphia**, after all. I mean, we’re talking about a sports town where fans threw snowballs at Santa Claus and threw batteries at Eagles quarterback Doug Pederson — the same Doug Pederson, in fact, who is now the Eagles coach. So if Morelli and his crew are deliberately blowing the whistle more against the Eagles than their opponents, who cares? This type of denigration couldn’t be offered to a more deserving team.

### Constant Change

I’m also old, cranky, and cynical. Whatever.

My frustration is not the my-flight-was-delayed-three-times-then-eventually-cancelled-and-there-are-no-more-flights-to-Cleveland-till-tomorrow-morning type. It’s not even the can’t-believe-my-boss-is-making-me-go-to-Cleveland kind of frustration. More like the why-aren’t-there-the-same-number-of-hot-dogs-and-buns-in-a-pack variety. So it’s a First World problem, to be sure, but still annoying. I’ll explain more in a moment.

But first, how ’bout a math problem to get us started?

If you make a purchase and pay with cash, what’s the probability that you’ll receive a nickel as part of your change?

Sure, if you want to get all crazy about this, then we can take all the fun out of this problem by stating the following assumptions:

- You only pay with paper currency. If you paid with coins, then the distribution of coins you’d receive as change would likely vary quite a bit.
- You never use 50¢ coins. Honestly, they’re just too obscure.
- Transaction amounts are uniformly distributed, so that you’re just as likely to receive 21¢ as 78¢ or any other amount.
- Cashiers don’t round because they dislike pennies. So, if you’re supposed to get 99¢ change, the cashier doesn’t hand you a dollar and say, “Don’t worry about it.” Instead, you actually get 99¢ change.

But stating assumptions is a form of mathematical douchebaggery, isn’t it? (As an aside, check out the definition of douchey that’s returned when you do a search. Sexist, anyone?) I prefer problems with no assumptions stated; let folks make their own assumptions to devise a model. If you and I get different answers because of different assumptions, no worries. Maybe we both learn something in the process.

Anyway, where was I? Oh, yeah…

Understanding the solution to that problem is a precursor to the issue that’s causing me frustration. I’ll give the solution in a minute, so pause here if you want to solve it on your own, but let me now allow the proverbial cat out of its bag and tell you why I’m frustrated.

At our local grocery store, there’s a **coin counting machine** that will count your change, sort it, and spit out a receipt that you can take to the customer service desk to exchange for paper currency. Walk in with a jar full of change, walk out with a fistful of fifties. Pretty nifty, right? Except the machine charges a ridiculous 8.9% fee to perform this service. No, thank you.

My bank used to have a similar coin counting machine, and if you deposited the amount counted by the machine into your account, there was no fee. The problem is that everyone was doing this to avoid the grocery store fee, so the machine broke often. The bank finally decided the machine wasn’t worth the maintenance fees and got rid of it. Strike two.

Which brings me to my current dilemma. One Saturday morning every month, we now spend 30 minutes counting coins and allocating them to appropriate wrappers. Which is fine. The problem, however, is that we run out of quarter and penny wrappers way faster than we run out of nickel or dime wrappers. Which brings me to the real question for the day:

Since pennies, nickels, dimes, and quarters are not uniformly distributed as change,

why the hell does every package of coin wrappers contain the same number for each coin type?

The Royal Sovereign Assorted Coin Preformed Wrappers is the best-selling collection of coin wrappers on Amazon, and it provides 54 wrappers for each coin type. They also offer a 360‑pack with 90 wrappers for each coin type; Minitube offers a 100‑pack with 25 wrappers for each coin type; and Coin-Tainer offers a 36‑pack with 9 wrappers for each coin type. But what **no one** offers, so far as I can tell, is **a collection of coin wrappers with a distribution that more closely resembles the distribution of coins that are received as change**.

Whew! It feels good to finally raise this issue for public consideration.

So, the question that I really wanted to ask you…

Given the distribution of quarters, dimes, nickels, and pennies that are received in change, and given the number of coins needed to fill a coin wrapper — 40 quarters, 50 dimes, 40 nickels, and 50 pennies — how many of each wrapper should be sold in a bundled collection?

To answer this question, I determined the number of coins of each type required for every amount of change from 1¢ to 99¢. The totals yield the following graph:

The number of pennies is nearly five times the number of nickels. And there are nearly twice as many quarters as dimes.

But I realize that’s a **theoretical** result that may not match what happens in practice, since this assumes that the amounts of change from 1¢ to 99¢ are uniformly distributed (they aren’t) and that cashiers don’t round down to avoid dealing with pennies (they do). In fact, when I made a purchase of $2.59 yesterday, instead of getting one penny, one nickel, one dime, and one quarter as change, the cashier gave me one penny, **three** nickels, and one quarter, in what was clearly a blatant attempt to skew my data.

So for an **experimental** result, I counted the pennies, nickels, dimes, and quarters in our home change jar. The results were similar:

The ratio of pennies to nickels is closer to three, but the ratio of quarters to dimes is still roughly two.

Using a hybrid of the theoretical and experimental results, and accounting for the fact that only 40 quarters and nickels are needed to fill a wrapper whereas 50 pennies and dimes are needed, it seems that an appropriate ratio of coin wrappers would be:

**quarters : dimes : nickels : pennies :: 17 : 8 : 6 : 19**

Okay, admittedly, that’s a weird ratio. Maybe something like **3:2:1:4**, to keep it simple. Or even 2:1:1:2. All I know is that 1:1:1:1 is completely insane, and this nonsense has got to stop.

Hello, Royal Sovereign, Minitube, and Coin-Tainer? Are you listening? I’ve completed this analysis for you, free of charge. Now **do the right thing**, and adjust the ratio of coin wrappers in a package accordingly. Thank you.

Wow, that was a long rant. Sorry. If you’ve made it this far, you deserve some comic relief.

How many mathematicians does it take to change a light bulb?

Just one. She gives it to a physicist, thus reducing it to a previously solved problem.If you do not change direction, you may end up where you are heading. – Lao Tzu

The only thing that is constant is change. – Heraclitus

Turn and face the strange ch-ch-ch-changes. – David Bowie

A Buddhist monk walks into a Zen pizza parlor and says, “Make me one with everything.” The owner obliged, and when the pizza was delivered, the monk paid with a $20 bill. The owner put the money in his pocket and began to walk away. “Hey, where’s my change?” asked the monk. “Sorry,” said the owner, “change must come from within.”

As for the “probability of a nickel” problem that started this post, here’s my solution.

For change amounts from 1¢ to 25¢, there are ten values (5‑9 and 15‑19) for which you’ll receive a nickel as part of your change.

This pattern then repeats, such that for change amounts from **25 n + 1** to

**25**, where

*n*+ 25*n*is the number of quarters to be returned, you’ll receive a nickel when the amount of change is

**25**, where

*n*+*k**k*∈ {5, 6, 7, 8, 9, 15, 16, 17, 18, 19}. For 0 ≤

*n*< 4, there are 40 different amounts of change that will contain a nickel, so the probability of getting a nickel as part of your change is 40/100, or 40%.

### Probability, the Playoffs, and the Indianapolis Colts

With one week left in the NFL season, Dan Graziano had this to say about the Indianpolis Colts’ chances of making the playoffs:

Indianapolis can still win a third straight AFC South title. Really, it can. All it needs is to win and then have

the Texans, Bengals, Chargers, Jets, Saints, Chiefs, Patriots and Browns all lose. The league will throw in the partridge in a pear tree.

If the Colts win and the Houston Texans lose, both would be 8-8, and the first four tiebreakers for deciding which team makes the playoffs – record against one another, record against divisional opponents, record against common opponents, and record within the conference – would not be enough to decide who makes the cut.

It then comes down to strength of victory and strength of schedule. And for things to play out in the Colts’ favor, a lot of things have to go their way.

Fox Sports referred to this as long shot, comparing it to a recent win by a horse who was 200-to-1:

Sure, going 9 for 9 here looks dim, but long shots come in every once in a while.

Are the Colts’ odds as good as that horse’s? Seems not.

Using a simplistic model, assume that each of the nine necessary outcomes are equally likely. That alone would put the Colts’ odds at 511-to-1. (Since 2^{9} = 512.)

But it’s not that simple. The following chart from 538.com gives the probability of each team winning their game this weekend:

The good news is that the Colts have an 81% chance of winning their game against Tennessee. The bad news is that it seems unlikely that any of the Texans, Bengals, Chiefs, or Patriots will lose, let alone all four of them. So putting all those numbers together, the Colts’ chances of making the playoffs are:

0.81 × 0.20 × 0.22 × 0.84 × 0.54 × 0.69 × 0.15 × 0.18 × 0.77 = 0.0002 = 0.02%

or, more precisely, about **4,311-to-1**. That’s more than a long shot; that’s an extended-to-an-unfathomable-distance shot.

The Colts are in the unenviable position of Lloyd Christmas in *Dumb and Dumber*, “So, you’re telling me there’s a chance…”

### Dreidel is Not Fair

Is dreidel fair?

The rules of dreidel are straightforward. At the beginning of each round, players put one coin into the pot. (For young kids, the “coins” are actually chocolate pieces in the shape of a coin and wrapped in gold foil. This is known as *geld*, and as far as I’m concerned, chocolate is a currency to kids.) Players then take turns spinning the dreidel, and a reward is earned based on which of the four Hebrew letters appears on top when the dreidel stops spinning:

**Nun:**nothing.**Hey:**half the pot.**Gimel:**all of the pot.**Shin:**put one in.

Play continues clockwise, with each person spinning the dreidel until Gimel occurs and all coins are removed from the pot. At that point, everyone antes another coin, and a new round starts with the next player.

Officially, a player is out of the game when she or he has no coins left to contribute to the pot, and the game ends when one person has all the coins. But practically speaking, the game often ends much earlier, because players get bored and quit or, in the case of very young kids, the game lasts beyond bedtime and the children are pulled away by their parents.

No matter how the game ends, though, **it’s not fair**.

The following table is courtesy of Paul J. Nahin (*Will You Be Alive 10 Years from Now?*, Princeton University Press, 2014, p. 81). It shows the amount, over the long run, that each player will win during a dreidel game.

Player |
||||||

1 |
2 |
3 |
4 |
5 |
||

Numberof Players |
2 |
1.143 | 0.857 | |||

3 |
1.361 | 0.956 | 0.680 | |||

4 |
1.617 | 1.102 | 0.757 | 0.524 | ||

5 |
1.900 | 1.267 | 0.855 | 0.580 | 0.398 |

In other words, the first player has a significant advantage over the others. In a game of five players who start with 10 coins each, the **first player will finish the game with 19 coins**, on average, whereas the **fifth player will finish with just 4 coins**. That’s if the game ends early. If played until one person gets all the coins, then the first player is **five times** more likely to win than the fifth player.

This disparity in odds is likely the reason that an unofficial rule of dreidel is that the youngest player goes first, the second-youngest player goes second, and so on.

The word *dreidel* is Yiddish and means “to turn around.” Because the dreidel is, after all, a top.

This fact is not lost on comedian Lewis Black, who has some thoughts on the matter.

Happy Chanukah!