Coin Flips for Cash: Newsletter Challenge (August 2015)

You nailed it Del. You answered precisely what they asked for. People can expect anything, regardless of the statistics.

This is another poorly phrased challenge question. The progression in the size of the pot is poorly defined, too. Does it double or increment after successive flips. There is also nothing saying that once one wins that the game is over. A TTH flip sequence can be followed by another flip of heads or tails. A TTHH flip could pay out 3$ or 8$(with a spare coin toss not part of the game), or 4$, 5$, 9$ or 24$ if the fourth head toss counts in the game.

I don't see a problem with the phrasing. It says "you flip a coin until you get a head..." so it's clear to me that when you get a head the game is over. And you get 1$ every time you flip, including the flip of heads that stops the game.

But what if the bet into the game was 1 Italian Lire? The definition of the game is so obtuse that almost any outcome is possible.

Step right up. Step right up! Everybody is a winner.

Well, I can EXPECT any amount. That's why people gamble!

I don't give a flip. Just gimme all the cash you got!

What does "and so on" mean? Does the payoff double each time or increase by $1?

If it doubles, half the time you get $1, 1/4 the time you get $2, 1/8 the time you get $4, ... = 1/2 + 2*1/4 + 4*1/8 + ...

In that case, your expected return is $0.50 * (1+1+1+1...) = $infinity. I don't know if that makes sense, but it seems that is what the math says.

$1.66

I don't think there is a "correct" answer. It might average $1.66, but then that wouldn't be the maximum to expect. If it's possible to flip 5 tails in a row 2% of the times the game is played, then $6.00 might be the maximum to expect. It all depends on whether it's a reasonable expectation, or approaching infinity.

"Expectation" in probability is the average outcome expected over a large number of plays. In this game, there is no limit, but the probability of reaching an additional throw reduced by 50% each throw. You are guaranteed 1 throw, or $1. Your expectation on the second throw is $1 from the first throw, and $1 times the probability that you threw a tails on the first throw, i.e., $1 + ($1 x .5)= $1.5. The is the beginning of the "geometric series", 1 + 1/2 + 1/4 + 1/8..., otherwise expressed as S = 1 + 1/2+1/2^2+1/2^3+...+1/2^n, in this case where n=>Infinity. In one of what I have always considered to be one of the most clever proofs in mathematics, this can be reduced to s = 1/(1 - 1/2) = 2. So the expected outcome over an unlimited number of throws is $2 per game.

With a Matlab run of 10 million games, the likely payout comes out to be about 25 dollars. Even with this large number of games, the effect of an occasional lucky player can greatly affect the average, and some runs averaged 50-60 dollars. Shown above are the payouts, the cumulative average payout, and a histogram of the payouts.

Great analysis BUT....

10 million games is more than anyone could ever play in a life time!

I'd like to see the results for meaningful runs of games Let's say 1 or 2 hours of play at say 5 flips a minute.

Dunno how it will effect the result, but it should be easy for you to run.

Del

In #9, I assumed the payout was $1, $2, $4, $8, ... (The problem statement was ambiguous). If you assume the payout is $1, $2, $3, $4, ... the it is much more defined. The expected return is

P=$1/2 + $2/4 + $3/8 + $4/16 + $5/32 .... = $2.

Agreed the problem statement was ambiguous. I think it should say "the most likely dollars you make every time you play this game?" The maximum dollars you can make is clearly infinite, though with zero probability.

In your analysis I believe you're ignoring the $1 you get even if you get heads on first throw. There's a 50% chance of getting $2 by getting heads on 2nd throw. But if you get tails on 2nd throw you're still in the game, so the most likely dollars must be > 2.

I make it 1 + (1 + 1/2 + 1/2² + 1/2³ + 1/2⁴........) = 1 + 2 = $3 for most likely score.

Please ignore most of that - faulty logic . I was forgetting you're out of the game with just $1 if you get heads on 1st throw. So I agree with your $2 (and george hanson # 7)

I'm surprised the article you linked doesn't give the general case of sum to n terms S_n when q > 1.

S_n = 1 + q + q²......q^n-1 (multiplying the whole thing by a constant makes a trivial change)

Multiplying by q, q*S_n = q + q² + q³.....qⁿ

Subtracting, (1 - q)*S_n = 1 - qⁿ so S_n = (1 - qⁿ)/(1 - q)

When q < 1, q^∞ = 0 and S_∞ = 1/(1 - q)

While I agree with Del the Cat's entry, (previous GA) I have noticed that the puzzle don't refer to any loss situation, plus the statement, (every time you play the game), suggesting the number of games that can be played? So I must go along with Del the Cat's entry, based on the fact that there is no loss situation, therefore only one game at a time is played. Therefore the expectation is a 50/50 situation for every time you play the game. The 50/50 situation being an average of the number of spins over at the start of each game, the larger the number of game played, the closer the average is to $3.

Other answerers have commented on the wording in this question. I'll refer to the ambiguous use of "every". What is the maximum I can (reasonably) expect to make EVERY time I play? $1. Theoretically, about half the time I will make $1, will never make less, and will frequently make more than $1 but not quite half the time, being mortal. It is clearly unreasonable to expect the latter to happen on every play.

there's no limit the number of tails the coin can fall!

the coin can fall on head, tail and rim! the statistical value the coin falls on tail is nearly 50%!

The most important words in the challenge question are "every time" and "make".

Regarding "every time": One way to interpret this is that in x number of games, what is the smallest amount you won in any one of those x games. Therefore, your expectations should be based on the number of time you plan to play.

If you only play one game, you have 100% chance of winning at least $1, a 50% chance of winning $2, and a 25% chance of winning $3 or more. Therefore, if you only plan to play one game, you could probably expect to win $2 if you are optimistic.

If you only plan to play two games, you have a 100% chance of winning a minimum of $1 for both game and a 25% chance of winning $2 for both games. Therefore, if you play two games, you have a less than 50% chance of winning more than $1 "every time".

If you play more than two games, your chance of winning more than $1 "every time" is even less.

Another interpretation is related to the meaning of "make". The fair fee to play this game is $2. This the average return based on the probability. Therefore, if you pay $2 to play, you would expect to break even. If it costs less to play, you would expect to make the difference between $2 and the fee. If it costs more than $2 to play, you would expect to lose, the difference between the fee and $2.

Jim

Maybe I'm missing something but, if I flip the coin and receive five "tails" before receiving a "heads", wouldn't I win $6?

Yes, but that's not a probable outcome. Several posters agree that it should say "the most likely dollars you make every time you play this game?" and that the answer is $2. We should be getting the "official" answer soon!