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Suppose Mark and Paul are sitting on a table, and Mark starts tossing an unbiased and fair coin. He tosses it for 99 times, and he gets 99 consecutive tails. At this point Mark asks Paul: "let's bet $100 on the next toss, do you want to pick tail or head?"

Question: In order to maximize his expected return, should Paul pick tail, head, or it doesn't matter?

Daniel Scocco
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4 Answers4

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If, as you said, it's an unbiased and fair coin, then it has an equal chance of coming up heads as tails on the 100th flip, by definition, because that's what an unbiased fair coin is: it is a coin that has an equal chance to come up heads or tails on any flip.

On the other hand, if you see a coin come up tails 99 times in a row you had best revisit your assumption that it is an unbiased fair coin. It would be foolish to bet on it coming up heads after that.

Addendum: Here's the reference you asked for: Gambler's fallacy

MJD
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  • Right, but over 100 tosses the expected result would be 50 tails and 50 heads, so if you already have 99 tails, the next toss is more likely to be head, ain't this the case? – Daniel Scocco May 01 '12 at 13:42
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    @DanielS No, it is not. The coin has no memory. It does not look back on its life and decide to come up heads because it has come up tails many times in a row. It cannot think. The expected result over 100 flips is 50 heads and 50 tails only if you have no other information about it. If you know that the first 99 flips were tails, then the expected result after 100 flips is 99$\frac12$ tails and $\frac12$ head. – MJD May 01 '12 at 13:43
  • No, Daniel, that isn't the case. Given that the first 99 tosses came up heads, the expected number of heads after the next toss is $99 \frac 1 2$. While probabilities should even out in the long run, there is nothing that "forces" it in the direction of evening out. – Thomas Andrews May 01 '12 at 13:45
  • @Mark Dominun, makes sense. But let me update the question with a similar problem. Please give me your take on it. – Daniel Scocco May 01 '12 at 13:45
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    @DanielS Don't update the question. Someone has answered your question. Ask a new question. (It makes it hard to keep questions and answers in sync if you keep changing your question.) – Thomas Andrews May 01 '12 at 13:46
  • @Thomas, ok I'll ask on another thread. – Daniel Scocco May 01 '12 at 13:51
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    @DanielS Before you ask about the Monty Hall problem you might try a search for "Monty Hall"; the problem has been discussed here many times. – MJD May 01 '12 at 13:56
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    @DanielS: in response to your first question, the concept you need to understand is conditional probability (http://en.wikipedia.org/wiki/Conditional_probability). You're not asking for the probability that the coin comes up tails $100$ times; you're asking for the probability that the coin comes up tails $100$ times conditional on it already having come up $99$ times, which is $\frac{1}{2}$. – Qiaochu Yuan May 02 '12 at 15:15
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    @DanielS: perhaps the following will be useful. There's no difference between flipping the coin $100$ times in a row and flipping $100$ identical copies of the coin at once, say in $100$ different cities around the planet. If you look at $99$ of the coins and they're all tails, then maybe you'd intuitively expect the last one to be heads. But you could've chosen to visit the cities in any order. If you'd visited the last one first, what would you have expected it to be, and how can the order in which you visit the cities possibly matter (if you actually believe that the coin is unbiased)? – Qiaochu Yuan May 02 '12 at 15:23
  • Well, you want to bet tails because it will almost certainly repeat the same pattern for the 100th flip. But, likewise, you want to bet heads because it has to happen eventually anyway, if its an unbiased and fair coin. I recommend you flip a coin to decide which way to bet. – CogitoErgoCogitoSum Feb 10 '13 at 21:19
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The restriction is that the coin is fair. There does not seem to be a restriction that the flipping is fair.

With practice I used to be able to get the same result from coin flips - mostly heads or mostly tails. One needs to start with the coin in the smae head up or tail up position, but it is not difficult to make the flipping unfair.

user1475
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To elaborate on Mark's answer. Let's say you don't necessarily believe that the coin is unbiased, but instead believe it has probability $p$ of coming up heads.

You degree of belief in the value of $p$ can be specified by providing the parameters $a$ and $b$ of a beta distribution. Intuitively, they correspond to the number of times you've seen heads and tails come up already. A typical 'default belief' is given by $a=b=1$, which essentially expresses maximum ignorance - you believe that the true value of $p$ is distributed uniformly between 0 and 1.

The benefit of doing this is that there is a simple method of updating your belief about the distribution of $p$ whenever you see a new coin toss - you simply increment $a$ by 1 whenever you see a head, and increment $b$ by one whenever you see a tail.

The mean of the distribution is simply $a/(a+b)$ and its variance is $ab/[(a+b)^2(a+b+1)]$. If you see 99 tails in a row, then your new values of $a$ and $b$ are $a=1$ and $b=100$, giving an expected value for $p$ of

$$E(p) \approx 0.01$$

$$\mathrm{StDev}(p) \approx 0.01 $$

So with a very high degree of certainty, you believe that the true value of $p$ is 0.01, and certainly isn't much more than 0.04. You would therefore be very naive to bet on heads coming up on the next toss (unless you were given very favorable odds!)

Chris Taylor
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Due to the law of large numbers, the proportion of heads to tails seen with an unbiased coin will converge to $\frac12$ as the number of flips tends towards infinity. If you were to make the bet before 99 coins have been tossed, then it would be apparent that the probability of seeing all tails is $(\frac12)^{99}$. If you have already achieved this then the probability of seeing a subsequent tail is $\frac12^{100}-\frac12^{99}$, it is not about the coin having a 'memory' but rather the probability of observing 99 events involving the coin with identical values. It would be the same if you decided to flip the coin 100,000 times and chose to observe it 0.1% of the time. Even if the average proportion of tails to heads of the 100,000 were 0.5, the probability of observing 99 consecutive tails would still be $(\frac12)^{100}-(\frac12)^{99}$. The same would also be true if you selected a new coin every time.

Nigel Sotelo
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