What Is the Probability of Getting Heads on All 12 Tosses?

If you toss a fair coin 12 times, the probability that every toss lands heads is the probability of one specific sequence: HHHHHHHHHHHH. Because each toss has a one-half chance of heads and the tosses are independent, multiply those probabilities together.

The result is $1/4096$, approximately $0.0244\%$. This assumes the coin is fair and that the outcome of one toss does not affect another.

Why the probabilities multiply

For a fair coin, each individual toss has two equally likely outcomes: heads or tails. So the chance of heads on one toss is $1/2$.

The event “heads on all 12 tosses” requires heads on the first toss and the second toss, and so on through the twelfth. For independent events, the probability that all occur is the product of their individual probabilities:

In decimal form, this is $0.000244140625$. As a percentage, multiply by 100:

So the chance is about 0.0244%, or roughly one time in 4096 sets of 12 tosses.

A counting check

You can verify the result by counting possible sequences. Each toss has two possible outcomes, and a sequence of 12 tosses therefore has

possible outcome sequences, assuming a fair coin and independent tosses. They are equally likely. Only one sequence consists entirely of heads, so the probability is one favorable sequence out of 4096:

For example, HHTHTTTHTHHH is one possible sequence, but it does not meet the requirement because it contains tails. Only HHHHHHHHHHHH counts as “heads every time.”

If the coin is biased

The answer changes if the probability of heads on a single toss is not $1/2$. If that probability is $p$, and tosses are independent, then

For instance, if a coin has a 60% chance of heads on each toss, then the chance of 12 consecutive heads is $0.6^{12} = 0.002176782336$, or about $0.218\%$. A biased coin does not change the multiplication method; it changes the probability assigned to each toss.

Independence matters too. If outcomes are influenced by earlier tosses, you cannot automatically raise a single-toss probability to the twelfth power. You would need the relevant conditional probabilities—the chance of heads at each step given what happened before—and multiply those instead.

Common mistakes

• Using $12/2$ or $1/24$. The event requires heads on every toss, not heads on one selected toss. The probabilities for all 12 required outcomes multiply.
• Counting many favorable sequences. There is only one sequence that has heads on all 12 tosses. If the question instead asked for exactly 6 heads, many sequences would qualify, and a binomial calculation would be needed.
• Confusing “all heads” with “at least one head.” These are different events. “All heads” means no tails occur; “at least one head” includes nearly every sequence except the one consisting entirely of tails.
• Reporting the decimal as a percentage without converting. The probability $0.00024414$ corresponds to about $0.0244\%$, not $0.00024414\%$.

If “12 coins” means 12 separate fair coins tossed once each, rather than one coin tossed 12 times, the calculation is identical under the same fairness and independence assumptions: all 12 must show heads, so the probability is again $1/4096$.