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# Coin Toss

Although even coin tossing could be the subject of bets (we'll show you how), it is more commonly used as an element of chance for various decision-making situations. For example, determining who will kick off in football or who will endure a round of drinks at a social event. More complex bets involve more than just the outcome of a single toss, bringing probability into play.

Coins have two sides— heads and tails. Thus, the outcome of a coin toss boils down to two possibilities. From the players' perspective, the results are contradictory, as one player's win is the other's loss. Both players, however, share an equal chance: colloquially, their odds are `50 to 50`.

The probability of landing on one side of the coin is `½ = 0.5`. In percentage terms, it's `50 %`, and in betting odds, it's `1 to 1`.

If two players were betting the same amount, say \$10, in the long run, it would be a so-called fair bet. Each player could win \$10 with a 0.5 probability or lose \$10 with the same chance. The expected value of the bet would be zero: `0.5 × \$10 + 0.5 × (-\$10) = 0`.

## Probability in Coin Tosses from Various Perspectives

Now let's explore the probability of the same side of the coin appearing multiple times or, for example, getting heads exactly 5 times, at least 8 times, etc.

In this context, coin tossing could be the subject of a proposal for a specific bet with a certain payout ratio, a proposal made by one player to another. The player accepting the bet (actually, the one proposing it too) should be able to calculate whether it's advantageous for them or not.

### Probability of the Same Side Appearing Multiple Times in a Row

Table 1: Probability of the same side appearing multiple times in a row
Same Side of the Coin Mathematical Probability In % 1 to …
0.5000000000 50.00% 2
0.2500000000 25.00% 4
0.1250000000 12.50% 8
0.0625000000 6.25% 16
0.0312500000 3.13% 32
0.0156250000 1.56% 64
0.0078125000 0.78% 128
0.0039062500 0.39% 256
0.0019531250 0.20% 512
10× 0.0009765625 0.10% 1,024

Probability, that we will toss one side of a coin, is `0.5`. The probability of tossing the same side of the coin twice in a row is `0.5 × 0.5`, three times in a row `0.5 × 0.5 × 0.5`, and so on, generally 0.5 raised to the number of successful coin tosses. For example, tossing the same side of the coin 10 times in a row would likely happen only once in about a thousand attempts (precisely in 1024 attempts): the probability of such an event is `0.510 = 0.0009765625`, i.e., in the form of odds 1 ÷ 0.0009765625 = 1024.

### Probability of the same side of the coin falling several times in a certain number of attempts

We might also be interested in the probability that a certain side of the coin will fall exactly that many times or at least that many times in a certain number of attempts. Generally k times out of n attempts (k ≤ n).

For example, in the following two tables, calculations are provided for 10 coin tosses (n = 10). Both tables express the same thing, but the one below expresses the probability in the format of 1 to … Then, we can calculate the probability that a certain side of the coin doesn't fall at all (k = 0), falls exactly 1 time (k = 1), up to exactly 10 times (k = 10).

If you're interested in the procedure or want to calculate it yourself, you can use the following formula called Bernoulli's scheme. More details and explanatory examples can be found under the link.

Let's just briefly add that is the so-called combinatorial number, `p` is the probability of success, `(1 – p)` is the probability of failure.

Let's also look at the difference between exactly and at least. For example, what is the probability that out of 10 coin tosses, heads will come up exactly 5 times or at least 5 times. Exactly means that heads must come up exactly 5 times (no fewer, no more), while at least means that it comes up 5 times or more – we simply calculate by adding the probabilities that heads will come up exactly 5 times + exactly 6 times + … + exactly 10 times.

Table 2: Probability of the same side of the coin falling several times out of 10 tosses
Number of Successes (k) Probability (exactly k successes) Probability (at least k successes)
0 0.0009765625
1 0.0097656250 0.9990234375
2 0.0439453125 0.9892578125
3 0.1171875000 0.9453125000
4 0.2050781250 0.8281250000
5 0.2460937500 0.6230468750
6 0.2050781250 0.3769531250
7 0.1171875000 0.1718750000
8 0.0439453125 0.0546875000
9 0.0097656250 0.0107421875
10 0.0009765625 0.0009765625

In the following table, probabilities are converted into the format of 1 to … (betting odds), which is apparently clearer than a purely mathematical decimal notation (probability from definition takes values from 0 to 1, where 0 is an impossible event, and 1 is a certain event).

We can observe how the probability of tossing a coin is symmetric around the most probable value. If we toss a coin 10 times, then most probably, one side of the coin will come up 5 times: `0.2461` or `1 to 4.06`. The probability that this side of the coin will come up 4 times or 6 times is the same: `0.2051` or `1 to 4.88`. Similarly, the probability that the same side of the coin will come up 3 times or 7 times, etc. This is because the probability of success and failure is `0.5:0.5`.

Note about betting odds: since the odds are the inverse of the probability, it follows that the higher the odds, the lower the probability, and vice versa.

Table 3: Probability of the same side of the coin falling several times out of 10 tosses (in the format of 1 to …)
Number of Successes (k) Probability (exactly k successes) Probability (at least k successes)
0 1 to 1,024
1 1 to 102 1 to 1.000978
2 1 to 23 1 to 1.0109
3 1 to 8.53 1 to 1.058
4 1 to 4.88 1 to 1.21
5 1 to 4.06 1 to 1.61
6 1 to 4.88 1 to 2.65
7 1 to 8.53 1 to 5.82
8 1 to 23 1 to 18
9 1 to 102 1 to 93
10 1 to 1,024 1 to 1,024

## Suggestion for a Bet – Example with Coin Tossing

Imagine someone approached you with the following offer for a bet. "I will toss a coin. If I get the same side of the coin that I choose at the beginning (say, heads), at least 8 times out of 10 throws, you pay me \$20. If I fail, I will pay you \$1 for each unsuccessful attempt, i.e., if heads come up only 7 times or fewer out of 10 attempts." Would you accept such a bet? Alternatively, a more precise question might be whether you would be willing to accept such a bet in the long run?

The stakes could be higher but in the same ratio, for example, \$2,000 to \$100. Perhaps you would think that getting the same side of the coin 8 times (or more) out of 10 throws is not something your opponent will easily achieve, and that you will watch them struggle and receive \$100 for each unsuccessful attempt (which makes up 10 throws). However, you would be mistaken.

One tiny detail is missing, and that is the probability of winning and losing. In general, we should be interested in what we can win and with what probability, and what we can lose and with what probability. This approach is called expected value (often denoted as EV – Expected Value). It is a certain average result – a weighted mean, where the weights are probabilities.

Looking at Table 2, we find that the probability of getting the same side of the coin at least 8 times is `0.0546875` (or approximately `5.47%`, if you will). The probability of failing is `1 – 0.0546875 = 0.9453125` (or approximately `94.53%`). Then, the one who accepts the bet may lose \$20 with a probability of 0.0546875 and win \$1 with a probability of 0.9453125:

`EV = 0.0546875 × (-\$20) + 0.9453125 × \$1 = -\$0.1484375`.

This means that a player who accepts the bet would lose `almost 15 cents` for every invested dollar in the long run, more precisely 14.84 cents. Thus, the disadvantage of this player would be approximately `14.84%`, and this number would simultaneously be the advantage of the one who proposed the bet. Long-term acceptance of such a bet would be significantly disadvantageous. You can compare it with other casino or gambling games – a ranking by (dis)advantage for the player.

The entire example, or evaluation of the bet, could be solved in another way, possibly more simply and quickly, because we already have all the calculations available. The probability of getting the same side of the coin at least 8 times out of 10 throws is `0.0546875`. Converted into odds, it's about `1 ÷ 0.0546875 = 1 to approximately 18` (see Table 3, the number has more decimal places not displayed).

We can interpret this as getting heads or tails at least 8 times out of 10 throws will probably happen `once in 18 attempts`. This is also a fair ratio where neither of the players would achieve profit (or loss). It's clear that if you are willing to pay \$20 instead of \$18, you put yourself at a disadvantage. However, it works the other way around as well. If you convince the other player that you will pay them less than \$18 for each successful attempt, the long-term advantage would shift to your side.

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Based on the original Czech article: Házení mincí.