Exploring the Intriguing World of Probability with a Simulation of Coin Flips

Written By Janice Doughtrey

Probability often presents us with fascinating scenarios that challenge our understanding of chance and randomness. One such intriguing concept is the outcome of coin flips, especially when different types of coins are involved. Imagine a situation where you have three distinct types of coins: one that always lands on heads (c1), one that always lands on tails (c2), and a fair coin that has an equal chance of landing on heads or tails (c3). What would happen if you were to flip these coins in a series of trials?

To discuss this topic expansively, let’s first understand the scenario and the mathematics behind it. We have three types of coins:

1. c1: Always lands on heads.
2. c2: Always lands on tails.
3. c3: A fair coin (50% chance of heads or tails).

We are looking at the probability of getting a head on the third flip (H3), given that the first two flips (H1 and H2) are heads. This is a conditional probability problem, and we’re applying Bayes’ Theorem here.

Understanding the Theorem:

Bayes’ Theorem allows us to update our prior beliefs with new evidence. In this case, our prior belief is the probability of each coin being chosen, and the new evidence is the first two flips resulting in heads.

Calculation

1. Probability of H1H2H3: This is the probability of getting heads three times in a row.
• For c1, this is certain (100% or 1), since it always lands on heads.
• For c2, it’s impossible (0%), as it always lands on tails.
• For c3, the probability is 1/8 (since each flip has a 1/2 chance of heads, and they are independent: 1/2 x 1/2 x 1/2 = 1/8).

Therefore, the sum of probabilities for H1H2H3 over all coins, C, is:

P(H1H2H3 | C) = 1 x 1/3 + 0 x 1/3 + 1/8 x 1/3 = 9/24.

Probability of H1H2: The probability of getting heads in the first two flips.

• For c1, it’s certain (1).
• For c2, it’s impossible (0).
• For c3, the probability is 1/4 (since each flip has a 1/2 chance of heads: 1/2 x 1/2 = 1/4).

Therefore, the sum of probabilities for H1H2 over all coins, C, is:

P(H1H2 | C) = 1 x 1/3 + 0 x 1/3 + 1/4 x 1/3 = 5/12.

Conditional Probability P(H3|H1H2): This is the probability of getting heads on the third flip given that the first two flips are heads.

By applying Bayes’ Theorem, P(H3 | H1H2) is calculated as:

P(H3 | H1H2) = P(H1H2H3) / P(H1H2) = 9/24 / 5/12 = 9/10.

Interpretation:

This result indicates that if you know the first two flips were heads, there’s a 90% chance that the next flip will also be heads. This high probability is because one of the coins always lands on heads, skewing the overall probability despite the fair coin being in the mix.

50 Versions of 10 Coin Flips:

If we were to simulate this scenario 50 times with 10 coin flips each, we would likely see a significant number of sequences with long runs of heads, especially if c1 or c3 is frequently chosen. The sequences would be less predictable with c3, but with c1, long runs of heads would be guaranteed. With c2, we would see runs of tails.

To explore this question, let’s consider a hypothetical simulation of 50 sets of 10 coin flips. This exercise not only offers insights into basic probabilities but also reveals how varying conditions can significantly alter outcomes. Such a simulation, while simple in its mechanism, can illuminate the surprising nature of random events and the probabilities that govern them.

Now, let’s delve into what we might expect from this simulation, considering the unique characteristics of each coin type.

• Sequence 1: HHHTHTHTTH
• Sequence 2: TTHHTTTHHH
• Sequence 3: HTHHTTTHTH
• Sequence 4: HTTHHHTHTH
• Sequence 5: TTHHTTHHTT
• Sequence 6: HTHTHHTTHH
• Sequence 7: TTHHHTTTHT
• Sequence 8: HTHHTHTHHT
• Sequence 9: TTHTHHTHHT
• Sequence 10: HTHTHTHTHT
• Sequence 11: TTHTHTHTHH
• Sequence 12: HTHTHHHTT
• Sequence 13: TTHHHTTHH
• Sequence 14: HTHHHTHH
• Sequence 15: TTHHHTHH
• Sequence 16: HTHHHTHT
• Sequence 17: TTHHHTHT
• Sequence 18: HTHHHTTH
• Sequence 19: TTHHHTTH
• Sequence 20: HTHHHHTT
• Sequence 21: TTHHHHTT
• Sequence 22: HTHHTHTH
• Sequence 23: TTHHTHTH
• Sequence 24: HTHTHTTH
• Sequence 25: TTHTHTTH
• Sequence 26: HTHTHTTTH
• Sequence 27: TTHTHTTTH
• Sequence 28: HTHTHTTHT
• Sequence 29: TTHTHTTHT
• Sequence 30: HTHTTTTTH
• Sequence 31: TTHTTTTTH
• Sequence 32: HTHTTTTHT
• Sequence 33: TTHTTTTHT
• Sequence 34: HTHTHHHTH
• Sequence 35: TTHTHHHTH
• Sequence 36: HTHTHHHHT
• Sequence 37: TTHTHHHHT
• Sequence 38: HTHTHHHTT
• Sequence 39: TTHTHHHTT
• Sequence 40: HTHTHTHTH
• Sequence 41: TTHTHTHTH
• Sequence 42: HTHTHTHTH
• Sequence 43: TTHTHTHTH
• Sequence 44: HTHTHTHTH
• Sequence 45: TTHTHTHTH
• Sequence 46: HTHTHTHTH
• Sequence 47: TTHTHTHTH
• Sequence 48: HTHTHTHTH
• Sequence 49: TTHTHTHTH
• Sequence 50: HTHTHTHTH

Coin Flip Simulation

As you can see, there is a wide variety of sequences possible, even with just 10 coin flips. Some sequences, such as HHHHHHHHHH and TTTTTTTTT, are very unlikely, while others, such as HTHTHTHTH, are more likely. The probability of any given sequence depends on the probability of each coin landing on heads or tails.

This exercise in probability illustrates how conditional probabilities can significantly differ from simple independent event probabilities. It also shows how the presence of one skewed variable (like a biased coin) can greatly influence overall outcomes. This principle can be applied in various fields, from gambling strategies to predictive models in finance and science.

In real-world scenarios, understanding the underlying conditions and variables is crucial for making accurate predictions and decisions. Just like in this coin flip problem, real-life events often depend on several factors, and discerning these factors is key to understanding the probabilities at play.