A game of luck

Probability Level 3

Mark was flipping a fair 4-sided dice (1,2,3,4), the rules to his game were simple:

If the number he flipped is the same number as before, he would continue on and give himself a point to his score (not on first turn)
If the number was different from the previous, he would stop and record his score
The first turn cancels out the second rule and the score is set to one after the first roll/flip

What score with a coin (2 sides) under the same conditions, on his initial attempt (and may or may not losing on the next attempt), would match the probability of getting a score of 14 on his initial attempt (and may or may not losing on the next attempt) with the 4 sided dice?

The answer is 27.

1 solution

Jordan Cahn
Mar 15, 2019

You should specify the probability of getting a score of at least 14, since the probability of getting exactly 14 is different.

That being said, the probability of getting at least 14 is simply $\dfrac{1}{4^{13}}$ , since you need to repeat your first flip 13 times. The probability of getting a score of at least $n$ with the coin is $\dfrac{1}{2^{n-1}}$ , for the same reasons. Thus, we need to solve $\begin{aligned} \frac{1}{4^{13}} &= \frac{1}{2^{n-1}} \\ \frac{1}{2^{26}} &= \frac{1}{2^{n-1}} \\ n-1 &= 26 \\ n &= \boxed{27} \end{aligned}$

If we stipulated a score of exactly 14, the probability would become $\dfrac{1}{4^{13}}\times\dfrac{3}{4} = \dfrac{3}{4^{14}}$ since we must flip something different on the 15th flip. Similarly, the probability of getting a score of exactly $n$ with the coin is $\dfrac{1}{2^{n-1}}\times \dfrac{1}{2} = \dfrac{1}{2^n}$ . Thus, we must solve $\begin{aligned} \frac{3}{4^{14}} &= \frac{1}{2^n} \\ \frac{3}{2^{28}} &= \frac{1}{2^n} \\ 2^n &= \frac{2^{28}}{3} \\ n &= 28 - \log_2(3) \\ n &\approx 26.4 \end{aligned}$ Thus, there is no number of flips that will give precisely the same probability, but 26 is closest.

https://www.wolframalpha.com/input/?i=4%2F((1%2F4)%5E13)+%3D+2%2F((1%2F2)%5Ex)
Would this work? It gives the correct answer but I'm not sure if theres a fallacy in the way I'm setting it up here

Edit: Oh I see yours also accounts for the possibility of continuing to flip coins even after youve hit your target score

Kyle T - 2 years, 2 months ago

I'm not sure how you arrived at that equation, so I can't really tell if it's a valid way to set it up. Simplifying each side yields $4^{14} = 2^{x+1}$ , which does indeed give $x=27$ , but I can't really say if your approach will work generally without knowing what your approach was.

Jordan Cahn - 2 years, 2 months ago

I was trying to think of it like a discrete math problem where we count the total number of strings that work divided by the total number of possible strings, so the 4 in the numerator is because we could start with any number 1-4 and the (1/4)^13 in the denominator was because we'd need to roll the same number 13 times, similarly on the other side of the equation we only have heads or tails hence the 2 in the numerator and the (1/2)^x, solving for x should tell you how many flips it would take to reach that same probability.
I like your solution better

Kyle T - 2 years, 2 months ago

Very nice solution, I fixed the question just by appending "(and may or may not losing on the next attempt)"

Joshua Crawford - 2 years, 2 months ago

A game of luck

The answer is 27.

1 solution

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