Spin spinner with n numbers twice (win if you hit same number twice), or flip n coins (win if all heads)?
This is a tough one, so make sure you justify your reasoning with mathematics
4 thoughts on “Would You Rather…”
In the roulette wheel bet, do you have to pick the number in advance of the first spin? I assume so. In that case, this is really “cute” in that assuming that n must be greater than 1, things vary for the first four values of n. Starting with that fourth value (n = 5), it will always be better to pick the coin flipping option. For n = 2, the chances are equal, as is the case for n = 4. For n = 3, the roulette wheel is the better choice. Thereafter, 1/n^2 will always be greater than 1/(2^n).
In the roulette wheel bet, do you have to pick the number in advance of the first spin? I assume so. In that case, this is really “cute” in that assuming that n must be greater than 1, things vary for the first four values of n. Starting with that fourth value (n = 5), it will always be better to pick the coin flipping option. For n = 2, the chances are equal, as is the case for n = 4. For n = 3, the roulette wheel is the better choice. Thereafter, 1/n^2 will always be greater than 1/(2^n).
In the roulette wheel bet, do you have to pick the number in advance of the first spin? I assume so. In that case, this is really “cute” in that assuming that n must be greater than 1, things vary for the first four values of n. Starting with that fourth value (n = 5), it will always be better to pick the coin flipping option. For n = 2, the chances are equal, as is the case for n = 4. For n = 3, the roulette wheel is the better choice. Thereafter, 1/n^2 will always be greater than 1/(2^n).
In the roulette wheel bet, do you have to pick the number in advance of the first spin? I assume so. In that case, this is really “cute” in that assuming that n must be greater than 1, things vary for the first four values of n. Starting with that fourth value (n = 5), it will always be better to pick the coin flipping option. For n = 2, the chances are equal, as is the case for n = 4. For n = 3, the roulette wheel is the better choice. Thereafter, 1/n^2 will always be greater than 1/(2^n).
Cool one. Depends on the n. Is (1/n)^2 > (1/2)^n? Here’s some python code to hack away at some small sample sizes: https://dl.dropboxusercontent.com/u/3646828/would_you_rather.py
Cool one. Depends on the n. Is (1/n)^2 > (1/2)^n? Here’s some python code to hack away at some small sample sizes: https://dl.dropboxusercontent.com/u/3646828/would_you_rather.py