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I was watching Veritasium's Would You Take This Bet? video. In a part of the video Derek asks people whether they would accept the bet in the case of each true guess for flipping the coin the person would win $10$ dollars and for each true guessing this $10 $ dollar will increase twofold as $10+20+40$... etc. But for each false guess the person betting will lose $10$ dollars. So in video he tells that probability of losing money for $100 $ times of guessing is $1/2300$. I tried to find the this probability by myself. I mean it is obvious the probability of losing money in these circumstances but I couldn't find the same conclusion as Derek.

So I tried to find the minimum number of trues guesses that would make the person at lose in final situation.

For $10$ of his guesses are true $10\times ((1-(2^7))/(1-2))=1270\rightarrow$ Gaining money $10\times 93=930 \rightarrow$ Losing Money.

For $9 $ of his guesses are true $10\times ((1-(2^6))/(1-2))=630 \rightarrow$ Gaining money $10\times 94=940 \rightarrow$ Losing Money.

In order to lose money, the person has to be false at least $ 6$ of his guesses.

pbinom(6,size = 100, prob = 0.5) = 1.00298e-21

This the result that I found. Where did I make a mistake?

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You have not linked to the video, presumably https://youtu.be/vBX-KulgJ1o?t=190 starting around 3:10 up to 4:40 but I do not think it quite says what you describe. The bet sizes do not change through the process.

Instead it is a fair coin with favourable bets (win $+20$, lose $-10$). So if you have $100$ bets, you will lose money with if your side comes up $33$ or fewer times out of $100$ but win overall with your side coming up $34$ or more times, since $20 \times 33 -10 \times 67 =-10 <0$ while $20 \times 34 -10 \times 66 =+20 > 0$.

That makes the probability of losing overall $\sum \limits_{k=1}^{33} {100 \choose k} 2^{-100}$ which you can find in R with pbinom(33,100,1/2) giving about $0.00043686$ or about $\frac{1}{2289}$.

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  • $\begingroup$ Wow, I would have understood from the expected value that was found. Thanks for your response. But is my result true in the case of money is multiplied by 2 for each true guessing? $\endgroup$ Mar 12, 2023 at 6:13
  • $\begingroup$ If you were to say that with your side coming up $6$ times you would get $10+20+40+80+160+320 -10\times 94 = -310<0$ and coming up $7$ times $10+20+40+80+160+320 +640-10\times 93 = +340>0$ then yes, pbinom(6,100,1/2) giving about $10^{-21}$ would be the equivalent calculation. $\endgroup$
    – Henry
    Mar 12, 2023 at 12:55

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