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Assume a number between 1-100 (inclusive) is chosen randomly. You then attempt to guess the number. On each guess, if you didn't get the exact number, you're told whether the guess is higher or lower than the true number. If you get the answer on your first guess you receive ${$}5$, on your 2nd guess ${$}4$, 3rd guess ${$}3$, 4th ${$}2$, 5th ${$}1$, 6th ${$}0$, 7th $-{$}1$.

I'm wondering what the EV of this game is. I calculated it to be ${-$}0.1339$ but not sure whether that's correct. My method was $\frac{1}{100}*5 + \frac{99}{100}*\frac{2}{100}*4...$

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  • $\begingroup$ Please show how you arrived at that answer. $\endgroup$
    – Gonçalo
    Commented Jul 8 at 5:18
  • $\begingroup$ I did (1/100)*5+(99/100)*(2/100)*4+(99/100)*(98/100)*(4/100)*3 +... $\endgroup$
    – james
    Commented Jul 8 at 5:52
  • $\begingroup$ You should put that calculation in the original post, so that it's part of the permanent record. Comments are ephemeral and may be deleted without warning. It's unlikely in this case, but it's still best to put it in the question proper. $\endgroup$
    – Brian Tung
    Commented Jul 8 at 7:01
  • $\begingroup$ Henry (in the comments beneath my answer) pointed out that you have the first guess giving $5$, the second guess giving $4$, the third guess giving $3$, but the fourth guess gives only $1$ (and the remaining guesses drop by $1$ each). Is that drop of $2$ intentional, or was it an oversight? If it's intentional, I'll change my answer. $\endgroup$
    – Brian Tung
    Commented Jul 8 at 16:09
  • $\begingroup$ @BrianTung fixed the typo and edited the original post to include my original process. $\endgroup$
    – james
    Commented Jul 8 at 17:36

1 Answer 1

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You can get one number (either $50$ or $51$) on your first guess, then either of two numbers on your second guess, then any of four numbers on your third guess, and so forth, except that your seventh guess will nail any one of only $37$ numbers (since that's all that are left). So assuming that the secret number is selected uniformly at random from $1$ through $100$ inclusive, the expected number of guesses is

\begin{align} E(\text{guesses}) & = \frac{1}{100}\cdot 1 + \frac{2}{100}\cdot 2 + \frac{4}{100}\cdot 3 + \frac{8}{100}\cdot 4 + \frac{16}{100}\cdot 5 + \frac{32}{100}\cdot 6 + \frac{37}{100}\cdot 7 \\ & = \frac{580}{100} \\ & = 5.8 \end{align}

The value is $\$6$ minus the number of guesses in dollars, so it should be $\$0.20$. Maybe I misunderstood the set-up? Perhaps you could edit your post with how you obtained your result?

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  • $\begingroup$ but isn't the likelihood of getting it on guess 2 not 2/100 but (99/100)*(2/100) since we have to assume you actually make it to guess 2 first. I guess linearity of expectation applies here? But I must be missing something because (99/100)*(98/100)*(96/100)*(92/100)*(84/100)*(68/100)*(36/100) = 0.17 suggesting that there's a 17% chance of needing an 8th guess (even though that's not possible since 2^7 > 100). $\endgroup$
    – james
    Commented Jul 8 at 5:36
  • $\begingroup$ No, it would be $\frac{99}{100} \cdot \frac{2}{99} = \frac{2}{100}$; if you condition on missing it with the first guess, then you must reflect that there are only $99$ choices remaining. $\endgroup$
    – Brian Tung
    Commented Jul 8 at 6:46
  • $\begingroup$ @Henry: Ahh, good catch! I'll prompt the OP to see if that was in fact a typo or their intention. $\endgroup$
    – Brian Tung
    Commented Jul 8 at 16:08
  • $\begingroup$ @BrianTung yea I was making the mistake of not accounting for the previous numbers I'd already seen. silly mistake. thanks for the great answer. $\endgroup$
    – james
    Commented Jul 8 at 17:33
  • $\begingroup$ @Henry: that was a typo. Thanks for pointing it out. I've edited the original post $\endgroup$
    – james
    Commented Jul 8 at 17:34

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