Timeline for Expected value of highest card before first ace
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 18, 2023 at 23:52 | history | edited | RobPratt |
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| Sep 18, 2023 at 19:43 | vote | accept | legionwhale | ||
| Sep 18, 2023 at 19:00 | answer | added | Mike Earnest | timeline score: 2 | |
| Sep 18, 2023 at 16:36 | comment | added | David K | @SomeCallMeTim Yes, you can set the "highest card" to $0$ explicitly in the problem statement, as has been done. It was "implicitly" setting it to $0$ that I found questionable. (You could explicitly set the "highest cared" to any value you like in that case, but $0$ works out especially neatly.) | |
| Sep 18, 2023 at 16:05 | history | edited | legionwhale | CC BY-SA 4.0 |
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| Sep 18, 2023 at 16:04 | comment | added | legionwhale | @DavidK If there is no highest card, then it is natural to let $H$ take the value of $0$. I discussed this in my earlier comment, and noted that my solution implicitly assumed this. Thus, as Tim and 2661923 point out, it is not impossible for the expected value in Lourrran's analogy to be less than $2$. But I will add this to the post, as it (understandably) seems to have caused confusion. I would also like to remark that no one has yet answered the two specific questions I asked -- though, of course, it would be relevant to know if my solution is incorrect. | |
| Sep 18, 2023 at 15:41 | comment | added | user2661923 | @DavidK True. However, suppose that you arbitrarily construe the case of the first card being an Ace to represent that the maximum high card is $~0.~$ Further suppose that the sample space of equally likely choices has value $~A~$ for the number of ways that the first card can be an Ace and value $~B~$ for the number of ways that the first card can be other than an Ace. Then, by permitting the first card to be an Ace, you are simply scaling the final result by the factor $~\dfrac{B}{A+B}.$ | |
| Sep 18, 2023 at 15:29 | comment | added | David K | @SomeCallMeTim It's unclear to me what you mean by "weight $0$ in the sum implicitly". If no cards are drawn before the first ace, the value of the highest card is undefined. Yet the probability of that case is positive. This makes the entire sum undefined. | |
| Sep 18, 2023 at 14:48 | comment | added | SomeCallMeTim | The solution does allow for the first card to be an ace. Indeed, you calculate $\Bbb{P}(S_n)$ correctly, but you seem to have forgotten that $\sum_{j\leq 1}\Bbb{P}(S_j)$ is $\frac{1}{13}$, and should be interpreted as the probability that the first card is an ace. As these have weight $0$ in the sum implicitly, you adjust the average correctly. This also explains your answer @Lourrran, you get the correct result when allowing the ace to be the first card. | |
| Sep 18, 2023 at 13:08 | comment | added | Lourrran | I have some doubts with your result. If we consider a smaller game, only 12 cards, only aces and 2 and 3, I adapt your formula, and I obtain 1.8333 which is obviously wrong. | |
| Sep 18, 2023 at 12:56 | comment | added | legionwhale | @Henry I thought the impact of the event would further be deprecated by the weighting of the expected value, but in hindsight it does have an effect of order $1 \times 10^{-1}$. If we say that the highest card before $A$ has value $0$ if it doesn't exist factor it in, we get $\mathbb{P}(H \le 2) = \mathbb{P}(H = 2) - \mathbb{P}(H = 0)$ so we would need to subtract $\frac{2}{13}$ from the final expected value. Bizzarely enough, it seems the original solution already accounted for this unintentionally. | |
| Sep 18, 2023 at 12:50 | history | edited | legionwhale | CC BY-SA 4.0 |
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| Sep 18, 2023 at 9:28 | comment | added | Henry | The probability that an ace is the first card is $\frac4{52}$, which is not really negligible. But you can ask for the conditional expected value of the highest card before first ace given that the first card is not an ace. | |
| Sep 18, 2023 at 9:26 | comment | added | Henry | "We can imagine any permutation of the deck": $52!$ ways. "We can imagine any permutation of the deck (ignoring permutation of the aces)": $52!/4!$ ways, though the earlier number may be easier to handle | |
| Sep 18, 2023 at 3:57 | history | asked | legionwhale | CC BY-SA 4.0 |