I have ten cards, Ace,2,3,4,5,6,7,8,9,10. The value of the Ace is 1.They’re shuffled, then dealt one by one, face up. For the first card, you automatically win $10. For the next 9 cards, if the card face-up is greater than every previous card, you win $10 more. Further, the game does not end when you draw a card that is not a winning card and continues until all of the cards are gone. What is the expected winning amount?
Expected winning amount!!
- Thread starter Ctiger
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As always, use indicator random variables. For (k=2:10) let (X_k=1) if the (k)-th card is greater than all the previous ones, (X_k=0) otherwise. The expected winnings are then (10+10\sum_{k=2}^{10}P[X_k=1]).
For the (k)-th card to be greater than all the previous ones, it must have a value (v\geq k) and the cards before it must come from the set (\{1,2,...,v-1\}). For a fixed (v) the probability of this happening is (\frac{(v-1)(v-2)\cdots (v-k+1)}{10\cdot 9\cdots (10-k)}\cdot \frac{1}{9-k}=\frac{_{v-1}P_{k-1}}{_{10}P_k}=\frac{(k-1)!}{_{10}P_k}\binom{v-1}{k-1}). So the probability of the (k)-th card being greater than all the previous ones is (\frac{(k-1)!}{_{10}P_k}\sum_{v=k}^{10}\binom{v-1}{k-1}=\frac{(k-1)!}{_{10}P_k}\binom{10}{k}=\frac{(k-1)!}{k!}=\frac{1}{k}). Plugging this into the expression above, we get that the expected winnings are
(10+10\sum_{k=2}^{10}\frac{1}{k}=10\sum_{k=1}^{10}\frac{1}{k})
For the (k)-th card to be greater than all the previous ones, it must have a value (v\geq k) and the cards before it must come from the set (\{1,2,...,v-1\}). For a fixed (v) the probability of this happening is (\frac{(v-1)(v-2)\cdots (v-k+1)}{10\cdot 9\cdots (10-k)}\cdot \frac{1}{9-k}=\frac{_{v-1}P_{k-1}}{_{10}P_k}=\frac{(k-1)!}{_{10}P_k}\binom{v-1}{k-1}). So the probability of the (k)-th card being greater than all the previous ones is (\frac{(k-1)!}{_{10}P_k}\sum_{v=k}^{10}\binom{v-1}{k-1}=\frac{(k-1)!}{_{10}P_k}\binom{10}{k}=\frac{(k-1)!}{k!}=\frac{1}{k}). Plugging this into the expression above, we get that the expected winnings are
(10+10\sum_{k=2}^{10}\frac{1}{k}=10\sum_{k=1}^{10}\frac{1}{k})
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For the n-th draw the expectation to win is 1/n. Since the number of all possible outcomes is C_10_n * n! and the number of outcomes satisfying the condition of winning is C_10_n * (n-1)! { the greatest value must be the last in the set of values } then after dividing we get 1/n. The expected winning amount will be 10*(1+1/2+1/3+......+1/10) = approximately $29.2
How would the solution of this question change if the game ended when a winning card was not drawn? i.e. what is the expected winnings of pulling series of increasing cards of any length from 1 to 10?
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I suppose that the same approach can be used here as well.
For the n-th draw there are C_10_n * n! possible set of values. Since in order to reach the n-th draw we must have an increasing sequence, only one set per n! satisfies our condition. Thus we have only C_10_n * 1 satisfying outcomes and therefore the probability is 1/(n!). The expected winnings is 10(1+1/2!+1/3!+.... +1/10!)= approx 10*(e-1)=$17.2
For the n-th draw there are C_10_n * n! possible set of values. Since in order to reach the n-th draw we must have an increasing sequence, only one set per n! satisfies our condition. Thus we have only C_10_n * 1 satisfying outcomes and therefore the probability is 1/(n!). The expected winnings is 10(1+1/2!+1/3!+.... +1/10!)= approx 10*(e-1)=$17.2
i just find it weird that if we consider the case of a very long game, and randomly picking integers (instead of from a A234...10 deck), then the first game has an unbounded expected value since the harmonic series diverges, and the second game converges to that weird e-thing. i guess it's just because we can always keep picking numbers that are bigger than everything else before....but we are guaranteed to mess up if we need everything in order?! that can't be right. i can't see why that should be...do you see? can you explain or give a link? 
the first game--spiced up with increasing winnings/prizes as the game progresses, penalties for not winning a turn, choice to end game whenever you want, etc--could be a fun game in the Price Is Right
the first game--spiced up with increasing winnings/prizes as the game progresses, penalties for not winning a turn, choice to end game whenever you want, etc--could be a fun game in the Price Is Right
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Well as you were trying to point out, it's much more likely that you'll be hitting occasional peaks (points where you've gotten a card greater than all previous ones) than get cards that keep increasing in value. In fact, unless your first cards are A, 2, 3, ... in the second scenario you're bound to have to stop playing as soon as you get an A (if not earlier), whereas in the second scenario the game is guaranteed to go on all the way to the last card. I think that's a satisfying enough explanation
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The largest of the first i cards has an equal chance of being in any of the first i positions, so the probability that it's in the i-th position is 1/i.
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