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Question

There are 100 blank cards- on each card, Alpha writes any positive integer he wants to. He then presents the deck to you (and you can play this game as many times as you want with this deck) and you draw the top card, look at the card, draw another card, etc. At any point you can stop the process claiming the card you just drew was the highest card in the deck. If you are correct, you win \$5, if you're wrong, you have to pay \$1

Would you take up the offer? If so, what's your strategy?

What if the deck is shuffled directly after Alpha writes the numbers and shuffled between each game? What would you do if you were Alpha?

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Maybe draw and memorize every number that gets higher. Like if you draw a 60, memorize it, then if you draw a 70, memorize it and go through the whole deck and go through again until you reach the highest number again?

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but would you take the bet? Can you guarantee to win money off of it? That's the question ;D

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but would you take the bet? Can you guarantee to win money off of it? That's the question ;D

That depends on if Alpha would let me go back through the cards or not

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That depends on if Alpha would let me go back through the cards or not

Take the bet and go through the deck and remember higest number, holding onto the last card. hence you lose..

Then take the bet again this time picking the "winning" card.. this time you win 5\$

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hehe, that's a waaaay better and waaay simpler answer than the site that made it gave. And yours works regardless of shuffling or w/e

Nice job Taliesin!!!

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it's here- though Taliesin's answer is so direct and simple I can't but help think I'm missing something out of the problem...

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Take the bet and go through the deck and remember higest number, holding onto the last card. hence you lose..

Then take the bet again this time picking the "winning" card.. this time you win 5\$

Yeah I thought of that right after I posted that but I'm so preoccupied with the Yak Hunt I didn't feel like changing it =P

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hehe, that's a waaaay better and waaay simpler answer than the site that made it gave. And yours works regardless of shuffling or w/e

Nice job Taliesin!!!

It would't work though if he has two packs of cards though. After the first he will just pull out the second.

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It would't work though if he has two packs of cards though. After the first he will just pull out the second.

I think that the way the OP works on the original site is that you don't get two times through the same deck. You have to run each deck blind.

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The solution posted on the other site is this: You flip over the first 50 cards, and remember the highest number - call it X. Then, start flipping over cards until you see one higher than X, and declare that to be the highest.

Their reasoning (although they state it poorly) is this: the strategy above is guaranteed to win if the second highest card is in the first 50, and the highest card is in the second 50. The probability of the second highest card being in the first 50 must be 1/2. The probability of the highest being in the second 50 also would then be 50/99 (they call it 50%, but close enough). Multiplying those two together, you get a probability of winning about 1/4.

My objection to this is that the solution assumes the order is determined randomly, without any bias - in other words, without the dealer choosing the order for the cards. There is no mention of randomness, though, in the OP. I would love to give this bet, as a dealer, if I could choose the order. I would specifically construct decks to beat this strategy, and clean up - especially if each person could only play once.

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If you can only cycle through the deck once...

Keep track of the number of cards flipped and the highest card up to that point. If the 96th card is higher than any others, say it's the highest. For each card higher that follows, say the same. The most you can lose (\$4) is still less than your winnnings (\$5). Then again, the odds of this being the case are quite unlikely anyways.

But who would want to do all of that work to win only \$5???

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So it's a different deck each time, and I see where they're coming from in their solution... hmm.

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Even if you shuffle it, it should always have the same high card. If you add new cards each time, lets start with 50 maybe, thats a whole new problem.

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The problem of finding largest number with unknown distribution is an old widely published classic. But let me restate this problem here, as it may be something more original.

So Alpha prepares a deck of 100 cards by writing a different positive number on each. She then presents you with the cards one by one in truly random order. You can reject a card and then it is tossed away, or you can make a guess that the number on the card is the highest. When you have guessed correctly, or when the entire deck is gone, the game is over.

If your visit to Alpha falls on an odd day, the rules are as following:

For each incorrect guess you pay \$1. If you go through the entire deck without guessing the highest number, she will have you for dinner. (Alphas have strange diets, I hear.)

On an even day, game has a different character. You pay \$20 for each incorrect guess. When you have guessed correctly, Alpha pays \$100 and lends you her limo to get home. If you went through the entire deck failing to make the correct guess, she looks the other way and lets you escape unharmed.

QUESTION 1: what's the average amount of money you expect to lose on an odd day with the best strategy possible?

QUESTION 2: devise the best strategy to maximaize your win on an even day. Calculate the average expected win with the best strategy.

Question 1 shouldn't be all that difficult to solve.

Question 2 is another matter.

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If your visit to Alpha falls on an odd day, the rules are as following:

For each incorrect guess you pay \$1. If you go through the entire

deck without guessing the highest number, she will have you for dinner.

(Alphas have strange diets, I hear.)

QUESTION 1: what's the average amount of money you expect

to lose on an odd day with the best strategy possible?

0\$.

make no guesses and enjoy your dinner.

Unless "have you for dinner" is in the Dr. Lechter sense.

Either way tho, you pay \$0.

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The problem of finding largest number with unknown distribution is an old widely published classic. But let me restate this problem here, as it may be something more original.

So Alpha prepares a deck of 100 cards by writing a different positive number on each. She then presents you with the cards one by one in truly random order. You can reject a card and then it is tossed away, or you can make a guess that the number on the card is the highest. When you have guessed correctly, or when the entire deck is gone, the game is over.

If your visit to Alpha falls on an odd day, the rules are as following:

For each incorrect guess you pay \$1. If you go through the entire deck without guessing the highest number, she will have you for dinner. (Alphas have strange diets, I hear.)

On an even day, game has a different character. You pay \$20 for each incorrect guess. When you have guessed correctly, Alpha pays \$100 and lends you her limo to get home. If you went through the entire deck failing to make the correct guess, she looks the other way and lets you escape unharmed.

QUESTION 1: what's the average amount of money you expect to lose on an odd day with the best strategy possible?

QUESTION 2: devise the best strategy to maximaize your win on an even day. Calculate the average expected win with the best strategy.

Question 1 shouldn't be all that difficult to solve.

Question 2 is another matter.

With the odd day....The best strategy would be to only visit on even days. But assuming you did visit, the only strategy you can safely use is to guess the first card, and then guess again every time you get a new maximum valued card - the payoff for missing the highest card is essentially minus infinity (being eaten), so you can take no chances.

For the even days... one winning strategy would be to play it like the other problem. Let 37 cards go past and take one guess with our odds of 37% - that would give us an expected pay out of about 0.37*100+0.63*-20 = 24.4

Whether or not that's an optimal strategy is another matter.

Edited by armcie
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...The best strategy is to make no guesses and enjoy your dinner.

Unless "have you for dinner" is in the Dr. Lechter sense.

Either way tho, you pay \$0.

Your best strategy priorities must be alligned as following:

1. Escape with your life (do not get eaten).

2. Save money. Or rather know in advance how much to bring with you. That is find your average loss if you absolutely have to guess the card.

(Alpha does not pay you on odd days, just lets you go, if you win.)

With the odd day....The best strategy would be to only visit on even days. But assuming you did visit, the only strategy you can safely use is to guess the first card, and then guess again every time you get a new maximum valued card - the payoff for missing the highest card is essentially minus infinity (being eaten), so you can take no chances.

....

- Alphas have strange calendars too. You never know whether it's her even or odd day. Nothing's there to stop Alpha from having few odd, or even days in a row.

- That's the strategy, I would adopt. So what's the average loss?

This problem is different from picking largest number with unknown distribution. Here you get multiple guesses, until the deck is depleted.

Good thoughts like the above need not go inside spoilers. Spoilers are for long solutions with lots of numbers.

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