Dween
Dween
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October 22nd, 2014 at 9:12:50 AM permalink
On a televised lottery game show, 3 contestants play a game in which the object is to choose the highest number from 1-100 from a random lot.

Player 1 is brought into the studio while the other contestants remain in isolation. She picks a card from a rack of 100 cards, and a scanner reveals the value of their card. She may keep the card, or choose another to try to improve. NOTE: This is NOT like the Showcase Showdown on the Price is Right; They do not add the numbers, but may simply replace the number they originally chose. This is then the secret target number to beat.

Player 2 is brought into the studio. She is not told the secret target number. She chooses a card, and its value is revealed. She may replace her number if she wishes. Whoever has the higher number continues, while the other player is eliminated.

Player 3 is brought into the studio, and the process repeats once more. Whoever has the higher number of the two wins the prize.

At what number should each player stop, or choose again?
Does any player have an advantage over another?
-Dween!
onenickelmiracle
onenickelmiracle
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October 22nd, 2014 at 9:52:46 AM permalink
I could be wrong, but I don't see an advantage going last if you don't know the score to beat.
I am a robot.
jml24
jml24
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October 22nd, 2014 at 10:55:27 AM permalink
OK here's my stab at it:

Initially the median value of the deck is 50.5.

Player 1 has a 50% chance of getting a number higher than 50. If she does, she should stop. If she does not, she should pick again because the majority of the remaining cards are better than her card.

If player 2 assumes player 1 followed proper strategy there is a 50% chance the deck has one high card removed leaving a median value of 50. The other 50% of the time it has two cards removed. The first removal was a low card leaving the median at 51. The second removal was random so we have no additional information so still must assume the median is 51. So half the time median is 50, half the time median is 51 so the second player follows the same strategy as the first.

Extending this logic, player 3 follows the same strategy. So no player has an advantage.

I have a feeling there is a hole in my logic somewhere. Let the big brains chime in.
dwheatley
dwheatley
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October 22nd, 2014 at 11:16:34 AM permalink
Because player 1 knows they are competing against 2 people in the end, not just 1, they probably need to shoot higher than 50.

Just 1 redraw allowed per person, right?
Wisdom is the quality that keeps you out of situations where you would otherwise need it
jml24
jml24
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October 22nd, 2014 at 11:29:43 AM permalink
Quote: dwheatley

Because player 1 knows they are competing against 2 people in the end, not just 1, they probably need to shoot higher than 50.



All three players are competing to be the highest of the three. The fact that player one and two face off first doesn't change anything. It would be the same if all three revealed their cards simultaneously.
kenarman
kenarman
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October 22nd, 2014 at 12:13:12 PM permalink
Quote: jml24

All three players are competing to be the highest of the three. The fact that player one and two face off first doesn't change anything. It would be the same if all three revealed their cards simultaneously.



I am not sure that is right. If we assume player one has drawn a card in the 50's with his first draw I think he should assume that it will be beaten. There are a possible 4 draws taking place after him the odds of none of these 4 having a value above 60 is quite small. With out allowing for the drawn cards my rusty math gives me .6x.6x.6x.6 = .1296. This logic is probably also true for the earlier players.
Be careful when you follow the masses, the M is sometimes silent.
SOOPOO
SOOPOO
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October 22nd, 2014 at 12:14:22 PM permalink
Quote: Dween



Player 3 is brought into the studio, and the process repeats once more.



Ambiguous wording. After one of the first two players is eliminated, does the winner start over again and compete with player 3 anew?

Or is the winning number selected in the first round by either player 1 or 2 the number player number 3 must beat?

Assuming it is the second scenario, then just shooting for a number over 50 is a losing approach.... if you get 51 on your first pull and stand your odds of winning against two competitors is small.

I redraw at 83 or below, stand at 84 or above. Seems about right.
Dween
Dween
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October 22nd, 2014 at 12:30:27 PM permalink
Only one redraw is allowed per person, if they choose to redraw. They may of course stay with their initial draw.

There IS replacement of cards between players, even between draws of a replacement card. An inattentive contestant could feasibly pick the same number twice. In one video example, two players both chose the same card on their initial draw. I am assuming a tie is possible, and they would have a playoff.

The show is called Make Me Rich, a quarterly show from the Michigan Lottery.
Here is the video:
From the start (with contest intros)
From the instructions
-Dween!
Dween
Dween
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October 22nd, 2014 at 12:32:32 PM permalink
Sorry for the ambiguity.

Player 3 will be trying to beat the number previously drawn and held by player 1 or 2, whichever is higher. Each contestant gets one shot at picking a number.
-Dween!
jml24
jml24
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October 22nd, 2014 at 12:36:05 PM permalink
Quote: kenarman

I am not sure that is right. If we assume player one has drawn a card in the 50's with his first draw I think he should assume that it will be beaten. There are a possible 4 draws taking place after him the odds of none of these 4 having a value above 60 is quite small. With out allowing for the drawn cards my rusty math gives me .6x.6x.6x.6 = .1296. This logic is probably also true for the earlier players.



I am not sure either, but player one is not certain to face four draws. What is certain is that when any player redraws after pulling a number greater than 50, he has a greater probability of lowering his number than raising it.

If no redraws are allowed, this is a simple game where each player has 1/3 probability of winning. I contend that a redraw only makes sense if it has a positive expectation of increasing your number.

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