Simple Math Problem

[Haupt_234]

Need help on a solution here: 4 balls numbered 1,2,3, and 4 are throwin into a hat, one being the best and 4 being the worst. The ball chooser cannot see the balls nor feel that one is different from the other. They are all the same size/shape. Assuming chooser 1,2,3 and 4 pick consecutively, does chooser 1 have any advantage over chooser 2,3, or 4?

A quick explanation/ formula would be greatly appreciated. Thanks guys.

Haupt_234

[Cerril]

Simply no, unless there are other factors. The explanation is that there are as many ways for 1,2,3,4 to be ordered (i.e. 1234, 1342, 4231, etc.) for any given number to be first as any other.

Now if there were some givens, like ball one is 10% larger, ball 2 is 5% larger, and chooser 1 has a chance of noticing this blind... but all other things being equal each chooser has the same chance of getting any given ball in the dark.

[FrankLu99]

[ QUOTE ]
Need help on a solution here: 4 balls numbered 1,2,3, and 4 are throwin into a hat, one being the best and 4 being the worst. The ball chooser cannot see the balls nor feel that one is different from the other. They are all the same size/shape. Assuming chooser 1,2,3 and 4 pick consecutively, does chooser 1 have any advantage over chooser 2,3, or 4?

A quick explanation/ formula would be greatly appreciated. Thanks guys.

Haupt_234

[/ QUOTE ]

this seems like a simple game theory problem which can be best explained through a probablity tree.

Suppose there are two balls, 1 and 2 & two players A and B. Player 1 has no advantage over player 2. Its a 50%-50% that A will get the best ball. Although player B has no choice in the matter, his odds are the same.

Suppose there are 3 players A,B and C.
1/3 chance A will choose the best ball
2/3 chance A will not, where there is a 1/2 chance B will choose the best ball and 1/2 chance C will be left with the best ball.
So C has the same chance as A of getting the ball (1/2 *2/3) = 1/3

This type of reasoning can be expanded out to 4 players.

[kpux]

No one chooser has any advantage over the other. They all have an equal shot, 25%, to choose the best ball, basically because the order they choose simply does not matter.

Now, the situation would change a bit if the choosers announced whether they had picked the best ball to the remaining choosers after they chose. Since the knowledge the choosers have changes, their odds would too.

The logic is somewhat akin to drawing to, say, a flush or straight in Hold 'Em. In a full ring game, if you have two spades and there are two on the flop, you know 5 of the 52 cards in the deck cannot come out. So 9 remaining spades out of 47 cards gives you about a 33% chance to make your flush by the end of the hand. However, knowing your opponents hole cards would change the amount of knowledge you have and therefore your odds. If you knew 4 of your opponents' hole cards were not spades, you have a much better chance of making your flush, 47 - 8 = 39. 9/39 cards left gives you almost even money to hit your flush.

The same thing with the balls. You don't know what ball the other chooser has picked so you are basically operating under the condition that all four are still in there and have an equal shot of being picked. The same way you would analyze a draw in Hold 'Em by "pretending" the rest of the deck hasn't been dealt.

This is all coming from a math major who was a little too bored at work. /images/graemlins/tongue.gif

[felson]

kpux, it doesn't matter whether anyone announces whether they have the best ball. If you're player 2, 1/4 of the time player 1 will announce that he has the best ball. 3/4 of the time, player 1 won't have it. Either way, that knowledge won't affect your strategy.

The players have no decisions to make and the ball distribution is uniform. Therefore, no player can have an advantage over another.

[kpux]

If the first chooser announces "I have the best ball", then choosers 2, 3, and 4 all have a 0% chance of getting the best ball. My point was that in a scenario where the players talk to each other in such a manner, the odds of the game change after each player picks, but only during the game. The odds at the beginning of the game are still the same, because nobody has picked yet. Much in the same way that if you hold a flush draw, if the player to your right folds two of your suit, your odds of getting your flush are obviously decreased. But before you knew this info, your odds of hitting the flush were the same. Or at least, you would play the hand under the assumption that you will make your flush about 33% of the time.

[stinkypete]

[ QUOTE ]
So 9 remaining spades out of 47 cards gives you about a 33% chance to make your flush by the end of the hand.

This is all coming from a math major who was a little too bored at work. /images/graemlins/tongue.gif

[/ QUOTE ]

you're pretty far off on that probability for a math major :P

[StogeyMike]

Don't forget that the drawer has TWO chances to make his flush - the turn and the river. The math would then be 9/47 + 9/46, or about 38.7% chance to hit - poster's estimate of 33% is roughly correct.

I think you were considering the chances of hitting it with 1 card to come.

No offense.

[aloiz]

[ QUOTE ]
Don't forget that the drawer has TWO chances to make his flush - the turn and the river. The math would then be 9/47 + 9/46, or about 38.7% chance to hit - poster's estimate of 33% is roughly correct.


[/ QUOTE ]

This is not quite correct. To figure out the odds that you hit the flush by the river calculate the odds that you don't make the flush and subtract that from 1.

1 - 38/47 * 37/46 =~ 35%.

Or you could calcualte as follows. C(9,1)*C(39,1)/C(47,2) + C(9,2)/C(47,2) =~ 35%.

Another valid way (what I assume you were intending to do) would be calculate P(hit flush on turn) + P(don't hit flush on turn) * P(hit flush on river | don't hit flush on turn) = 9/47 + 38/47 * 9/46 =~ 35%

aloiz

[LinusKS]

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Now, the situation would change a bit if the choosers announced whether they had picked the best ball to the remaining choosers after they chose.

[/ QUOTE ]

Announcing doesn't change the other players' odds.

If I draw the best ball, the other players' chances are the same whether or not I announce it.

[daryn]

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Now, the situation would change a bit if the choosers announced whether they had picked the best ball to the remaining choosers after they chose.

[/ QUOTE ]

Announcing doesn't change the other players' odds.

If I draw the best ball, the other players' chances are the same whether or not I announce it.

[/ QUOTE ]


sigh.


can we assume that if the first guy who picks says "i picked #1" that he is telling the truth? let's assume that, otherwise this is stupid.

anyway, kpux is right, if the first guy goes and then draws the #1 ball, and announces it, then the rest have a 0% chance of selecting the #1 ball. i'm aware, and i'm sure kpux is aware that this will only happen 1/4 of the time. it doesn't matter. he is stipulating that WHEN IT HAPPENS, the chances of the other guys all drop to 0%.

i could liken it to the collapse of a quantum wave function, but i don't want to get too technical here /images/graemlins/grin.gif



it's like on espn when they show the percentages, they are skewed a little because they take into account cards they have seen which have been folded or are in play in other players' hands.

[drudman]

[ QUOTE ]
[ QUOTE ]

Now, the situation would change a bit if the choosers announced whether they had picked the best ball to the remaining choosers after they chose.

[/ QUOTE ]

Announcing doesn't change the other players' odds.

If I draw the best ball, the other players' chances are the same whether or not I announce it.

[/ QUOTE ]

Man, just think about what you said. Go all-in with two cards to come and a four flush. You have a 35% chance of making it... until the next card is dealt! At which point, your odds of having a flush by the end have changed to either 17% or 100%. That turn being dealt face up is akin to announcing what Player 1 pulled from the hat. If the turn and river were dealt simultaneously, it would be akin to each player keeping secret his pull until the last pull had been made.