Classic Type Game Theory Problem

[David Sklansky]

I just thought of this problem recently when a player mistakingly exposed his pat lowball hand too soon in a triple draw game against an all in opponent.

I realized that the situation rephrased in rigorous terms is a simply stated classic type game theory problem, perhaps never addressed before. Just in case that's so, we'll call it the Sklansky Exposed Pat Hand Problem. It goes like this:

Player A and Player B are both dealt a real number from zero to one. Higher number wins. No betting except for antes. Player A looks at his downcard and decides whether to keep it or replace it. If he replaces it he gets that second card face down. After Player A acts, Player B has the same option. And of course his decison will be based partially on what A did. But the thing about this game is that Player B's first card is face up. So B knows that A's decision to replace was based on what A saw.

If there is $100 in the pot and both players play perfectly what is the EV for both players? What is the optimum strategy?

I'm going to put this question in the Poker Theory, Probability, and Science Math and Philosophy Forums at the same time.

[Shandrax]

Let's replace a number > 0.50 with 1 and a number < 0.50 with 0. The reason to do this is that if you redraw with a number higher than 0.50, it is more likely that you will make it worse.

Under this assumption and with "optimal" play for both sides I got an EV of 0!

If player A has 1 he stands (EV: +12.5%) and if he has 0 he redraws (EV: -12.5%). That play has a better expectation than trying to stand/bluff with 0 (overall EV: -25%).

Player B always stands with 1 and always redraws with 0.

[daryn]

what if A gets .457890435087582038946078?

[Shandrax]

I have edited it in the hope that it makes sense now.

Hmmm... what is Player B's move if Player B has 0.85 showing and Player A decides to keep his card?

[Shandrax]

Oh lol, I overlooked that B's number is showing. Geez...especially since it's called "exposed" hand problem.

Let's say that A's inital holding is a and B's initial holding is b.
Then A redraws if:
1. b<.5 and a<.5
2. b>.5 and a<1-sqrt(2*(1-b))

B redraws iff b<.5.

Now assuming the initial pot is 1, the value of the game should be

Player A has the best information in this game, and therefore will have an overall advantage. B only has the information of his own card, and what A does , but A could be bluffing on any given act. a1 = player A's initial card, a2 = replacement if taken, same for b1 and b2.

Based on B's only reliable information (his own card), he will replace with b1 < .5 (since he is most likely to improve), and stand with b1 > .5 (since a replacement will most likely result with b2 < b1)

If a1 < .5, A must replace.
If .5 < a1 < b1, A must replace since he has no chance of winning otherwise (since B will not replace).

Even though A will have an uphill battle drawing against a b1 > .5, at least he has the fortune of knowing whether or not he MUST draw. B in turn will draw sometimes unecessarily and give up a win. I wil have to think harder for winning percentages.

[Shandrax]

Well, of course player A has the advantage. Player B has to decide below which number he will always draw and above which number he will always stand. Let's say just as an example, he always draws < 0.25 and always stands > 0.75 and inbetween it depends on what A does.

A on the other side has to bluff a certain amount of times. Overall it's this all about the Nash equilibrum and very complicated. No idea how you calculate this.

[Gabe]

If I was forced to play this game right now without thinking about it:

If I was A, and B was <.50 I would draw if I was <.50. If B was <.75 I would stand on anything >.50.

If I was B, I would stand on anything >.75.

However, after I give it some thought the over/under maybe somewhere between .66 and .75.

This strategy might be totally out to lunch. Hopefully, I'll have time to give this some thought.

[Jellyroll]

If both players played straight forward, with no "bluffing", then both players should play as follows:

Player A:

- should always draw if his card is less than Player B's or his card is less than .5, otherwise he should stand (with a card > .5 and beating player B).

If Player A is behind, he should draw for obvious reasons. Although, in the situation where both A and B have less than .5, but A is in the lead, Player A should redraw anyway. If A stands with a card under .5, B will draw anyway, assuming he's behind, most likely improving his hand.

Player B:

- should always redraw if player A stands (it shows B is beaten)

- should only redraw, when Player A redraws, if his hand is worse than average ( < .5)

Playing like this will result in neutral EV on both sides.

However, the difficulty comes in when you have a situation where Player A has a hand like .70 and Player B has a hand like .75. In this situation, it would be advantageous for Player A to get Player B to redraw every time. There is a greater chance of Player B drawing to a lower number than A's card, rather than A drawing to beat Player B.

If Player A "bluff stands" too frequently, Player B would profit from always standing too. Also, if Player A never bluffs, Player B should also always draw. So, it seems as though you have to follow bluffing frequency.

I am assuming that in this case that frequency would be ((1 - [player b's card]) * 50) percent. So, if Player B has .90, and Player A has stood with a range of .5 to 1, he will have B beat 80% half the time, and will be behind 20% the time. The correct bluffing frequency would be 20%.

So, if Player A is bluffing 20% of the time here. Player B will need to redraw 20% of the time in this situation in order to not call too often or too little (and be exploited).

Although, it should also be noted, that it would be advantageous for Player A to bluff when he has the highest range of hands.

For example:

If Player A has .60 and Player B has .50 - .59, he should be bluffing around 20% of the time. It would be best if he bluffed with the upper 20% range of his hands (like .58 - .59) and redrawing with the lower half. This maximizes his chance of still winning when Player B mistakingly redraws.

I may be completely off on this, but maybe some of my ideas are on. Maybe someone can verify? I'm not sure how you'd go about calculating the EV of this though. If I had to guess, I'd say they were still equal EV.

JellyRoll

[Xhad]

OK, so work was really slow today so I think I pretty much figured this out except the exact amounts of EV earned by each player in the game (except, as I was typing it up I caught a damn algebraic error I need to fix). I'm too tired to figure that out at this point, but I decided I'd post what I have so far just in case there's an error someone could catch.

I broke the entire process down into levels of thinking to make it easier to follow and to refer back to previous sections.

Throughout this post “a” will be used to mean A's hand, and “b” will be used to mean B's hand. Note that each player's hand is also that player's probability of winning against a random hand, which makes a lot of the math simple.

LEVEL 1

The first thing that jumped out at me was that if A ever draws, B's play is totally automatic since A now has a random hand and there is no future betting.

-The EV of drawing against a random hand is $50.
-The EV of standing pat against a random hand is $100*b

Therefore, .5 is an EV neutral hand, all worse hands should draw and all better hands should stay.

Note also that there is no reason for B to ever stay pat with a worse-than-random hand whether A stays pat or not. This is because B always rates to improve on the draw in this case so the draw is worth it even if A might be standing pat on a bad hand for deception purposes.

LEVEL 2

Because there is no reason for B to deviate from the above and A knows B's hand, this means that A is always given the option to make his EV $50 if B's hand is worse than random and his EV (1-b)$100 if B's hand is better than random.

At this point it's best to think in terms of whether A's hand is better than B's, and whether either hand is better than a random hand:

a > b > .5 (ABR): A should clearly stand pat. If B stands pat A will automatically win, and if B draws A figures to win a majority of the time, but if A gives himself a random hand he will make his hand worse a majority of the time and B will always stand pat when he does so.

a > .5 > b (ARB): B will always draw here, which means his hand is essentially random, and A's hand is already better than random, which means standing pat gives A more than half-equity in the pot. Drawing will result in B also drawing making the hand a push for both players. A must stand pat.

.5 > b > a (RBA): B will always draw, and A's hand is worse than a random hand so A must also draw.

.5 > a > b (RAB): Same reasoning as above. A must draw.
b > a > .5 (BAR): Here it gets a little trickier. The EV of A drawing is equal to $100(1-b) since B will stay in this case. The EV of A standing pat is $100(probability of B taking the draw)(a) since for this trick to work, B not only has to take the draw but also end up with a hand worse than what A started with. Here is a situation that has the potential for deceptive play.

b > .5 > a (BRA): Similar situation to above, though more likely to result in drawing as the right play since you rate to improve your hand if you draw.


Here I figured it would be best if I just assumed a course of action and had B react to it, so I assumed A would draw in both of the above cases.

LEVEL 3

If B catches on to the fact that A's strategy is as above, then A's cards being “hidden” is useless because B will always know when he is beat; he will never be ahead if A stands pat. With this in mind, B always chooses to draw if A stands.

LEVEL 4

Going back to cases BAR and BRA from Level 2 we can find a way to sometimes stand pat with losers (which I'll call “snowing” for simplicity even though it's not quite the same thing) but still turn a profit if B continues to always draw. The reason is because if a and b are close in value, the chances of B making a worse hand than A already stayed with can be greater than the chances of A outdrawing B. A can't just stand pat every time he gets a losing hand though because sometimes he rates to lose even if B falls for his trick.

If we assume that B always draws if A stays, then the EV of each action is as follows:

Snow: EV = $100(a) since a is the probability that a's hand will beat a random hand.

Draw: EV = $100(1-b) by similar reasoning.

So for a snow to be +EV, $100(a) must be greater than $100(1-b), which means that a + b must be greater than 1. Right now B's strategy is perfectly vulnerable to snows, so we'll now assume A starts snowing at every correct opportunity.

(to be continued)

[Xhad]

LEVEL 5

Now B has to defend against this tactic, lest he continue to bust his .95's every time A snows with a .60. To recap A's strategy:

ARB: Always stand pat.
ABR: Always stand pat.
RBA: Always draw.
RAB: Always draw.
BAR: Stay if A + B > 1, otherwise draw.
BRA: Stay if A + B > 1, otherwise draw.

B is not concerned about the cases where A draws anymore because play there is automatic. The concern is whether to draw when A stays, or stay hoping to catch a snow.

The first thing to do is put A on a hand range if he stands pat. His hand range for snows is 1-b to b (note that b always has a hand > .5 here, so 1-b will always be less than b). His hand range for legitimate hands is b to 1. This gives a hand range of 1-b to 1.

This would mean that the probability of a snow is [b – (1-b)]/[1-(1-b)], which simplifies to 2 – 1/b. So EV of staying pat is $100 (2-1/b) since B always wins if he stands pat against a snow.

But of course this isn't the whole story. Sometimes B will draw against a snow only to still be ahead after the draw, and sometimes B will fail to catch up when A had a legitimate hand.

So what we next need to do is figure out A's average hand. Since we already know his range is (1-b) to 1, this would mean that his average hand is [1-(1-b)]/2, or 1-(.5)b.

So the EV of drawing would be $100 (1 – A's average hand) (which is $50b), since that is the probability of winning if we draw.

All this would mean that to stay pat, the following would have to be true:

$100(2-1/b) > $50b

Which simplifies to:

-1/2 b^2 + 2b – 1 > 0

Solving the quadratic forumula yields 2 – sqrt(2) or 2 + sqrt(2), the latter is not a legal hand so 2 – sqrt(2) (or a little under .60) is the minimum snow-catching hand.

[Jerrod Ankenman]

[ QUOTE ]
I just thought of this problem recently when a player mistakingly exposed his pat lowball hand too soon in a triple draw game against an all in opponent.

I realized that the situation rephrased in rigorous terms is a simply stated classic type game theory problem, perhaps never addressed before. Just in case that's so, we'll call it the Sklansky Exposed Pat Hand Problem. It goes like this:

Player A and Player B are both dealt a real number from zero to one. Higher number wins. No betting except for antes. Player A looks at his downcard and decides whether to keep it or replace it. If he replaces it he gets that second card face down. After Player A acts, Player B has the same option. And of course his decison will be based partially on what A did. But the thing about this game is that Player B's first card is face up. So B knows that A's decision to replace was based on what A saw.

If there is $100 in the pot and both players play perfectly what is the EV for both players? What is the optimum strategy?


[/ QUOTE ]

First, let's dispense with the easy and obvious strategies.

Call the exposed card y.
If y < .5, then if A draws, B should draw.
If y > .5, then if A draws, B should be pat.

Hopefully these are clear.

Suppose there was a strategy such that A could have +EV with a hand less than .5 by patbluffing against y < .5 Then Y could foil such a strategy by drawing. Hence optimally:

If y < .5 and A is worse than .5, A and B should draw.
If y < .5 and A is better than .5, A should be pat and B should draw.


That's two cases, and we haven't even done any algebra yet!

Ok, now the interesting case: y > .5.

Now A should obviously call with hands that are better than y, since by doing so he locks up equity equal to his hand's value. He may or may not "pat-bluff."

If he does do so, however, since there is always a showdown, it's dominated for him to do so with any hands that are weaker than hands he would throw away and draw.

Hence A's strategy is:

(draw region) -> (patbluff region ?) -> y -> (valuepat region)

The only relevant threshold here is between the draw region and the patbluff region. Call this value x.

B doesn't have regions, only a mixed strategy of drawing and being pat to make A indifferent to patbluffing. Call B's % of hands that draw z. B will select this value such that A is indifferent to drawing or patbluffing at x.

At x, here are A's equities (in pots):

<draw> = (1-y)
<pat> = (z)(x)

1-y = zx, or
z = (1-y)/x (1)

Now let's consider A's play. A will choose x such that Y is indifferent to being pat or drawing when A is pat.

Y's EVs:
<draw>: (1-x)/2 (draw equity against the range [x,1])
<pat>: y-x (wins when A patbluffs, otherwise 0)

(1-x)/2 = y-x
x = 2y-1

Now there's a lower bound to the validity of this equation. For example, consider y = .55. Then x = .10! This seems intuitively wrong, and it is. The reason is that for some values of x, Y is no longer indifferent, because the card space is limited to [0,1]. We can also see this in z.

z <= 1
(1-y)/x <= 1
((1-x)/2)/x <= 1
x >= 1/3

So that provides a floor for x; that is, A never pat-bluffs with a hand worse than 1/3.

So we have the third case:

If y > .5, then A is pat with hands better than 2y-1, or 1/3, whichever is higher, and draws otherwise. If A draws, Y is pat. Otherwise Y draws with (1-y)/x of his hands, where x is the cutoff value above.

Finding the value of the game is left as a simple exercise for the reader. The cases where A draws are simple, and you can just assume Y always draws (since he's indifferent to drawing or not).

Thanks David, I needed a simple problem to distract me today.

Jerrod Ankenman

[David Sklansky]

So have you seen this problem before? If not, don't forget to name it after me in you next article.

"Thanks David, I needed a simple problem to distract me today."

Jerrod Ankenman

OK smartypants. Suppose with B's card face up, they each have TWO chances to switch. B's second card (if he switches on the first round) is face up or face down. Feel free to do whichever problem you find "simpler".

[Darryl_P]

I think you made a mistake. You said Y's EV when pat is simply y-x. Shouldn't it be (y-x)/(1-x) ? That could explain the sqrt's that some others have in their solutions.

[BB King's]

... your solution in the SMP-Forum. I really don’t follow your calculation – my fault ?!

How do you respond when B’s upcard is 0.580 ?

You can see my solution in the SMP-Forum.

BB King’s ... Formerly known as the pokerplayer formerly known as Jack

[BB King's]

I really like the idea – you get a problem named after you – do you see why ?

But I really think you should check the solution !

BB King’s ... Formerly known as the pokerplayer formerly known as Jack

[ QUOTE ]
If y < .5 and A is better than .5, A should be pat and B should draw.

[/ QUOTE ]

This doesn't seem right. If y = .0001 & A has .0002, surely A would draw.

[Jerrod Ankenman]

[ QUOTE ]
So have you seen this problem before? If not, don't forget to name it after me in you next article.

"Thanks David, I needed a simple problem to distract me today."

Jerrod Ankenman

OK smartypants. Suppose with B's card face up, they each have TWO chances to switch. B's second card (if he switches on the first round) is face up or face down. Feel free to do whichever problem you find "simpler".

[/ QUOTE ]

I did actually make a mistake in my solution, pointed out by BB King. This changes the thresholds but we can just resolve the equations in the same way.

The problem with two streets is harder; I don't have time to work on it right now, but the distributions of hands on street 2 are non-uniform and a little more complex.

I have not seen this type of problem particularly before, but it's part of the class of [0,1] problems that Bill and I have spent a lot of time solving, so I have experience with these types of problems.

If anyone was offended by my "simple problem" banter, I hope you realize that I just like to get on David's case about the game theory problems he poses. It's clearly not a particularly difficult problem, but you can learn a lot by solving these types of things.

Jerrod

[Jerrod Ankenman]

[ QUOTE ]
[ QUOTE ]
If y < .5 and A is better than .5, A should be pat and B should draw.

[/ QUOTE ]

This doesn't seem right. If y = .0001 & A has .0002, surely A would draw.

[/ QUOTE ]

Er, .0002 is not better than .5.

Jerrod

[BBB]

As was pointed out by someone else, I think you made a mistake in B's EV when pat. I fixed the your post, and found a strategy boundary (for a different reason) at y=0.586, if I figured correctly.

[ QUOTE ]

Ok, now the interesting case: y > .5.

Now A should obviously call with hands that are better than y, since by doing so he locks up equity equal to his hand's value. He may or may not "pat-bluff."

If he does do so, however, since there is always a showdown, it's dominated for him to do so with any hands that are weaker than hands he would throw away and draw.

Hence A's strategy is:

(draw region) -> (patbluff region ?) -> y -> (valuepat region)

The only relevant threshold here is between the draw region and the patbluff region. Call this value x.

B doesn't have regions, only a mixed strategy of drawing and being pat to make A indifferent to patbluffing. Call B's % of hands that draw z. B will select this value such that A is indifferent to drawing or patbluffing at x.

At x, here are A's equities (in pots):

<draw> = (1-y)
<pat> = (z)(x)

1-y = zx, or
z = (1-y)/x (1)

Now let's consider A's play. A will choose x such that B is indifferent to being pat or drawing when A is pat.

B's EVs:
<draw>: (1-x)/2 (draw equity against the range [x,1])
&lt;pat&gt;: <font color="red">(</font>y-x<font color="red">)/(1-x)</font> (wins when A patbluffs, otherwise 0<font color="red">. Note that A's hand must be on the interval [x,1].</font>)

(1-x)/2 = <font color="red">(</font>y-x<font color="red">)/(1-x)</font>
<font color="red">(1-x)^2 = 2(y-x)
1-2x+x^2 = 2y-2x
1+x^2 = 2y
x^2 = 2y-1
x = sqrt(2y-1)</font>

Now there's a lower bound to the validity of this equation. For example, consider y = .55. Then x = <font color="red">.31</font>! <font color="red">This is clearly wrong; note that if A stands pat at .31, then if B stands pat, A loses, and if B draws, A has .31 equity, whereas if A draws, B will stand, and A has .45 equity. B is still indifferent, since he does not know A's card, but A has made a -EV decision in order to give B a neutral EV decision.</font>

[/ QUOTE ]

In order to avoid such situations, A must draw whenever A's holecard is &lt; 1-y. Setting y-1 equal to sqrt(2y-1) yields y = 2-sqrt(2), or about 0.586.

Thus, the cutoff x can only be sqrt(2y-1) when y &gt; 0.586. For lower values of y, a higher cutoff must be established, since A is obligated to draw when his holecard is &lt; 1-y.

So, for 0.5 &lt; y &lt; 0.586, B must draw if A stands pat, since A's patbluffing cutoff will be higher than what it would need to be to make B's decision EV neutral. Thus, A should stand pat with any card larger than 0.5 if 0.5 &lt; y &lt;0.586.

EDIT: Also, if 0.5 &lt; y &lt;0.586, A should stand pat with a holecard &gt; 1-y, since B will then draw, and A's equity will be his holecard value (A's equity will be 1-y if he draws, since B will then stand).

So, to summarize:

PLAYER A:
If y &lt; 0.5, stand pat if holecard is better than 0.5, draw if worse than 0.5
If 0.5 &lt; y &lt; 0.586, stand pat if holecard is better than y-1, draw if worse than y-1.
If y &gt; 0.586, stand pat if holecard is better than sqrt(2y-1), draw if worse than (2y-1)

PLAYER B:
If A draws, stand pat if y &gt; 0.5, draw if y &lt; 0.5
If A stands pat, draw if y &lt; 0.586, randomly draw z = (1-y)/sqrt(2y-1) of the time if y &gt; 0.586.

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
If y &lt; .5 and A is better than .5, A should be pat and B should draw.

[/ QUOTE ]

This doesn't seem right. If y = .0001 &amp; A has .0002, surely A would draw.

[/ QUOTE ]

Er, .0002 is not better than .5.

[/ QUOTE ]

Doh. I misread that. I thought it said if A was better than y. :| My bad.

[BBB]

Oops, I meant:

So, to summarize:

PLAYER A:
If y &lt; 0.5, stand pat if holecard is better than 0.5, draw if worse than 0.5
If 0.5 &lt; y &lt; 0.586, stand pat if holecard is better than <font color="red">1-y</font>, draw if worse than <font color="red">1-y</font>.
If y &gt; 0.586, stand pat if holecard is better than sqrt(2y-1), draw if worse than (2y-1)

PLAYER B:
If A draws, stand pat if y &gt; 0.5, draw if y &lt; 0.5
If A stands pat, draw if y &lt; 0.586, randomly draw z = (1-y)/sqrt(2y-1) of the time if y &gt; 0.586.

[/ QUOTE ]

[Trantor]

[ QUOTE ]
I just thought of this problem recently when a player mistakingly exposed his pat lowball hand too soon in a triple draw game against an all in opponent.

I realized that the situation rephrased in rigorous terms is a simply stated classic type game theory problem, perhaps never addressed before. Just in case that's so, we'll call it the Sklansky Exposed Pat Hand Problem. It goes like this:

Player A and Player B are both dealt a real number from zero to one. Higher number wins. No betting except for antes. Player A looks at his downcard and decides whether to keep it or replace it. If he replaces it he gets that second card face down. After Player A acts, Player B has the same option. And of course his decison will be based partially on what A did. But the thing about this game is that Player B's first card is face up. So B knows that A's decision to replace was based on what A saw.

If there is $100 in the pot and both players play perfectly what is the EV for both players? What is the optimum strategy?

I'm going to put this question in the Poker Theory, Probability, and Science Math and Philosophy Forums at the same time.

[/ QUOTE ]

A variation on old theme?

* Just for fun a gambler approaches you and asks you if you want to make an even bet with him. Sure you say, being a gambler yourself. Now he says he's going to pick two different random numbers from a distribution not known to you. He will then write the two numbers of two small pieces of paper and put each one in each of his enclosed hands. You then pick a hand and he will reveal the number in that hand. You then place your money on which hand you think contains the larger number. He will match your money, betting on the other hand, the numbers will then be revealed and whoever was right keeps the sum of the money. You only get one shot at this, and the bet is $10,000. The question is, is there a strategy to beat 50:50 odds and if so how?

http://www.azillionmonkeys.com/qed/hints.html

[BillChen]

I believe this is right. Also notice that the threshold is x = sqrt(2) - 1, where if y = 1-x, x stands pat at 0.414. The value of r shows up a lot.

Bill Chen

[Darryl_P]

If there is some nonzero probability that the distribution has finite bounds, then you can achieve positive EV by choosing an arbitrary number and deciding to go with the first number if it's higher, otherwise switching.

[BBB]

[ QUOTE ]

PLAYER A:
If y &lt; 0.5, stand pat if holecard is better than 0.5, draw if worse than 0.5
If 0.5 &lt; y &lt; 0.586, stand pat if holecard is better than 1-y, draw if worse than 1-y.
If y &gt; 0.586, stand pat if holecard is better than sqrt(2y-1), draw if worse than (2y-1)

PLAYER B:
If A draws, stand pat if y &gt; 0.5, draw if y &lt; 0.5
If A stands pat, draw if y &lt; 0.586, randomly draw z = (1-y)/sqrt(2y-1) of the time if y &gt; 0.586.

[/ QUOTE ]

Darryl P posted in the SM&amp;P forum that Player A's EV is .5062 for this strategy. (I later calculated the same answer, so I believe he is correct.).

[punter11235]

[ QUOTE ]
Player A and Player B are both dealt a real number from zero to one.

[/ QUOTE ]

This game doesnt exist. Because its impossible to deal real numbers out of all real numbers. The set of cards must be finite.

Best wishes

[David Sklansky]

"This game doesnt exist. Because its impossible to deal real numbers out of all real numbers. The set of cards must be finite."

For those who don't get it, he only means the above in an irrelvant technical way due to the way infinity is handled. If the question stipulated that the cards ranged from zero to one. in jumps of one billionth, and I asked for the answer to four decimal places, the answer would be just what was calculated.

Meanwhile one wonders why the point is brought up at all. Could it perhaps be a gut reaction to an inability to do the problem even if it was rephrased in a way to meet the objection?

[alThor]

[ QUOTE ]
"This game doesnt exist. Because its impossible to deal real numbers out of all real numbers. The set of cards must be finite."

For those who don't get it, he only means the above in an irrelvant technical way due to the way infinity is handled. If the question stipulated that the cards ranged from zero to one. in jumps of one billionth, and I asked for the answer to four decimal places, the answer would be just what was calculated.

Meanwhile one wonders why the point is brought up at all. Could it perhaps be a gut reaction to an inability to do the problem even if it was rephrased in a way to meet the objection?

[/ QUOTE ]

Don't even validate (by response) this insanity that is occurring on the Theory and Science boards, with people claiming there is "no such thing as a uniform distribution on [0,1]." It is pure nonsense. The people saying this are either trolling, or beyond reason.

Your initial phrasing of the question (and all previous times you've implicitly invoked this uniform distribution) need no explanation, and are correct.

Here is one of many reputible sites that discusses this elementary distribution:
http://mathworld.wolfram.com/UniformDistribution.html

And for those that think continuous distributions don't exist in the real world, perhaps they can use their wisdom to describe a discrete distribution that tells us the timing of the decay of radioactive particles. Good luck with that.

alThor

P.S. As has been said countless times on the probability board, there is no uniform distribution on [0,infinity]. I often wonder if people are confusing this with the current case.

[punter11235]

Its important, because you need number N (number of cards) which is important part of the solution. You can rephrase this problem to : "there is N cards dealt uniformly at random etc..." and then this problem make sense.
Uniform distribution does exist but it doesnt change the fact that picking real numbers uniformly at random is impossible.

Best wishes

[punter11235]

[ QUOTE ]
P.S. As has been said countless times on the probability board, there is no uniform distribution on [0,infinity]. I often wonder if people are confusing this with the current case.

[/ QUOTE ]

alThore, uniform distribution over [0;1] does exist but it is not possible to use it to pick up numbers from [0;1] at random. You can construct statement as : "probability of choosing number from 0.25 to 0.75 is such and such" but you cant calculate probability of choosing any particular number. This is why I said that you need finite set. It is important because if you try any calculations you need the number of all cards in the game. Hopefully the solution the solution may be written as simple function of N and then you can extend N as high as you wish.

Best wishes

[Vincent Lepore]

Hi Gabe,

I don't know the correct strategy but I'll bet you should not Draw if b is =&lt;.65 Like I said I don't know the right answer but I wanted to say hello.

Vince