Cute Game Theory Problem [Archive] - Two Plus Two Older Archives

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09-02-2001, 03:59 PM

This question is not really germane to real world play. It is however a good test as to whether you really fully understand how to do and apply this game theory stuff.

Tom Weideman is playing Paul Pudaite head up something. They have both anted $50. The $100 pot has not yet increased and the last face down card is coming. But all their other cards are for some reason exposed. With that last face down card to come, Paul has a non improvable hand and Tom has exactly a 47% chance to beat it with his last unexposed card.(I know that 47% is not precisely possible in real life but don't worry about it.)

On this next to last betting round, the limit is $100. $100 is also the limit on the last betting round. Both players have lots of money in front of them. Both players have no tell on the other but they are both fully conversant with game theory and know that the other one is also.

Paul is first to act on the next to last betting round. How will this next to last round be played?

(A) Check-Check

(B) Check-Bet-Call

(C) Check-Bet-Checkraise-Call

(D) Bet-Call

(E) Bet-Raise-Call

(F) Something Else (If so, What?)

Remember that all cards except the last one are exposed and that both players are expert game theorticians and know the other guy is too. Anybody is allowed to answer this question except Tom or Paul since they might be biased. Please answer only with a letter or in the case of F, a curt description. There is only one right answer.

09-02-2001, 05:32 PM

09-02-2001, 05:43 PM

09-02-2001, 09:36 PM

Though I can't help but think I'm missing something...

What makes this question cute?

09-02-2001, 11:38 PM

It "E, 100%" when Tom acts first.

09-03-2001, 12:34 AM

C, 100%

09-03-2001, 01:19 AM

its in eye of the beholder, i suppose

but if my math is right, the fact that Tom's optimal strategy is to bet into a what both sides know is a check-raise is kind of neat

09-03-2001, 01:37 AM

C is actually, the right answer, I went through the math again and got +29.5 for $100 in the pot going into the river, +23.75 for $300 going into the river, and +25.83 for $500 going into the river.

09-03-2001, 04:54 AM

I know I'm not supposed to answer, but I'm going to break the rules anyway. The answer is F:

I'd take the money in the pot and write Paul a check for $20.83 so we can get on with our lives.

Tom

09-03-2001, 05:43 AM

Louie,

I have been up all night and will not to attempt this problem now, but at first glance it does seem a bit difficult to imagine why any sort of raising would take place...

'til later;-)

M

09-03-2001, 06:54 AM

>>I have been up all night and will not to attempt this problem now, but at first glance it does seem a bit difficult to imagine why any sort of raising would take place... <<

And yet...

Hint: Why do you think David thinks this problem is "cute"?

Tom Weideman

09-03-2001, 09:09 AM

Actually more intersting than funny !

<HTML> <HEAD> <meta http-equiv="Content-Type" content="text/html; charset=windows-1252"> <TITLE>cute2</TITLE> </HEAD> <BODY>

<Table border> <TR ALIGN="center" VALIGN="bottom"> <TD>bets</TD> <TD>bluff</TD> <TD>wins</TD> <TD>cost</TD> <TD>pot</TD> <TD>res</TD> </TR> <TR ALIGN="center" VALIGN="bottom"> <TD>0</TD> <TD>0,5</TD> <TD>0,705</TD> <TD>0</TD> <TD>100</TD> <TD>29,5</TD> </TR> <TR ALIGN="center" VALIGN="bottom"> <TD>1</TD> <TD>0,25</TD> <TD>0,5875</TD> <TD>100</TD> <TD>300</TD> <TD>23,75</TD> </TR> <TR ALIGN="center" VALIGN="bottom"> <TD>2</TD> <TD>0,166667</TD> <TD>0,548333</TD> <TD>200</TD> <TD>500</TD> <TD>25,83333</TD> </TR> <TR ALIGN="center" VALIGN="bottom"> <TD>3</TD> <TD>0,125</TD> <TD>0,52875</TD> <TD>300</TD> <TD>700</TD> <TD>29,875</TD> </TR> <TR ALIGN="center" VALIGN="bottom"> <TD>4</TD> <TD>0,1</TD> <TD>0,517</TD> <TD>400</TD> <TD>900</TD> <TD>34,7</TD> </TR> <TR ALIGN="center" VALIGN="bottom"> <TD>5</TD> <TD>0,083333</TD> <TD>0,509167</TD> <TD>500</TD> <TD>1100</TD> <TD>39,91667</TD> </TR> <TR ALIGN="center" VALIGN="bottom"> <TD>6</TD> <TD>0,071429</TD> <TD>0,503571</TD> <TD>600</TD> <TD>1300</TD> <TD>45,35714</TD> </TR> </Table>

 </BODY> </HTML>

The situation is stabilized with 2 bets. The bluff-ratio is =1/6 giving the draw-hand 54,83% wins/outs. It will cost 200 to see the river an there will be 500 in the pot.

EV(Paul)= (1-0,5483333)*500 - 200 = 25,83.

09-03-2001, 11:08 AM

09-03-2001, 11:58 AM

if there is a seat open.

JG

09-03-2001, 09:19 PM

09-03-2001, 11:44 PM

C

This is a restating of the "Prisoner's Dilemna" with a touch more poker relevance and a couple of extra boxes in the matrix.

09-04-2001, 08:32 AM

If this is a timed test, I've taken 30 seconds of thought and say

B.

If this is not a timed test and more ego is on the line for getting the right answer than for answering quickly, I reserve the right to retreat and "do the math".

--JMike

09-04-2001, 09:18 AM

I couldn't concentrate on work until "doing the math" and came up with a different answer:

C.

Tom bets to drive Paul's EV down and Paul check-raises to get a little bit of it back. Tom just calls, because each dollar that gets into the pot at this stage increases Paul's EV.

--JMike p.s. Derisive commentary from "Divad S." and the guy who switched his answer to a previous game theory problem can be taken as expected and read.

09-04-2001, 11:18 AM

The Prisoners Dilema was where each is made an offer, the results of which are dependant on the other's choice, and they both do better if they choose option 1 but each does better if he chooses option 2?

I agree with your answer, but don't see the relationship, mainly because Poker is a zero sum game but that dilema is not.

- Louie

09-04-2001, 11:47 AM

09-04-2001, 06:58 PM

Nope, you miscalculated (twice, apparently). You might want to recheck this number in your computations:

pot equity = 0.47 + (1/6)(0.53)

I get 55.83%, not the 54.83% that you got. This 1% difference for the $500 pot accounts for the $5 difference in our answers.

Tom Weideman

09-05-2001, 12:57 AM

It's seems to me that you are bluffing to much:

(0,53/6)*600 - 0,47*100 = 6

I like this one better:

(0,47/6)*600 - 0,47*100 = 0

09-05-2001, 11:38 AM

David,

Precisely what information does each player have after the last card(s?) is dealt?

How can anyone attempt this problem without knowing the answer to the above question?

Also,

does Paul act first on the final round?

Are the bets in fixed units of $100?

Maybe there is a unique common sense interpretation, (I admit there is on the last two questions) but I'd like to see it made explicit. Thanks.

Dirk(MildManneredMathMan)

09-05-2001, 01:59 PM

09-05-2001, 02:16 PM

will try it again.

09-05-2001, 02:36 PM

[1] The last card is down and neither player sees his opponent's card, neither projects tells, so neither player has information on the other player, other than of course that he received his last card. [2] On the river: surely the known pat hand will check, so the order of betting doesn't matter. [3] Yes, bets are $100.

- Louie

09-05-2001, 04:08 PM

This does not quite answer my main question. We are told that on the 2nd last round both Tom and Paul know that Tom has exactly 47% chance of having the best hand. But what exactly do the players know on the last card. One way I can see to interpret this is that after the last card, but before any action that might give further information:

EITHER: Tom will be 100% certain he has the best hand, but Paul will have no new information (this will happen 47% of the time)

OR: Tom will be 100% certain he has the worst hand, but Paul will have no new information (this will happen 53% of the time)

Is this the right interpretation? (Presumably we are to suppose that Paul's card does not affect his 53% estimate of being the best at the end.)

Dirk(MildManneredMathMan)

09-05-2001, 06:01 PM

C. very cute.

09-05-2001, 08:32 PM

Preamble: I hope I didn't switch the names Paul Tom below...

Before the last card, all cards are face up. Paul's hand is currently "best" but cannot improve (equivalent to a small straight) and Tom's hand is currently "worse" but has a 47% chance to outdraw Tom (equivalent roughly to flush plus big multi-card straight draw).

Paul doesn't need to look at his last card, it cannot help. When Tom looks, he will know for sure whether or not he has the best hand. He will certainly bet the 47% of the time he "outdraws" Paul, and will also bluff at some additional optimum frequency.

- Louie

09-05-2001, 11:08 PM

I think I see what Dirk is getting at, cause TECHNICALLY, the knowlege of Paul's last card would mean a TINY bit to Paul, as he might then know the whereabouts of one of Tom's outs.

This would alter the conditional probability that Tom is bluffing by a TINY amount, so he could have a slightly higher optimal calling frequency when he caught one of Tom's outs and a slightly lower optimal calling frequency when he didn't.

I think this detail is unimportant though. It'd be pretty farfetched for it to affect the actual answer, C, although it might tweak the EV a bit.

09-06-2001, 03:23 AM

I thought I posted a reply to this already, but now I don't see it, so here it is again...

Jack's right, I goofed. Paul would have never accepted that paltry check. That will teach David to use my name in an example.

I still hope Geary sits down in our game, though.

Tom Weideman

09-06-2001, 08:35 AM

Thanks Louie. I will try the problem now (after I teach). I think David meant us to ignore the effect of what Paul knows from his last card. Essentially the game is: ante, then 1 round of betting, then Tom (only) gets one card (which only he sees) from a 100 card deck where 47 say `you win' and 53 say `you lose' and that's the only card anyone gets, and then there's a second round of betting.

Dirk(MildManneredMathMan)

09-06-2001, 12:29 PM

C: Louie is right (also in scenario where Tom bets first, where it would go bet raise call).

In both case the fair deal would be Tom takes the antes and gives Paul $25.83 (+1/3c).

After the first round, Tom would most prefer the pot to be $300, but he cannot force that (regardless of who bets first), and he prefers a pot of $500 to $300 (or $100).

(No randomized play is needed.)

Dirk(MildManneredMathMan)

09-06-2001, 08:02 PM

....I'm not doubting that Louie is probably right, just reporting my first impression above.

I have managed to come up with different answers from myself (!) on several occasions, so obviously I could benefit from more practice with things like this.

As an interesting aside, it might be worthwhile to determine what ideal pot size Tom would like, not only in this example, but in ANY example; that is, a general equation rather than a specific one.

09-07-2001, 08:41 AM

What does LOL stand for?

(I would like to see more about this theory too.)

09-07-2001, 06:10 PM

Laugh Out Loud

09-09-2001, 05:51 AM

Is the key to this the optimal bluffing percentage for Tom on the final round?

If so, how do you find this and then reason back?

09-10-2001, 08:40 AM

Do it in general for given pot size T.