Hypothetical Heads Up Gambling Situation

[thechainsaw]

I did understand but didnt really express myself well.

Lets change the 10 to a 1 and ask what hand you would play. Well any 50% hand is better than a coin flip, so the correct answer is any hand of 50% or better i.e Q5 and above. QED.

Lets now say you have 2 chances, what hands do you play? Well on the 2nd trial, you would take the strategy above. On the first trial you can shorten your calling range, but to what I dont know - perhaps you would want a hand in the top third, my gut tells me. Maybe not.

You can continue the extrapolation until you get 10 goes, and I'm again guessing that on the first go you will need a hand in the top 1/10th, etc.

Does that make more sense, i'll try and work on a proof.

[AK-47]

thechainsaw : on the last hand, you can't choose your hand, you are forced to push

[BobAllinSki]

Proving the correct ranges is near enough impossible as a hand that is of a rank X wins with a % time Y, X & Y are not linearly related and are also not evenly distributed (X is, Y isnt)

It would be a much simpler game if you just both got dealt 1 card and the highest card wins, I'm sure a correct answer could be found for that as you know your % chance of winning, getting a bettter card, and that the chances of a better card are evenly distributed.

[jay_shark]

Before we answer the problem with 10 hands , we should find the solution for 2 hands and then proceed from there .

Since you are forced to shove on the second hand , you must take the gamble if your first hand is better than 50 % .

Here is my conjecture . If you play this game with n trials , then you should wait to shove with a hand in the top 1/(n-1)*100% of all hands . For n=10 , you should expect to see a hand in the top 11.11% of hands on your 9th trial .

top 11.1% of hands includes 66+ ,a7S,k10s+ a10+ kq

However , the hands you decide to shove all in with changes with each hand dealt . You have to loosen your shoving range .

[jay_shark]

Now this changes once you don't hit on the second hand .

For instance , if my second hand isn't in the top 11.1% of hands , then I should shove if my hand is in the top 1/8 or 12.5 % of all hands .

[AK-47]

Here, knowing the % winning chance of a given hand against any other RANDOM hand is enough.

[jay_shark]

Final answer :

On first deal you shove with the top 11.1%(1/9) of hands
On second deal " " " " " 12.5 %(1/8) of hands
On third deal " " " " " 14.2%(1/7) of hands
On fourth deal " " " " " 16.6% (1/6) of hands
.
.
.
On eighth deal you shove with the top 50 % (1/2) of hands

[thechainsaw]

[ QUOTE ]
Before we answer the problem with 10 hands , we should find the solution for 2 hands and then proceed from there .

Since you are forced to shove on the second hand , you must take the gamble if your first hand is better than 50 % .

Here is my conjecture . If you play this game with n trials , then you should wait to shove with a hand in the top 1/(n-1)*100% of all hands . For n=10 , you should expect to see a hand in the top 11.11% of hands on your 9th trial .

top 11.1% of hands includes 66+ ,a7S,k10s+ a10+ kq

However , the hands you decide to shove all in with changes with each hand dealt . You have to loosen your shoving range .

[/ QUOTE ]

I like the conjecture.

I think it can be proved inductively.

let n be the number of trials before a shove and P(n) your probability of winning.

if n = 0, P(0) = 50%

P(1) = .50*.50 + .50*.75 = .50(1.25) = 62.5%

and so on - i think it's possible to find a formula - basically if P(n-1) is favourable to P(n), fold it, otherwise shove

[roommate]

I'm liking jay's logic...

I know nicho's "twist" (I am his roommate) and I'm really excited to see how your answers change... But he has forbade me from twisting it myself... so for now I'm patient.

I like to work backwards myself...
on the 10th hand- I'll play 100% (OBV)
on the 9th- top 50%
on the 8th- top 1/3

etc... this is different than jay's... i say the first hand should be top 10%... because before you are dealt the first hand, in 10 hands you can expect 1 hand in the top 10%...

But, I guess once you are delivered the first hand you must choose between it and 9 more chances (which you are likely to get a top 1/9 hand)... so I guess taking a 10.5% hand is plus EV after the first hand is dealt... kinda tricky

so i change my answer to 1/9 (1st) 1/8 (2nd) etc... down to 1/2 (8th)... but what do you do on the 9th hand? folding ensures a "coinflip" on hand 10... so i guess you push with anything better than average... top 1/2...

good question...

it is still -EV to play this game against the luck-box

~rob

[Nichomacheo]

What does "top 10%" of hands mean to you all?