A relatively simple problem?

[poker-penguin]

OK, last night the following situation arose.

A guy offers to cut a deck of cards with me for $10 (high card takes it). I decline, since I'm relatively broke.

He calls me a chicken and offers to give me 2:1

The two options as I can see it are:

A) Cut a deck of cards, his $20 against my $10, high card takes it.

B) He cuts the deck, and I get two cuts to try and beat it. $10 to $10.

My gut feel is that B has a marginally higher EV (because if the second cut is required, there's one less loser in the deck) - I'd like a maths person to confirm that.

I also thought of an angle shoot where I would accept 2:1 odds and choose between A and B after his cut. My theory was he pulls < 8 I take A, otherwise B. Is that optimal?

[For the record, he backed down before we could do it]

[gaming_mouse]

Your EV for A) is:

-10*.5 + 20*.5 = 5

For B), we first compute the chance that you lose.

Note: (51 choose 2) = 1275
Also: The first term represent your oppo choosing 2, the second term 3, etc...

p = (1/13)*
( 0
+ (4 choose 2)/1275
+ (8 choose 2)/1275
+ (12 choose 2)/1275
+ (16 choose 2)/1275
+ (20 choose 2)/1275
+ (24 choose 2)/1275
+ (28 choose 2)/1275
+ (32 choose 2)/1275
+ (36 choose 2)/1275
+ (40 choose 2)/1275
+ (44 choose 2)/1275
+ (48 choose 2)/1275
)
= .3043

Now, the chance that you tie is (3/1275). The chance that you win on the first round is therefore:

1 - .3043 - (3/1275) = 0.693347059

The total chance that you win is therefore the sum of the infinite series:

0.693347059 + (3/1275)*0.693347059 + (3/1275)^2*0.693347059 +...

But we can just ignore everything after the first two terms:

0.694978464

Your EV for B) is therefore:

10*0.694978464 + (-10)*(1-0.694978464)= 3.899

So A) is clearly better.

gm

[gaming_mouse]

A mistake in the above. I calculated the chance of a tie incorrectly.

It should be:

((13 * 3) + (3 * 4) + (3 * 8) + (3 * 12) + (3 * 16) + (3 * 20) + (3 * 24) + (3 * 28) + (3 * 32) + (3 * 36) + (3 * 40) + (3 * 44) + (3 * 48)) / (13 * 1 275) = 0.0588235294

This changes the total chance of your win to approximately:

0.693347059 + 0.0588235294*0.693347059 + 0.0588235294^2*0.693347059 = 0.736531305

And your EV to:

10*0.736531305 + (-10)*(1-0.736531305)= 4.7306261

Closer, but A) is still better.

gm

[poker-penguin]

And that is why I will never be a professional proposition gambler.

I got that the EV of 20:10 was $5, but just figured the second one as 0.25 * -10 + 0.75 * +10 = (roughly 5) but removing a loser card would make it better.

Hmmm. /images/graemlins/confused.gif

[gaming_mouse]

[ QUOTE ]
And that is why I will never be a professional proposition gambler.

I got that the EV of 20:10 was $5, but just figured the second one as 0.25 * -10 + 0.75 * +10 = (roughly 5) but removing a loser card would make it better.

Hmmm. /images/graemlins/confused.gif

[/ QUOTE ]

You're shortcut method was not far off in this case: only by a few %.

There is always the chance I made an arithmetic error too, but I think I got it right.

gm

[poker-penguin]

If I can get people to agree to 50% advantages in my favour, then a couple of percent won't hurt me.

[Siegmund]

If you want an intuitive way to understand option B, think of it this way:

Three cards will be drawn from a deck. Two of them are yours, one of them is his. Each is equally likely to be the best card. Disregarding ties, you expect to win 2/3 of the time and lose 1/3 of the time. So, as a rough approximation, your EV should be close to 10*2/3 + -10*1/3 = +$3.33.

If you want an exact calculation: if he picks a card of rank i, there are 3 cards that tie him, 4(i-1) cards that beat him, and 52-4i cards the lose to him.

So, for a given i, you have 1275 possible draws, of which (52-4i)C2 of them lose, (55-4i)C2-(52-4i)C2 of them tie, and 1275-(55-4i)C2 of them win.

Adding up the 13 cases, there are 10556 ways to win, 975 ways to tie, and 5044 ways to lose. Exact EV = 10*(10556/16575) + -10*(5044/16575) = 848/255 ~ $3.325 if a tie is a tie, or 10*(10556/15600) + -10*(5044/15600) = 53/15 ~ $3.53 if a tie means you cut again.

[Paul2432]

[ QUOTE ]
Your EV for A) is:

-10*.5 + 20*.5 = 5

[/ QUOTE ]

I think we can simplify this. Hero will win 24/51, lose 24/51 and tie 3/51.

EV = 24/51*20 - 24/51*10 + 3/51*0 = 240/51 ~= 4.705882

[ QUOTE ]
For B), we first compute the chance that you lose.

Note: (51 choose 2) = 1275
Also: The first term represent your oppo choosing 2, the second term 3, etc...

p = (1/13)*
( 0
+ (4 choose 2)/1275
+ (8 choose 2)/1275
+ (12 choose 2)/1275
+ (16 choose 2)/1275
+ (20 choose 2)/1275
+ (24 choose 2)/1275
+ (28 choose 2)/1275
+ (32 choose 2)/1275
+ (36 choose 2)/1275
+ (40 choose 2)/1275
+ (44 choose 2)/1275
+ (48 choose 2)/1275
)
= .3043

Now, the chance that you tie is (3/1275). The chance that you win on the first round is therefore:

1 - .3043 - (3/1275) = 0.693347059

The total chance that you win is therefore the sum of the infinite series:

0.693347059 + (3/1275)*0.693347059 + (3/1275)^2*0.693347059 +...

But we can just ignore everything after the first two terms:

0.694978464

Your EV for B) is therefore:

10*0.694978464 + (-10)*(1-0.694978464)= 3.899

So A) is clearly better.

gm

[/ QUOTE ]

Nice analysis but I don't see why you need to do the part with the infinite series. You have three possible outcomes and each has a resolution.

EV = 10*0.636876471 - 10*0.3043 - 0 ~= 3.326

Paul

[gaming_mouse]

[ QUOTE ]

EV = 24/51*20 - 24/51*10 + 3/51*0 = 240/51 ~= 4.705882


[/ QUOTE ]

When you tie, your EV is not 0. You get to play again, and again you have the same advanctage as the first time. This is related to the infinite series question you had...

[ QUOTE ]
Nice analysis but I don't see why you need to do the part with the infinite series. You have three possible outcomes and each has a resolution.

[/ QUOTE ]

I was going under the assumption that if you tied, you would play again, and so on until there is a winner. Thus there are an infinite number of possible outcomes, but all but the first 3 contribute only negligibly to the final EV, because they are so unlikely. That is, we can just ignore the cases when there are 4 or more ties before we get a winner.

gm