Options - Related Question

Hey guys - got this question at work - would appreciate your help as my probability knowledge is a little rusty..

Basket with 5 balls - 3 black, 2 white... if you draw a black ball you have to pay 1 dollar, if you draw a white ball, you get a dollar... once a ball is drawn, it is not replaced, and you can stop drawing at ANY TIME.

What is the value to playing this game? My back of the envelope is coming to be about 17.5 cents - is that correct?

Thanks in advance and my apologies if this is not the right forum.

dibsy

[IggyWH]

I'm rusty on probability but I just wanted to chime in so when someone answers, they can explain this to me also.

Figuring out the probability of each outcome is easy. Since you can end at any given point, how do you figure out the exact value when you have that unknown of when someone would stop?

The only thing that would make sense is anytime you pick out 2 white balls, you'd stop. The same as if you picked out 3 black balls (if you went that far), you'd pick out the next 2 naturally.

How can you figure out the value when you don't know if someone would stop after picking out the first ball no matter what color it is or so on?

[BruceZ]

[ QUOTE ]
Hey guys - got this question at work - would appreciate your help as my probability knowledge is a little rusty..

Basket with 5 balls - 3 black, 2 white... if you draw a black ball you have to pay 1 dollar, if you draw a white ball, you get a dollar... once a ball is drawn, it is not replaced, and you can stop drawing at ANY TIME.

What is the value to playing this game? My back of the envelope is coming to be about 17.5 cents - is that correct?

Thanks in advance and my apologies if this is not the right forum.

dibsy

[/ QUOTE ]

Answer in white: <font color="white">
20 cents. If you draw white on the first ball, stop. If you draw black on the first ball, keep drawing until you are even or until there are no more balls.

EV = 2/5*1 + 3/5 *[-1 + E(2,2)]

where 2/5 is the probability of drawing white first, at which point we win 1 dollar. Clearly we stop at that point since we would be a 3:1 underdog on the next draw, and if that is black, we would be a 2:1 underdog on the next draw, and then all we can do is break even on the last two. E(2,2) is the EV of 2 white + 2 black that remain after you draw black first with probability 3/5, at which point we gain -1 dollar plus E(2,2). Now we need to evaluate E(2,2).

Note that E(2,2) can be no worse than 0 since you can always draw all 4 balls. If you draw white first with probability 1/2, then stop since if you continue, the EV of the remaining balls is 0 anyway (if you continue there is a 1/3 chance of drawing white first = 1, a 1/3 chance of black,black,white = -1, and a 1/3 chance of black,white = 0). If you draw black first with probability 1/2, so there are 2 white and 1 black remaining, then you can gain a white by drawing 2 whites in a row with probability 2/3 * 1/2 = 1/3. So E(2,2) = 1/2*1 + (1/2 *1/3) = 2/3, and

EV = 2/5*1 + 3/5*(-1 + 2/3) = 1/5 or 20 cents.</font>

[bobman0330]

My answer in white:
<font color="white">
A few subgames you might encounter:
2B, 1W: 1/3 you draw the white and quit, +1; 1/3 you draw the white second, 0; 1/3 you draw the white last, -1. EV=0.
2W, 1B: 1/3 you draw both whites, no black, +2; the rest are +1. EV = 4/3.
3B, 1W: 1/4, +1; 3/4 -&gt; 2B, 1W, which has EV 0, giving a total of -1. EV = -1/2.
2W, 2B: 1/2 -&gt; 2B, 1W, which has EV 0, giving a total of +1; 1/2 -&gt; 2W, 1B, EV = 4/3, giving a total of 1/3. EV = 2/3.

So. 40% of the time we go to 3B, 1W and stop at +1. 60% of the time, you get to 2B, 2W, and draw some more, ending at -1/3. Add them up and get +$.20
[color]

Thanks guys

[pzhon]

Recursive Mathematica code:

Clear[gamevalue];
gamevalue[black_, white_] :=
gamevalue[black, white] =
If[black == white == 0 || black &lt; 0 || white &lt; 0, 0,
Max[0, black/(black + white) (-1 + gamevalue[black - 1, white]) +
white/(black + white) (1 + gamevalue[black, white - 1])]]

Badly formatted output:

Table[gamevalue[i, j], {i, 0, 10}, {j, 0, 10}]

0 1 2 3 4 5 6 7 8 9 10
0 1/2 4/3 9/4 16/5 25/6 36/7 49/8 64/9 81/10 100/11
0 0 2/3 3/2 12/5 10/3 30/7 21/4 56/9 36/5 90/11
0 0 1/5 17/20 58/35 71/28 145/42 527/120 294/55 347/55 1041/143
0 0 0 12/35 1 113/63 93/35 235/66 202/45 777/143 492/77
0 0 0 0 4/9 47/42 21/11 199/72 428/117 705/154 182/33
0 0 0 0 1/15 127/231 284/231 3457/1716 8594/3003 5356/143 1333/286
0 0 0 0 0 61/396 563/858 4583/3432 13606/6435 8449/2860 9319/2431
0 0 0 0 0 0 697/3003 49/65 18457/12870 53683/24310 133097/43758
0 0 0 0 0 0 0 311/1040 10224/12155 74131/48620 106045/46189
0 0 0 0 0 0 0 0 7793/21879 84855/92378 26995/16796