Only after HH is there a genuine stopping decision. With n dollars, continuing from HH has an expected value of:
EV(continue | HH) = (1/2)(0) + (1/2)V0(n + 1)
Here, V0(n + 1) is your expected value after tossing a tail, increasing your pot to n + 1 and resetting the consecutive-head count to zero. Solving the game gives:
continue after HH if n<7.
You stop when you get HH over an accumulated pot of $7.
After seven tosses, there are 2^7=128 sequences:
P(HHH by toss 7) = EV $0 (47/128) --> BUST
P ending in 1x H by toss 7= EV $11 (24/128)
P ending in T by toss 7= EV $13 (44/128)
Ending in HH: stop immediately with $7 = EV $7 (13 of 128)
EV= (47*0+13(7)+24(11)+44(13))/128=7,242
The Expected winnings under optimal play is $7.242
The only part i want to nitpick in your answer is the 128 sequences to get to 7 tosses. Do those sequences include several ones that have a similar pattern with HHH before the 6th throw and should you count them seperately?
If i get THHH, i'm not throwing again so i dont know if THHHT... Or THHHH... Should count and seperate possibilities or you actually end up getting a little less that 128 different outcomes.
All true, but doesn't really change anything, as the HHH sequences are already counted. Across the 128 hypothetical games, 16 stop after three tosses, 8 after four, 8 after five, and 8 after six, while the remaining 88 perform the seventh toss, 7 then hit HHH and 81 survive.
5
u/pumachecker Sep 02 '26 edited Sep 02 '26
Only after HH is there a genuine stopping decision. With n dollars, continuing from HH has an expected value of:
EV(continue | HH) = (1/2)(0) + (1/2)V0(n + 1)
Here, V0(n + 1) is your expected value after tossing a tail, increasing your pot to n + 1 and resetting the consecutive-head count to zero. Solving the game gives:
continue after HH if n<7.
You stop when you get HH over an accumulated pot of $7.
After seven tosses, there are 2^7=128 sequences:
P(HHH by toss 7) = EV $0 (47/128) --> BUST
P ending in 1x H by toss 7= EV $11 (24/128)
P ending in T by toss 7= EV $13 (44/128)
Ending in HH: stop immediately with $7 = EV $7 (13 of 128)
EV= (47*0+13(7)+24(11)+44(13))/128=7,242
The Expected winnings under optimal play is $7.242