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Suppose you play the following game: You toss a fair coin. If you get heads, a hundred dollars are added to your reward. If you get tails, however, the game is stopped and you do not get anything at all. After each throw you can decide, whether you want to take the money or keep playing. When should you stop to play the game in order to get the maximum expected reward and why? What happens if the coin is biased and has an 80% chance of showing heads?

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    $\begingroup$ Clearly the answer depends on the outcome of the first three tosses. Note: you should explain how ties are handled. $\endgroup$ – lulu Dec 30 '19 at 0:27
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    $\begingroup$ As there are $6$ tosses in the original game, there is a chance that both players will guess right exactly $3$ times. That would be a tie. $\endgroup$ – lulu Dec 30 '19 at 0:33
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    $\begingroup$ Hint: after you have explained how ties are handled, you really only have two cases. Case $1$. The three coins all came up $H$. and Case $2$: Exactly one of the three came up $T$. (the other two cases follow by symmetry). In each case, just compute the winnings that each player expected to get, had the game continued to its conclusion. $\endgroup$ – lulu Dec 30 '19 at 0:39
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    $\begingroup$ Hmm, is there any way to do this question without ties? My initial thoughts were similar to @SohamKonar because I figured that we are looking at the moment right after 3 tosses, and as such, there are no ties with respect to the 3 tosses done. $\endgroup$ – probability_coin Dec 30 '19 at 0:53
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    $\begingroup$ See the problem of points for a historical discussion of how almost exactly the same problem - how to split the stakes if a game is interrupted - led to the beginnings of probability as a mathematical study. Spolier: the "obvious" way of splitting according to current score ratio is incorrect (by modern standards anyway!) $\endgroup$ – antkam Dec 30 '19 at 5:06
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I think this problem is conditional probability problem.
$ 3 $tosses have $4$ different status. If we see this situation at A :
Head may happens $0, 1, 2, 3$ times.
So we must calculate $4$ different cases.
For this we define some status.
If A wins we denote that event Upper letter "A".
If B wins we denote that event Upper letter "B".
If "Head" happens $0$ time, then we have to calculate
$P1(A/A=0)=P(A=3,B=3/A=0)=C^{3}_{3}*1/8$
∴A must have $50*1/8$dollars.

If "Head" happens $1$ time, then
$P1(A) = (A/A=1) = P(A=3,B=3/A=1) =C^{2}_{3}*1/8 = 3/8$
$P2(A)=P(A=4,B=2/A=1)=C^{3}_{3}*1/8$
∴In this case A must have $50*3/8+100*1/8$ dollars.
If "Head" happens $2$ times, then
$P1(A) = (A/A=2) = P(A=3,B=3/A=2) = C^{1}_{3}*1/8 = 3/8$
$P2(A)=P(A=4,B=2/A=2) +P(A=5,B=1/A=2) = C^{2}_{3}*1/8 + C^{3}_{3}*1/8 = 0.5$
∴In this case A must have $50*3/8+100*0.5$ dollars.
If "Head" happens $3$ times, then this means "Tail" happens 0 time.

So in this case A must have $100-50*1/8$ dollars.

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Ties matter because even though one player must be ahead after $3$ tosses, there's still a positive probability that, if the game were to continue, it would still end in a tie. The only fair way of dividing the prize, as illustrated by Ross Millikan in the example he treats in his first comment, is to give each player the expected amount of his or her ultimate winnings if the game were to be completed, given the current circumstances. The expression for those expected amounts will include a term which is equal to the (conditional) probability of a tie times the amounts the players would receive if that were to occur.

There are various ways ties could be resolved fairly. Here are some:

  • Split the $\$100$ equally between the two players, giving $\$50$ to each;
  • The player who won the first toss takes the entire $\$100$;
  • The first player to win four tosses takes the entire $\$100$, already indicated as a possibile method by Ross Millikan;
  • The player who wins the last toss takes $\$80$ and the other takes $\$20$.

For the second and fourth of these scenarios, the players' expected earnings given the situation after three tosses will be different from what it would be in the other two, so it's not really possible to give a satisfactory answer to your question without making some assumptions about what is supposed to happen in the case of ties.

Suppose the last of the above-listed methods is used to resolve ties, for instance. In that case, as lulu has noted in one comment, there are really only two possibilities to consider—the leading player is ahead $3$ to $0$ or is ahead $2$ to $1$.

If the leading player is ahead $3$ to $0$ then there are only two possible outcomes: the game ends in a tie, which will occur with probability $\ \frac{1}{8}\ $, and in which case the leading player will only get $\$20$, or the leading player wins, which will occur with probability $\ \frac{7}{8}\ $, and in which case, the leading player gets the entire $\$100$. The leading player's expected winnings are therefore $\ \frac{20}{8}+\frac{700}{8}=90\ $. In this case, therefore the $\$100$ should be split in the ratio $9:1$, with the leading player getting $\$90$, and the other getting $\$10$.

If the leading player is ahead $2$ to $1$, there are $4$ possible outcomes to consider, with the following probabilities and payouts to the the leading player: $$ \begin{array}{c|c|c} \text{outcome}& \text{probability}& \text{payout to leader}\\ \hline \text{leading player wins}&\frac{1}{2}&\$100\\ \hline \text{tie: leading player wins last toss} &\frac{1}{8}& \$80\\ \hline \text{tie: leading player loses last toss} &\frac{1}{4}& \$20\\ \hline \text{leading player loses} &\frac{1}{4}& \$0\\ \hline \end{array} $$ So the leading player's expected winnings in this case are $\ \frac{100}{2}+\frac{80}{8} +\frac{20}{4}$$=65\ $, and the $\$100$ should be split in the ratio $13:7$, with $\$65$ going to the leading player, and $\$35$ to the other.

For the other methods of resolving ties given above, the payouts to the leading player should be made according to the following table: $$ \begin{array}{c|c|c} \text{method of tie resolution}&\text{lead is $3$ to $0$}&\text{lead is $2$ to $1$}\\ \hline \$50\text{ to each player}&\$93.75 &\$68.75\\ \hline \$100 \text{ to winner of first toss}&\$100&\$87.50\text{ or } \$50\\ \hline \$100 \text{ to first player with four wins}&\$93.75 &\$68.75\\ \hline \end{array} $$ In the case of the second of the above-listed methods, a player leading $2:1$ after $3$ tosses should get $\$87.50$ if he or she won the first toss, but only $\$50$ otherwise.

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I would assume that the 100 dollars would be split proportionally to how many of the 3 completed flips each player guessed correctly.

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    $\begingroup$ No. Imagine the game were best $4$ of $7$ to eliminate ties and $A$ is ahead $3-1$ after four tosses. $B$ can only win with three wins, which is probability $\frac 18$, so should receive $\frac 18 \cdot 100=12.50,$ not $25$ Similarly, someone ahead $1-0$ should not get the whole stake. $\endgroup$ – Ross Millikan Dec 30 '19 at 0:59
  • $\begingroup$ @RossMillikan So you want to split the money based on the probability that each player will win if they continue the game from that point? That seems more reasonable, but we also do not have confirmation that they will play a 7th game to break a tie. $\endgroup$ – Soham Konar Dec 30 '19 at 1:04
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    $\begingroup$ Yes, each person should receive their expectation if the game were to continue. I was not assuming OPs game would flip again to break ties, I just wanted to show (in a similar game) that one should not split the way you suggested. I made the number of flips odd to avoid ties. This is similar to the story of how Pascal invented probability theory. $\endgroup$ – Ross Millikan Dec 30 '19 at 1:06
  • $\begingroup$ You should post your proposal/solution as an answer so that OP can accept it. $\endgroup$ – Soham Konar Dec 30 '19 at 1:08
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    $\begingroup$ lulu has hinted at it. I didn't want to step on that. $\endgroup$ – Ross Millikan Dec 30 '19 at 1:08

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