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Two gamblers put in an even money wager (they both bet $M/2) on the result of tossing a...
Topics: Math, Probability
Two gamblers put in an even money wager (they both bet $M/2) on the result of tossing a coin four times:
If more heads than tails occur the first player wins
If more tails than heads occur the second player wins
and if it is two each, then they play again
How should the prize money be divided between the two contestants if the game has to be called off before it is completed?
Write a set of rules for each scenario.
1 Answer | add yours
If at any stage the players are "neck-a-neck" then the chances of each winning are equal, so the prize money should be divided evenly.
i) After one toss, whichever player has had a success/earned a point has a greater chance of winning. Call this player player A. They only need two more successes to win, or one more to draw. The other player, B, needs three successes to win still, and two more to draw.
The chance that the player A (suppose it is a tail) goes on to win is equal to
P(TTT) + P(TTH) + P(THT) + P(HTT) = (1/2)(1/2)(1/2) x 4 = 4/8
The chance that the player A goes on to draw with player B is equal to
P(THH) + P(HTH) + P(HHT) = (1/2)(1/2)(1/2) x 3 = 3/8
Therefore, if play stops after the first toss, 4/8 of the prize money should go to player A. Also, half of the "draw money" should go to player A, ie (3/8)/2 = 3/16.
In total 8/16 + 3/16 = 11/16 of the prize money should go to player A.
ii) After two tosses, player A may have had 2 successes. The probability of them going on to win is
P(TT) + P(TH) + P(HT) = (1/2)(1/2) x 3 = 3/4
The probability of them going on to draw is P(HH) = (1/2)(1/2) = 1/4
Player A should then get 3/4 + (1/4)/2 = 6/8 + 1/8 = 7/8 of the prize money.
Alternatively, If player A has had 1 success, the probability of them going on to win or draw is equal to that of player B, as they have both had a success. In that case, player A should get 1/2 of the prize money.
iii) After three tosses we have the previous scenarios a) three tails b) two tails c) one tail ( d) no tails is equivalent to three heads - there is symmetry)
a) Player A has already won, so should get all the prize money
b) The chance player A goes on to win is P(T) = 1/2. The chance player A goes on to draw is P(H) = 1/2. Player A should then get 1/2 + (1/2)/2 = 3/4 of the prize money
c) Player A cannot win, but can draw. The chances of a draw are P(T) = 1/2.
Player A should then get (1/2)/2 = 1/4 of the prize money.
iv) If after four tosses, play stops and both players have two successes each, they should each receive half of the prize money.
If A needs tails T, if play stops after
i) T, A should get 11/16 of the prize money $M
ii) TT, A should get $(7/8)M
TH or HT, A should get $(1/2)M
iii) a) TTT, A should get all $M b) TTH, THT or HTT, A should get $(3/4)M
c) THH, HTH or HHT, A should get $(1/4)M
iv) TTHH, THTH, THHT, HTHT or HHTT, A should get $(1/2)M
Posted by mathsworkmusic on May 20, 2013 at 6:00 PM (Answer #2)
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