UncleEbenezer wrote:chas49 wrote:No thanks then
The sequence you chose gives you an advantage
I can't explain why though.
You are of course right. The odds are between 2:1 and 7:1 in Gengulphus's favour, depending on your initial choice. I'm not going to try and explain, either. At least, not unless well-boozed-up.
But, one for Gengulphus. Take the same game with three, four, five, six, seven or eight players. In each case, what position do you want your choice to be?
By no means a full answer but I was intrigued by UncleE's question. So I extended the Markov Chain model I used previously to a 3 player game. This meant I could play around. The updated spreadsheet is
here.
Take the case of 3 players where Player 1 chooses HHH and Player 2 chooses THH. In the 2 player game, Player 2's choice maximises Player 2's probability of winning. However, in the 3 player game Player 3 has 6 choices available and one of these, TTH, gives Player 3 a greater probability of winning than either of Players 1 and 2. I was a little curious about this and initially suspected my model was at fault. However, having looked in a little more detail, I don't think this is the case.
For these choices, the model indicates that if the game is still active after n tosses then Player 3 always has a greater probability of winning on the next toss than Player 2.
Consider such a game after 3 tosses. If the game has not yet ended then the possibilities for the first three tosses are
1. HHT
2. HTH
3. HTT
4. THT
5. TTT
Player 2 (THH) can win on the 4th toss from only the second of these possibilities, whilst Player 3 (TTH) can win from both the third and fifth possibilities. Since these five possibilities are equally likely, Player 3's probability of winning on the fourth toss is twice that of Player 2 (and Player 1 cannot win at all).
If the game is still active after the 4th toss, then there are 7 possibilities (10 obtained from the 5 above followed by H or T, respectively less the three of these 10 that bring the game to an end on the fourth toss). These 7 possibilites are
1. HHTH
2. HHTT
3. HTHT
4. HTTT
5. THTH
6. THTT
7. TTTT
Player 2 can win from 2 of these (no's 1 and 5), whilst Player 3 can win from 4 (no's 2, 4 , 6 and 7), so again Player 3's probability of winning on the fifth toss is twice that of Player 2
Player 3's comparative advantage over Player 2 increases to 5:1 if the game is still active after 5 tosses (I'll leave that one as an exercise!) and Player 3 retains the advantage for any subsequent tosses.
More generally(*), there are 8*7=56 possible combinations in which Players 1 and 2 can make their choices. However, from Player 3's perspective, these only amount to half this number since the order in which two possibilities are eliminated from Player 3's range of choices is immaterial as far as Player 3 is concerned. Of these 28 possibilities, Player 3 can obtain a greater probability of winning than either of the other two players in 24 of these cases.
However, this leaves 4 cases where this is not possible. Denoting these by the choices not available to Player 3 these 4 cases are:
1. HHT, THT
2. HHT, TTH
3. HTH, TTH
4. HTT, THH
For three of these(no's 1, 3 and 4) , Player 3 can achieve an "equal first" probability with another Player. For no 2, Player 3's best choice provides only an "equal last" probability with another player.
As for the 4, 5, 6 and 7 player game - who knows? (though I suspect shedding insight may well involve moving even further out of the realm of puzzles).
(*) Errors and omissions excepted!
modellingman